Probability current in the relativistic Hamiltonian …mborn29/rembielinski.pdfProbability current...

Probability current in the relativistic Hamiltonianquantum mechanics

Jakub Rembielinski

University of Lodz

Max Born Symposium, University of Wroc law, 28–30 June, 2011

Formulation of relativistic quantum mechanics: Salpeterequation

K. Kowalski and J. Rembielinski, Salpeter equation and probabilitycurrent in the relativistic Hamiltonian quantum mechanics (accepted forpublication in Phys. Rev. A (2011)); continuation of:

K. Kowalski and J. Rembielinski, The relativistic massless harmonicoscillator, Phys. Rev. A 81, 012118 (2010)

H =√

c2p2 + m2c4 + V (x), H → i~∂

∂t, x→ x, p→ −i~∇

The spinless Salpeter equation:

∂tφ(x, t) = [

√m2c4 − ~2c2∆ + V (x)]φ(x, t)

L.L. Foldy, Synthesis of covariant particle equations, Phys. Rev. 102,568 (1956)

Probability current 2/27

H =√

c2p2 + m2c4 + V (x), H → i~∂

∂t, x→ x, p→ −i~∇

∂tφ(x, t) = [

√m2c4 − ~2c2∆ + V (x)]φ(x, t)

H =√

c2p2 + m2c4 + V (x), H → i~∂

∂t, x→ x, p→ −i~∇

∂tφ(x, t) = [

√m2c4 − ~2c2∆ + V (x)]φ(x, t)

H =√

c2p2 + m2c4 + V (x), H → i~∂

∂t, x→ x, p→ −i~∇

∂tφ(x, t) = [

√m2c4 − ~2c2∆ + V (x)]φ(x, t)

Leslie L. Foldy (1919–2001)

Edwin E. Salpeter (1924–2008)Probability current 4/27

The Salpeter equation can be written in the form of theintegro-differential equation

i~∂φ(x, t)

∫d3yK (x− y)φ(y, t) + V (x)φ(x, t)

where the kernel is given by

K (x− y) = − 2m2c3

(2π)2~K2(mc

~ |x− y|)|x− y|2

The Salpeter equation presumes the Newton-Wigner localization schemeimplying the standard quantization rule x→ x, p→ −i~∇.Consequently, the Hilbert space of solutions to the Salpeter isL2(R3, d3x):

〈φ|ψ〉 =

∫d3xφ∗(x)ψ(x)

Therefore, we should identify |φ(x, t)|2 with the probability densityρ(x, t) satisfying the normalization condition:∫

d3x ρ(x, t) = 1

i~∂φ(x, t)

∫d3yK (x− y)φ(y, t) + V (x)φ(x, t)

K (x− y) = − 2m2c3

(2π)2~K2(mc

~ |x− y|)|x− y|2

〈φ|ψ〉 =

∫d3xφ∗(x)ψ(x)

d3x ρ(x, t) = 1

i~∂φ(x, t)

∫d3yK (x− y)φ(y, t) + V (x)φ(x, t)

K (x− y) = − 2m2c3

(2π)2~K2(mc

~ |x− y|)|x− y|2

〈φ|ψ〉 =

∫d3xφ∗(x)ψ(x)

d3x ρ(x, t) = 1

The Salpeter equation was discarded because of:

I its nonlocality

I the lack of manifest Lorentz covariance

However

I the nonlocality of the Salpeter equation does not disturb the lightcone structure

I the space L2(R3, d3x) of solutions to the Salpeter equation isinvariant under the Lorentz group

I the Salpeter equation possesses solutions of positive energies onlyand we have no problems with paradoxes occuring in the case of theKlein-Gordon equation

I the agreement of predictions of the spinless Salpeter equation withthe experimental spectrum of mesonic atoms is as good as for theKlein-Gordon equation

I the Salpeter equation is widely used in the phenomenologicaldescription of the quark-antiquark-gluon system as a hadron model

I its nonlocality

However

I its nonlocality

However

I its nonlocality

However

I its nonlocality

However

I its nonlocality

However

I its nonlocality

However

I its nonlocality

However

I its nonlocality

However

The Klein-Gordon equation[i~∂

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

The Klein-Gordon equation was accepted because of its manifest Lorentzcovariance. However its grave flaws are:

