The Quantum Theory ofAtoms and Molecules:

The Schrodinger equation

Hilary Term 2008Dr Grant Ritchie

An equation for matter waves?

De Broglie postulated that every particles has an associated wave of wavelength:

ph /=!

Wave nature of matter confirmed by electron diffraction studies etc (see earlier).

If matter has wave-like properties then there must be a mathematical function that is thesolution to a differential equation that describes electrons, atoms and molecules.

The differential equation is called the Schrödinger equation and its solution is called thewavefunction, Ψ.

The classical wave equation

2

2

22

21

tvx !

"!=

!

"!

We have seen previously that the wave equation in 1–d is:

Can this be used for matter waves in free space?

Try a solution: e.g. )(),( tkxietx

!"=#

Not correct! For a free particle we know that E=p2/2m.

An alternative….

tx !

"!=

!

"!#

2

2

)(),( tkxietx

!"=#

ti

xm !

"!=

!

"!# hh

2

22

2

Try a modified wave equation of the following type:

(α is a constant)

Now try same solution as before: e.g.

Hence, the equation for matter waves in free space is:

For )(),( tkxietx

!"=# ),(),(

2

22

txtxm

k!=! "h

hthen we have

which has the form: (KE) × wavefunction = (Total energy) × wavefunction

The time-dependent Schrödinger equation

),(2

2

txVm

pE +=For a particle in a potential V(x,t) then

and we have (KE + PE) × wavefunction = (Total energy) × wavefunction

titxV

xm !

"!="+

!

"!# hh

),(2 2

22

TDSE

Points of note:

1. The TDSE is one of the postulates of quantum mechanics. Though the SEcannot be derived, it has been shown to be consistent with all experiments.

2. SE is first order with respect to time (cf. classical wave equation).

3. SE involves the complex number i and so its solutions are essentially complex.This is different from classical waves where complex numbers are used implyfor convenience – see later.

The Hamiltonian operator

!=!""#

$%%&

'+

(

()=!+

(

!() HtxV

xmtxV

xm

ˆ),(2

),(2 2

22

2

22hh

)(ˆ2

ˆ)(

2ˆ

2

2

22

xVm

pxV

xmH x +=!

!"

#$$%

&+

'

'(=h

xipx!

!"= hˆ

LHS of TDSE canbe written as:

where Ĥ is called the Hamiltonian operator which is the differential operator thatrepresents the total energy of the particle.

Thus

where the momentum operator is

Thus shorthand for TDSE is:t

iH!

"!=" hˆ

Solving the TDSE

!+"

!"#=

"

!")(

2 2

22

xVxmt

ih

h

)()(),( tTxtx !="

Suppose the potential is independent oftime i.e. V(x, t) = V(x) then TDSE is:

LHS involves variation of ψ with t while RHS involves variation of ψ with x. Hencelook for a separated solution:

t

TiTxV

xT

m !

!=+

!

!" ##

#h

h)(

2 2

22then

Now divide by ψ T:t

T

TixV

xm !

!=+

!

!"

1)(

1

2 2

22

hh #

#

LHS depends only upon x, RHS only on t. True for all x and t so both sides must equal aconstant, E (E = separation constant).

Thus we have:

ExVxm

Et

T

Ti

=+!

!"

=!

!

)(1

2

1

2

22 #

#

h

h

Time-independent Schrödinger equation

Solving the time equation: h

hh

/)(1 iEt

AetTdtiE

T

dTE

dt

dT

Ti

!="!="=

This is exactly like a wave e-iωt with E = ћω. Therefore T(t) depends upon the energy E.

To find out what the energy actually is we must solve the space part of the problem....

The space equation becomes !!!!!

EHExVxm

==+"

"# ˆ)(2 2

22

or h

This is the time independent Schrödinger equation (TISE).

The TISE can be very difficult to solve – it depends upon V(x)!

Eigenvalue equations

The Schrödinger Equation is the form of an Eigenvalue Equation: !! EH =ˆ

where Ĥ is the Hamiltonian operator,

ψ is the wavefunction and is an eigenfunction of Ĥ;

E is the total energy (T + V) and an eigenvalue of Ĥ. E is just a constant!

)(2

2ˆˆˆ

2

2

xVdx

d

mVTH +!=+=

h

Later in the course we will see that the eigenvalues of an operator give the possibleresults that can be obtained when the corresponding physical quantity is measured.

