Today’s Outline - September 26, 2012csrri.iit.edu/~segre/phys405/12F/lecture_11.pdf · 2012. 9....
Transcript of Today’s Outline - September 26, 2012csrri.iit.edu/~segre/phys405/12F/lecture_11.pdf · 2012. 9....
Today’s Outline - September 26, 2012
• Hermitian operators
• Determinate states
• Eigenfunctions and eigenvalues
• Discrete and continuous spectra
• Statistical interpretation
Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided
Homework Assignment #05:Chapter 3: ??,??,??,??,??,??due Wednesday, October 3, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 1 / 10
Today’s Outline - September 26, 2012
• Hermitian operators
• Determinate states
• Eigenfunctions and eigenvalues
• Discrete and continuous spectra
• Statistical interpretation
Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided
Homework Assignment #05:Chapter 3: ??,??,??,??,??,??due Wednesday, October 3, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 1 / 10
Today’s Outline - September 26, 2012
• Hermitian operators
• Determinate states
• Eigenfunctions and eigenvalues
• Discrete and continuous spectra
• Statistical interpretation
Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided
Homework Assignment #05:Chapter 3: ??,??,??,??,??,??due Wednesday, October 3, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 1 / 10
Today’s Outline - September 26, 2012
• Hermitian operators
• Determinate states
• Eigenfunctions and eigenvalues
• Discrete and continuous spectra
• Statistical interpretation
Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided
Homework Assignment #05:Chapter 3: ??,??,??,??,??,??due Wednesday, October 3, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 1 / 10
Today’s Outline - September 26, 2012
• Hermitian operators
• Determinate states
• Eigenfunctions and eigenvalues
• Discrete and continuous spectra
• Statistical interpretation
Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided
Homework Assignment #05:Chapter 3: ??,??,??,??,??,??due Wednesday, October 3, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 1 / 10
Today’s Outline - September 26, 2012
• Hermitian operators
• Determinate states
• Eigenfunctions and eigenvalues
• Discrete and continuous spectra
• Statistical interpretation
Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided
Homework Assignment #05:Chapter 3: ??,??,??,??,??,??due Wednesday, October 3, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 1 / 10
Today’s Outline - September 26, 2012
• Hermitian operators
• Determinate states
• Eigenfunctions and eigenvalues
• Discrete and continuous spectra
• Statistical interpretation
Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided
Homework Assignment #05:Chapter 3: ??,??,??,??,??,??due Wednesday, October 3, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 1 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx
= 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition
, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition
, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉
= 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx
=�
��~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
~if ∗g
∣∣∣∣∞−∞−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx
= 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Vector properties: addition
The expectation value of an observ-able Q(x , p) is
since any observable must be realby definition, we can write
This holds for all wave functionsand operators which represent ob-servables have the property that:
Such operators are said to be her-mitian.
〈Q〉 =
∫Ψ∗QΨ dx = 〈Ψ|QΨ〉
〈Q〉 = 〈Q〉∗
〈Ψ|QΨ〉∗ = 〈QΨ|Ψ〉 = 〈Ψ|QΨ〉
〈f |Qg〉 = 〈Qf |g〉
Is p is hermitian?
〈f |pg〉 =
∫ ∞−∞
f ∗~i
dg
dxdx =
���~
if ∗g
∣∣∣∣∣∞
−∞
−∫ ∞−∞
~i
df ∗
dxg dx
=
∫ ∞−∞
(~i
df
dx
)∗g dx = 〈pf |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 2 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉
Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx
=
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx
= 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx
=
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx
= 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx
= f ∗g∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx
= −〈 ddx
f |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(− ip + mωx)†
=1√
2~mω(i
p + mωx
) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(i
p + mωx
)
= a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(ip
+ mωx
)
= a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(ip + mωx)
= a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(ip + mωx) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(ip + mωx) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉
= 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(ip + mωx) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉
= 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Problem 3.5
The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that
〈f |Qg〉 = 〈Q†f |g〉Q = Q†
Find the hermitian conjugates of x , i , d/dx , a+, and QR.
〈f |xg〉 =
∫f ∗(xg) dx =
∫(xf )∗g dx = 〈xf |g〉
〈f |ig〉 =
∫f ∗(ig) dx =
∫(−if )∗g dx = 〈−if |g〉
〈f | ddx
g〉 =
∫f ∗
dg
dxdx = f ∗g
∣∣∣∞−∞−∫ (
df
dx
)g dx = −〈 d
dxf |g〉
a†+ =1√
2~mω(−ip + mωx)† =
1√2~mω
(ip + mωx) = a−
〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 = 〈(QR)†f |g〉
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 3 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system.