I problems with probablistic interpretation (probability density can benegative)

I paradoxes such as “zitterbewegung” and Klein paradox connectedwith existence of negative energy solutions

Nonrelativistic version of the Klein-Gordon theory would be

i~∂ψ

2mψ → −~2 ∂

∂t2=

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

i~∂ψ

2mψ → −~2 ∂

∂t2=

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

i~∂ψ

2mψ → −~2 ∂

∂t2=

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

i~∂ψ

2mψ → −~2 ∂

∂t2=

Probability current

Probability current for the Klein-Gordon equation[i~∂

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

In the case of the Klein-Gordon equation the probability density is

ρKG =i~

(ψ∗∂ψ

∂t− ψ∂ψ

~V |ψ|2

)The corresponding probablity current is given by

jKG = − i~2m

(ψ∗∇ψ − ψ∇ψ∗)

Problems:

I ρKG can be negative

I difficulties with the limit m = 0

Probability current

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

ρKG =i~

(ψ∗∂ψ

∂t− ψ∂ψ

~V |ψ|2

jKG = − i~2m

(ψ∗∇ψ − ψ∇ψ∗)

Problems:

Probability current

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

ρKG =i~

(ψ∗∂ψ

∂t− ψ∂ψ

~V |ψ|2

jKG = − i~2m

(ψ∗∇ψ − ψ∇ψ∗)

Problems:

Probability current

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

ρKG =i~

(ψ∗∂ψ

∂t− ψ∂ψ

~V |ψ|2

jKG = − i~2m

(ψ∗∇ψ − ψ∇ψ∗)

Problems:

Probability current

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

ρKG =i~

(ψ∗∂ψ

∂t− ψ∂ψ

~V |ψ|2

jKG = − i~2m

(ψ∗∇ψ − ψ∇ψ∗)

Problems:

Probability current

∂t− V (x)

ψ(x, t) = (m2c4 − ~2c2∆)ψ(x, t)

ρKG =i~

(ψ∗∂ψ

∂t− ψ∂ψ

~V |ψ|2

jKG = − i~2m

(ψ∗∇ψ − ψ∇ψ∗)

Problems:

Probability current for the Salpeter equation

∂tφ(x, t) = [

√m2c4 − ~2c2∆ + V (x)]φ(x, t)

Probability density:ρ(x, t) = |φ(x, t)|2

Using the continuity equation

∂t+∇·j = 0

we derived the following formula on the probability current:

j(x, t) =c

(2π)3~6

∫d3pd3k

p + k√m2c2 + p2 +

√m2c2 + k2

ei(k−p)·x

~ φ∗(p, t)φ(k, t)

∂tφ(x, t) = [

√m2c4 − ~2c2∆ + V (x)]φ(x, t)

∂t+∇·j = 0

j(x, t) =c

(2π)3~6

∫d3pd3k

p + k√m2c2 + p2 +

√m2c2 + k2

ei(k−p)·x

~ φ∗(p, t)φ(k, t)

∂tφ(x, t) = [

√m2c4 − ~2c2∆ + V (x)]φ(x, t)

∂t+∇·j = 0

j(x, t) =c

(2π)3~6

∫d3pd3k

p + k√m2c2 + p2 +

√m2c2 + k2

ei(k−p)·x

~ φ∗(p, t)φ(k, t)

Properties:

I correct nonrelativistic limit

limc→∞

j = − i~2m

(φ∗∇φ− φ∇φ∗)

I good behaviour of the total current∫j(x, t) d3x = 〈φ|vφ〉

where v is the operator of the relativistic velocity

where p = −i~∇, and p0 = E/c =

√m2c2 + p2 =

√m2c2 − ~2∆.

These formula is not valid in the case of the Klein-Gordon equation.

Properties:

limc→∞

j = − i~2m

(φ∗∇φ− φ∇φ∗)

√m2c2 + p2 =

√m2c2 − ~2∆.

Properties:

limc→∞

j = − i~2m

(φ∗∇φ− φ∇φ∗)

√m2c2 + p2 =

√m2c2 − ~2∆.

Properties:

limc→∞

j = − i~2m

(φ∗∇φ− φ∇φ∗)

√m2c2 + p2 =

√m2c2 − ~2∆.