TISE for a free-particle

!!

Exm

="

"#

2

22

2

h

)/()()(),( hEtkxietTxtx

!±==" #

For a free particle V(x) = 0 and TISE is:

and has solutions

Thus the full solution to the full TDSE is:

m

kEwhereeore

ikxikx

2

22h

==!

"

Corresponds to waves travelling in either ± x direction with:

(i) an angular frequency, ω = E / ћ ⇒ E = ћ ω!

(ii) a wavevector, k = (2mE)1/2 / ћ = p / ћ ⇒ p = h / λ !

WAVE-PARTICLE DUALITY!

Interpretation of Ψ(x,t)

As mentioned previously the TDSE has solutions that are inherently complex ⇒Ψ (x,t)cannot be a physical wave (e.g. electromagnetic waves). Therefore how can Ψ (x,t)relate to real physical measurements on a system?

The Born Interpretation

dxtxPdxtxdxtxtx ),(),(),(),(2*

=!=!!

Ψ*Ψ is real as required for a probability distribution and is the probability per unitlength (or volume in 3d).

The Born interpretation therefore calls Ψ the probability amplitude, Ψ*Ψ (= P(x,t))the probability density and Ψ*Ψ dx the probability.

Probability of finding a particle in a small length dx at position x and time t is equal to

Expectation values

xxPxx

i

ii!"= )(

!!"

"#

"

"#

$=== dxtxxdxxxPxx2),()(

Thus if we know Ψ (x, t) (a solution of TDSE), then knowledge of Ψ*Ψ dx allows theaverage position to be calculated:

In the limit that δx→ 0 then the summation becomes:

!!"

"#

"

"#

$== dxtxxdxxPxx2222 ),()(Similarly

The average is also know as the expectation value and are very important in quantummechanics as they provide us with the average values of physical properties because inmany cases precise values cannot, even in principle, be determined – see later.

Normalisation

!!"

"#

"

"#

=$= 1),()(2dxtxdxxP

Total probability of finding a particle anywhere must be 1:

This requirement is known as the Normalisation condition. (This condition arisesbecause the SE is linear in Ψ and therefore if Ψ is a solution of TDSE then so is cΨwhere c is a constant.)

Hence if original unnormalised wavefunction is Ψ(x,t), then the normalisation integral is:

!"

"#

$= dxtxN22 ),(

And the (re-scaled) normalised wavefunction Ψnorm = (1/N) Ψ.

Example 1: What value of N normalises the function N x (x - L) of 0 ≤ x ≤ L?

Example 2: Find the probability that a system described by the function 21/2sin (πx) where0 ≤ x ≤ 1 is found anywhere in the interval 0 ≤ x ≤ 0.25.

Boundary conditions for Ψ

In order for ψ to be a solution of the Schrödinger equation to represent a physicallyobservable system, ψ must satisfy certain constraints:

1. Must be a single-valued function of x and t;2. Must be normalisable; This implies that the ψ → 0 as x → ∞;3. ψ (x) must be a continuous function of x;4. The slope of ψ must be continuous, specifically dψ (x)/dx must be

continuous (except at points where potential is infinite).

Ψ(x)

x

Ψ(x)

x

Ψ(x)

x

Ψ(x)

x

Stationary states

h/)()()(),( iEtextTxtx!

==" ##

Earlier in the lecture we saw that even when the potential is independent of time thewavefunction still oscillates in time:

Solution to the full TDSE is:

But probability distribution is static:

2//2)()()(*),(),( xexextxtxP

iEtiEt !!! =="=#+ hh

Thus a solution of the TISE is known as a Stationary State.

Summary

titxV

xm !

"!="+

!

"!# hh

),(2 2

22

dxtxPdxtxdxtxtx ),(),(),(),(2*

=!=!!

!!"

"#

"

"#

=$= 1),()(2dxtxdxxP

TDSE:

Born interpretation:

Normalisation:

TISE: !!!!!

EHExVxm

==+"

"# ˆ)(2 2

22

or h

h/)()()(),( iEtextTxtx!

==" ##

Boundary conditions on wavefunction: single-valued, continuous, normalisable,continuous first derivative.

It’s never as bad as it seems….