If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉
= 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉
= 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉
= 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉
≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q
and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
Determinate states
Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.
In this case, the standard deviation of Q must be exactly zero.
Since Q is hermitian, then Q − q must be as well.
σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0
The only function whose square in-tegral vanishes is zero
(Q − q)Ψ ≡ 0
QΨ = qΨ
This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.
Thus, determinate states are eigenfunctions of Q.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 4 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states.
But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
All about eigenfunctions...
An eigenfunction multiplied by a constant gives the same eigenvalue:
Q(aΨ) = q(aΨ)
Zero cannot be an eigenfunction
The collection of all eigenvalues of an operator is its spectrum
If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.
Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:
Hψ = Eψ
If the spectrum of an operator is discrete (eigenvaluse are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states
If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 5 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.
This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by
, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by
, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Example 3.1
Find the eigenfunctions and eigenvalues of Q ≡ i ddφ
First let’s see if Q is hermitian.This operator acts on functionsf (φ), where
f (φ+ 2π) = f (φ)
integrating by parts
The eigenvalue equation is
This is solved by, which uponapplication of the periodicboundary condition, gives
〈f |Qg〉 =
∫ 2π
0f ∗(idg
dφ
)dφ
= i f ∗g∣∣∣2π0−∫ 2π
0i
(df ∗
dφ
)g dφ
= 〈Qf |g〉
id
dφf (φ) = qf (φ)
f (φ) = Ae−iqφ
e−iq2π = 1, q = 0,±1,±2, . . .
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 6 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!
Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!
Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉
q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!
Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉
= q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!
Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉
→ q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!
Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉
q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉
= q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉
→ 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete
(can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Discrete spectra
Important properties of the eigenfunctions of hermitian operators
1. Eigenvalues are real
observables are real!Qf = qf
〈f |Qf 〉 = 〈Qf |f 〉q〈f |f 〉 = q∗〈f |f 〉 → q∗ = q
2. Eigenfunctions with distincteigenvalues are orthogonal
Qg = q′g
〈f |Qg〉 = 〈Qf |g〉q′〈f |f 〉 = q∗〈f |f 〉 → 〈f |g〉 = 0
Even with degenerate eigenvalues, it is possible to choose orthogonaleignefunctions by taking linear combinations.
3. The eigenfuctions of an observable operator are complete (can’t provethis so take it as an axiom following Dirac!)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 7 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable.
But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~
∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~
∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)
This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: orthonormality
For continuous spectra, the eigenfunctions are not normalizable and theproperties (reality, orthogonality and completeness) are not directlyprovable. But ...
Take the case of eigenfunctionsand eigenvalues of the momen-tum operator.
restricting only to real values ofp
choosing A = 1/√
2π~ will nor-malize this set of eigenfunctions
~i
d
dxfp(x) = pfp(x)
fp(x) = Ae ipx/~∫ ∞−∞
f ∗p′fp dx = |A|2∫ ∞−∞
e i(p−p′)x/~ dx
= |A|22π~δ(p − p′)
fp(x) =1√2π~
e ipx/~
〈fp′ |fp〉 = δ(p − p′)This is called Dirac orthonormality
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 8 / 10
Momentum eigenfunctions: completeness
This restricted set of eigenfunctions of momentum are also complete.
f (x) =
∫ ∞−∞
c(p)fp(x) dp =1√2π~
∫ ∞−∞
c(p)e ipx/~ dp
〈fp′ |f 〉 =
∫ ∞−∞
c(p)〈fp′ |fp〉 =
∫ ∞−∞
c(p)δ(p − p′) dp = c(p′)
While this subset of eigenfunctions of p are not In a true Hilbert space,they have quasi-normalizibility and can be used to build wavepackets whichdo represent physical states, albeit without a determinate momentum.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 9 / 10
Momentum eigenfunctions: completeness
This restricted set of eigenfunctions of momentum are also complete.