〈φ|vφ〉2 ≤ c2

I existence of the massless limit

j(x, t) =c

(2π)3~6

∫d3pd3k

|p|+ |k|ei

(k−p)·x~ φ∗(p, t)φ(k, t) (m = 0)

We have a remarkable formula

j(x, t) =

∫d3yd3zK (|y−x|, |x−z|)[φ∗(y, t)∇zφ(z, t)−φ(z, t)∇yφ

∗(y, t)]

K (|u|, |w|) = − im2c3

(2π)3~2

|u||w|1

|u|+ |w|K2

~(|u|+ |w|)

]The current can be also written as

j = − imc2

∞∑n=1

(2n − 3)!!

(2n)!!

)2n 2n−1∑k=0

(−1)k∇kφ∗∇2n−k−1φ

〈φ|vφ〉2 ≤ c2

j(x, t) =c

(2π)3~6

∫d3pd3k

|p|+ |k|ei

(k−p)·x~ φ∗(p, t)φ(k, t) (m = 0)

j(x, t) =

∗(y, t)]

K (|u|, |w|) = − im2c3

(2π)3~2

|u||w|1

|u|+ |w|K2

~(|u|+ |w|)

j = − imc2

∞∑n=1

(2n − 3)!!

(2n)!!

)2n 2n−1∑k=0

(−1)k∇kφ∗∇2n−k−1φ

〈φ|vφ〉2 ≤ c2

j(x, t) =c

(2π)3~6

∫d3pd3k

|p|+ |k|ei

(k−p)·x~ φ∗(p, t)φ(k, t) (m = 0)

j(x, t) =

∗(y, t)]

K (|u|, |w|) = − im2c3

(2π)3~2

|u||w|1

|u|+ |w|K2

~(|u|+ |w|)

j = − imc2

∞∑n=1

(2n − 3)!!

(2n)!!

)2n 2n−1∑k=0

(−1)k∇kφ∗∇2n−k−1φ

〈φ|vφ〉2 ≤ c2

j(x, t) =c

(2π)3~6

∫d3pd3k

|p|+ |k|ei

(k−p)·x~ φ∗(p, t)φ(k, t) (m = 0)

j(x, t) =

∗(y, t)]

K (|u|, |w|) = − im2c3

(2π)3~2

|u||w|1

|u|+ |w|K2

~(|u|+ |w|)

j = − imc2

∞∑n=1

(2n − 3)!!

(2n)!!

)2n 2n−1∑k=0

(−1)k∇kφ∗∇2n−k−1φ

ExamplesFree massless particle on a line

i∂φ(x , t)

√− ∂2

∂x2φ(x , t),

where we set c = 1.The solution is

φ(x , t) =

a + it

x2 + (a + it)2

Hence, we get the probability density

ρ(x , t) = |φ(x , t)|2 =2a

a2 + t2

(x2 − t2 + a2)2 + 4a2t2

and the probability current

j(x , t) =a

4πt2ln

(x + t)2 + a2

(x − t)2 + a2− ax

x2 − 3t2 + a2

(x2 − t2 + a2)2 + 4a2t2

i∂φ(x , t)

√− ∂2

∂x2φ(x , t),

φ(x , t) =

a + it

x2 + (a + it)2

ρ(x , t) = |φ(x , t)|2 =2a

a2 + t2

(x2 − t2 + a2)2 + 4a2t2

j(x , t) =a

4πt2ln

(x + t)2 + a2

(x − t)2 + a2− ax

x2 − 3t2 + a2

(x2 − t2 + a2)2 + 4a2t2

i∂φ(x , t)

√− ∂2

∂x2φ(x , t),

φ(x , t) =

a + it

x2 + (a + it)2

ρ(x , t) = |φ(x , t)|2 =2a

a2 + t2

(x2 − t2 + a2)2 + 4a2t2

j(x , t) =a

4πt2ln

(x + t)2 + a2

(x − t)2 + a2− ax

x2 − 3t2 + a2

(x2 − t2 + a2)2 + 4a2t2

i∂φ(x , t)

√− ∂2

∂x2φ(x , t),

φ(x , t) =

a + it

x2 + (a + it)2

ρ(x , t) = |φ(x , t)|2 =2a

a2 + t2

(x2 − t2 + a2)2 + 4a2t2

j(x , t) =a

4πt2ln

(x + t)2 + a2

(x − t)2 + a2− ax

x2 − 3t2 + a2

(x2 − t2 + a2)2 + 4a2t2

-20 -10 0 10 20

ΡHx, tL

Figure: The time evolution of the probability density related to the solution ofthe Salpeter equation for a free massless particle in one dimension. Theparameter a = 1.

