ELECTRODYNAMICS IN QUANTUM MECHANICS: GAUGETRANSFORMATIONS

Link to: physicspages home page.To leave a comment or report an error, please use the auxiliary blog.Reference: Griffiths, David J. (2005), Introduction to Quantum Mechan-

ics, 2nd Edition; Pearson Education - Problem 4.61.A gauge transformation in classical electrodynamics is possible if we

replace the scalar potential ϕ and the vector potential A by

ϕ′ = ϕ− ∂Λ

∂ t(1)

A′ = A+∇Λ (2)

where Λ is any function of position and time. These transformations (pro-vided both are made at the same time) leave the fields E and B unchanged,as we can see by direct substitution.

Using the modified potentials we get

E =−∇ϕ′− ∂A′

∂ t(3)

=−∇ϕ +∂∇Λ

∂ t− ∂ (A+∇Λ)

∂ t(4)

=−∇ϕ− ∂A∂ t

(5)

B = ∇×A′ (6)= ∇×A+∇×∇Λ (7)= ∇×A (8)

where in the last line we used ∇×∇Λ = 0 for any function Λ. Test stuff.In quantum mechanics, we would like to know if a gauge transformation

leaves solutions of the Schrödinger equation unchanged. If so, then quan-tum theory is gauge invariant with respect to electrodynamics. In fact, itturns out that under a gauge transformation, the wave function for a particleof charge q changes from Ψ to

Ψ′ = eiqΛ/h̄

Ψ (9)1

ELECTRODYNAMICS IN QUANTUM MECHANICS: GAUGE TRANSFORMATIONS 2

To prove this, we can start with the Schrödinger equation from the poston the electromagnetic force law and use the same notation for derivativesand vector components as in the last part of that post (we will indicate avector component by a subscript x, y or z and a derivative by a superscript.Thus Axz

y = ∂ 2Ay/dxdz and so on) we wish to show that if Ψ satisfies theequation

ih̄∂Ψ

∂ t=

12m

(−h̄2

∇2Ψ+

h̄q−i

(2A ·∇Ψ+Ψ∇ ·A)+q2|A|2Ψ

)+qϕΨ

(10)

then replacing Ψ by Ψ′= eiqΛ/h̄Ψ satisifes the same equation with the gaugetransformed potentials. That is:

ih̄∂Ψ′

∂ t=

12m

(−h̄2

∇2Ψ′+ ih̄q(2A′ ·∇Ψ

′+Ψ′∇ ·A′)+q2|A′|2Ψ

′)+qϕ′Ψ′

(11)The left hand side is

ih̄∂Ψ′

∂ t= ih̄

∂

∂ t(eiqΛ/h̄

Ψ) = ih̄ΨteiqΛ/h̄−qΛ

tΨeiqΛ/h̄ (12)

To work out the right hand side, we note that each term inside the largestset of parentheses has the form of a sum of three terms, with the first of thesethree terms consisting of factors involving the x components of a vector or xderivatives of functions, the second term involving y and the third involvingz. All three of these terms have the same form, so we will consider the xcomponent in detail. We will first calculate the two derivatives of Ψ′

Ψ′ = eiqΛ/h̄

Ψ (13)

(Ψ′)x =

(iqh̄

ΛxΨ+Ψ

x)

eiqΛ/h̄ (14)

(Ψ′)xx =

(iqh̄

Λxx

Ψ+iqh̄

ΛxΨ

x +Ψxx)

eiqΛ/h̄ +

(iqh̄

ΛxΨ+Ψ

x)

iqh̄

ΛxeiqΛ/h̄

(15)

Since every term in 11 is linear in either Ψ′ or one of its derivatives, therewill be a common factor of eiqΛ/h̄ which we can cancel off, so we will omitthis term from now on to simplify the calculation. Now we consider eachterm in the large parentheses in 11 in turn.

ELECTRODYNAMICS IN QUANTUM MECHANICS: GAUGE TRANSFORMATIONS 3

−h̄2∇

2Ψ′ =−iqh̄Λ

xxΨ−2iqh̄Λ

xΨ

x +q2(Λx)2Ψ+ terms in yand z

(16)

2ih̄qA′ ·∇Ψ′ = 2ih̄q(Ax +Λ

x)(iqΛxΨ/h̄+Ψ

x)+ terms in yand z (17)

=−2q2AxΛxΨ+2ih̄qAxΨ

x−2q2(Λx)2Ψ+2ih̄qΛ

xΨ

x + terms in yand z(18)

ih̄qΨ′∇ ·A′ = ih̄qΛ

xxΨ+ ih̄qAx

xΨ+ terms in yand z (19)

q2|A′|2Ψ′ = q2

ΨA2x +2q2

ΨAxΛx +q2

Ψ(Λx)2 + terms in yand z (20)

Adding up all these terms, we find that all terms involving Λ and itsderivatives cancel out, and we are left with

− h̄2Ψ

xx +2ih̄qAxΨx + ih̄qAx

xΨ+q2ΨA2

x + terms in yand z (21)

which is just the terms from the original equation before the gauge transfor-mation.

Finally, we note that the last term in 11 is

qϕ′Ψ′ = qϕΨeiqΛ/h̄−qΛ

tΨeiqΛ/h̄ (22)

and the last term cancels the last term in 12, so all references to Λ cancelout of 11 and the original Schrodinger equation is recovered.