f (x) =
∫ ∞−∞
c(p)fp(x) dp
=1√2π~
∫ ∞−∞
c(p)e ipx/~ dp
〈fp′ |f 〉 =
∫ ∞−∞
c(p)〈fp′ |fp〉 =
∫ ∞−∞
c(p)δ(p − p′) dp = c(p′)
While this subset of eigenfunctions of p are not In a true Hilbert space,they have quasi-normalizibility and can be used to build wavepackets whichdo represent physical states, albeit without a determinate momentum.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 9 / 10
Momentum eigenfunctions: completeness
This restricted set of eigenfunctions of momentum are also complete.
f (x) =
∫ ∞−∞
c(p)fp(x) dp =1√2π~
∫ ∞−∞
c(p)e ipx/~ dp
〈fp′ |f 〉 =
∫ ∞−∞
c(p)〈fp′ |fp〉 =
∫ ∞−∞
c(p)δ(p − p′) dp = c(p′)
While this subset of eigenfunctions of p are not In a true Hilbert space,they have quasi-normalizibility and can be used to build wavepackets whichdo represent physical states, albeit without a determinate momentum.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 9 / 10
Momentum eigenfunctions: completeness
This restricted set of eigenfunctions of momentum are also complete.
f (x) =
∫ ∞−∞
c(p)fp(x) dp =1√2π~
∫ ∞−∞
c(p)e ipx/~ dp
〈fp′ |f 〉 =
∫ ∞−∞
c(p)〈fp′ |fp〉
=
∫ ∞−∞
c(p)δ(p − p′) dp = c(p′)
While this subset of eigenfunctions of p are not In a true Hilbert space,they have quasi-normalizibility and can be used to build wavepackets whichdo represent physical states, albeit without a determinate momentum.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 9 / 10
Momentum eigenfunctions: completeness
This restricted set of eigenfunctions of momentum are also complete.
f (x) =
∫ ∞−∞
c(p)fp(x) dp =1√2π~
∫ ∞−∞
c(p)e ipx/~ dp
〈fp′ |f 〉 =
∫ ∞−∞
c(p)〈fp′ |fp〉 =
∫ ∞−∞
c(p)δ(p − p′) dp
= c(p′)
While this subset of eigenfunctions of p are not In a true Hilbert space,they have quasi-normalizibility and can be used to build wavepackets whichdo represent physical states, albeit without a determinate momentum.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 9 / 10
Momentum eigenfunctions: completeness
This restricted set of eigenfunctions of momentum are also complete.
f (x) =
∫ ∞−∞
c(p)fp(x) dp =1√2π~
∫ ∞−∞
c(p)e ipx/~ dp
〈fp′ |f 〉 =
∫ ∞−∞
c(p)〈fp′ |fp〉 =
∫ ∞−∞
c(p)δ(p − p′) dp = c(p′)
While this subset of eigenfunctions of p are not In a true Hilbert space,they have quasi-normalizibility and can be used to build wavepackets whichdo represent physical states, albeit without a determinate momentum.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 9 / 10
Momentum eigenfunctions: completeness
This restricted set of eigenfunctions of momentum are also complete.
f (x) =
∫ ∞−∞
c(p)fp(x) dp =1√2π~
∫ ∞−∞
c(p)e ipx/~ dp
〈fp′ |f 〉 =
∫ ∞−∞
c(p)〈fp′ |fp〉 =
∫ ∞−∞
c(p)δ(p − p′) dp = c(p′)
While this subset of eigenfunctions of p are not In a true Hilbert space,they have quasi-normalizibility and can be used to build wavepackets whichdo represent physical states, albeit without a determinate momentum.
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 9 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = ?
=
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = ?
=
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y)
=
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y)
→ 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy
= c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10
Position eigenfunctions
The same can be shown forthe position operator x
what eigenfunction has theproperty that multiplied bythe variable x is identical tomultiplying it by the constanty?
picking A = 1
and for completeness
so that
c(y) = f (y)
x gy (x) = y gy (x)
gy (x) = Aδ(x − y) =
∫ ∞−∞
g∗y ′(x)gy (x) dx
= |A|2∫ ∞−∞
δ(x − y ′)δ(x − y) dx
= |A|2δ(y − y ′)
gy (x) = δ(x − y) → 〈g∗y ′ |gy 〉 = δ(y − y ′)
f (x) =
∫ ∞−∞
c(y)gy (x) dy
=
∫ ∞−∞
c(y)δ(x − y) dy = c(x)
C. Segre (IIT) PHYS 405 - Fall 2012 September 26, 2012 10 / 10