-20 -10 0 10 20

jHx, tL

Figure: The time development of the probability current for a free masslessparticle moving in a line. The parameter a = 1.

The solution referring to the particle moving to the right (left)

φ±(x , t) =

±ix ∓ t ± ia

Therefore, the corresponding probability density and probability currentare

ρ±(x , t) = |φ±(x , t)|2 = ±j±(x , t) =a

(x ∓ t)2 + a2

-20 0 20 40

Ρ+Hx, tL

Figure: The behavior of the probability density ρ+(x , t) referring to the case ofthe free massles particle moving to the right. The stable maximum of theprobability density is going with the speed of light c=1.

Free massive particle on a line

i∂φ(x , t)

√m2 − ∂2

∂x2φ(x , t)

We have found the solution

φ(x , t) =

πK1(2ma)

a + it√x2 + (a + it)2

K1[m√x2 + (a + it)2]

The corresponding probability current is

j(x , t) =m2

πK1(2ma)

{[x2 − (a + it)2

x2 + (a + it)2K2[m

√x2 + (a + it)2]

− K0[m√x2 + (a + it)2]

]× a− it√

x2 + (a− it)2K1[m

√x2 + (a− it)2]

i∂φ(x , t)

√m2 − ∂2

∂x2φ(x , t)

φ(x , t) =

πK1(2ma)

a + it√x2 + (a + it)2

K1[m√x2 + (a + it)2]

j(x , t) =m2

πK1(2ma)

{[x2 − (a + it)2

x2 + (a + it)2K2[m

√x2 + (a + it)2]

− K0[m√x2 + (a + it)2]

]× a− it√

x2 + (a− it)2K1[m

√x2 + (a− it)2]

i∂φ(x , t)

√m2 − ∂2

∂x2φ(x , t)

φ(x , t) =

πK1(2ma)

a + it√x2 + (a + it)2

K1[m√x2 + (a + it)2]

j(x , t) =m2

πK1(2ma)

{[x2 − (a + it)2

x2 + (a + it)2K2[m

√x2 + (a + it)2]

− K0[m√x2 + (a + it)2]

]× a− it√

x2 + (a− it)2K1[m

√x2 + (a− it)2]

-10 -5 0 5 10

ΡHx, tL

Figure: The time development of the probability density ρ(x , t) = |φ(x , t)|2corresponding to the case of the free massive particle. The mass m = 0.5 anda = 1.

-20 0 20

jHx, tL

Figure: The plot of the probability current for a free massive particle moving ina line versus time, where m = 0.5 and a = 1.

Massless particle in a linear potential

i∂φ(x , t)

√− ∂2

∂x2φ(x , t) + xφ(x , t)

where we set c = 1 and ~ = 1. We have found the solution of the form

φ(x , t) =1

e−λ2 t

{1√λ+ i

e(−λt+ix)2

2(λ+i) erfc

[−λt + ix√

2(λ+ i)

+1√λ− i

e(λt−ix)2

2(λ−i) erfc

[λt − ix√2(λ− i)

where erfc(z) = 1− 2√π

0e−t

dt is the complementary error function.

Massless particle in a linear potential

i∂φ(x , t)

√− ∂2

∂x2φ(x , t) + xφ(x , t)

where we set c = 1 and ~ = 1. We have found the solution of the form

φ(x , t) =1

e−λ2 t

{1√λ+ i

e(−λt+ix)2

2(λ+i) erfc

[−λt + ix√

2(λ+ i)

+1√λ− i

e(λt−ix)2

2(λ−i) erfc

[λt − ix√2(λ− i)

where erfc(z) = 1− 2√π

0e−t

dt is the complementary error function.

-30 -20 -10 0 10

ΡHx, tL

Figure: The plot of the probability density ρ(x , t) = |φ(x , t)|2, referring to themassless particle in a linear potential. The parameter λ = 1. The classicalΛ-shaped dynamics of the maxima of the probability density is easily observed.

-2 -1 1 2t

Figure: The plot of the expectation value of the position operator versus time(solid line). The dotted line refers to the classical trajectory

-20 -10 0

jHx, tL

Figure: The plot of the probability current versus time.

Plane wave solutions

i~∂φ(x, t)

∂t=√m2c4 − ~2c2∆φ(x, t)

possesses plane wave solutions

φ(x, t) = Ce−i~ (Et−k·x)

where E =√m2c4 + k2c2 and C is a normalization constant. We have

j = ρv

where ρ = |φ|2 = |C |2, and v is the relativistic three-velocity given by

v =c2k

i~∂φ(x, t)

∂t=√m2c4 − ~2c2∆φ(x, t)

j = ρv

v =c2k

i~∂φ(x, t)

∂t=√m2c4 − ~2c2∆φ(x, t)

j = ρv

v =c2k

Massless particle in three dimensions

i∂φ(x, t)

∂t=√−∆φ(x, t)

We have obtained the following solution

φ(x, t) =(2a)

a + it

[r2 + (a + it)2]2

where r = |x|. Therefore the probability density is

ρ(x, t) = |φ(x, t)|2 =(2a)3

a2 + t2

[(r2 − t2 + a2)2 + 4a2t2]2

The probability current is given by

j(x, t) =

{− a3

2π2r2t3

32r2t4 + (r2 + t2 + a2)[3(r2 − t2 + a2)2 + 12a2t2 − 8r2t2]

[(r2 − t2 + a2)2 + 4a2t2]2

8π2r3t4ln

(r + t)2 + a2

(r − t)2 + a2

i∂φ(x, t)

∂t=√−∆φ(x, t)

φ(x, t) =(2a)

a + it

[r2 + (a + it)2]2

ρ(x, t) = |φ(x, t)|2 =(2a)3

a2 + t2

[(r2 − t2 + a2)2 + 4a2t2]2

j(x, t) =

{− a3

2π2r2t3

32r2t4 + (r2 + t2 + a2)[3(r2 − t2 + a2)2 + 12a2t2 − 8r2t2]

[(r2 − t2 + a2)2 + 4a2t2]2

8π2r3t4ln

(r + t)2 + a2

(r − t)2 + a2

i∂φ(x, t)

∂t=√−∆φ(x, t)

φ(x, t) =(2a)

a + it

[r2 + (a + it)2]2

ρ(x, t) = |φ(x, t)|2 =(2a)3

a2 + t2

[(r2 − t2 + a2)2 + 4a2t2]2

j(x, t) =

{− a3

2π2r2t3

32r2t4 + (r2 + t2 + a2)[3(r2 − t2 + a2)2 + 12a2t2 − 8r2t2]

[(r2 − t2 + a2)2 + 4a2t2]2

8π2r3t4ln

(r + t)2 + a2

(r − t)2 + a2

i∂φ(x, t)

∂t=√−∆φ(x, t)

φ(x, t) =(2a)

a + it

[r2 + (a + it)2]2

ρ(x, t) = |φ(x, t)|2 =(2a)3

a2 + t2

[(r2 − t2 + a2)2 + 4a2t2]2

j(x, t) =

{− a3

2π2r2t3

32r2t4 + (r2 + t2 + a2)[3(r2 − t2 + a2)2 + 12a2t2 − 8r2t2]

[(r2 − t2 + a2)2 + 4a2t2]2

8π2r3t4ln

(r + t)2 + a2

(r − t)2 + a2

0.0 0.5 1.0 1.5 2.0

ΡHr, tL

Figure: The time evolution of the probability density, where a = 1, showing thespreading of the wavefunction.

0 5 10 15 20

jHr, tL¤

Figure: The time development of the norm of the probability current, wherea = 1.

Probability current in the relativistic Hamiltonian …mborn29/rembielinski.pdfProbability current...

Documents

Transcript of Probability current in the relativistic Hamiltonian …mborn29/rembielinski.pdfProbability current...