Question

[Refer to Problem 4.59 for background.] In classical electrodynamics the potentials Aand$\mathit{\phi }$are not uniquely determined; 47 the physical quantities are the fields, E and B.

(a) Show that the potentials

${\mathbf{\phi }}^{\mathbf{\text{'}}}\mathbf{\equiv }\mathbf{\phi }\mathbf{-}\frac{\mathbf{\partial }\mathbf{\Lambda }}{\mathbf{\partial }\mathbf{t}}\mathbf{,}\mathbf{}{\mathbf{A}}^{\mathbf{\text{'}}}\mathbf{\equiv }\mathbf{A}\mathbf{+}\mathbf{\nabla }\mathbf{\Lambda }$

(where$\mathbf{\wedge }$is an arbitrary real function of position and time). yield the same fields as$\mathit{\phi }$and A. Equation 4.210 is called a gauge transformation, and the theory is said to be gauge invariant.

(b) In quantum mechanics the potentials play a more direct role, and it is of interest to know whether the theory remains gauge invariant. Show that

${\mathit{\Psi }}^{\mathbf{\text{'}}}\mathbf{\equiv }{\mathit{e}}^{\mathbf{i}\mathbf{q}\mathbf{\Lambda }\mathbf{/}\mathbf{\hslash }}\mathit{\Psi }$

satisfies the Schrödinger equation (4.205) with the gauge-transformed potentials${\mathit{\phi }}^{\mathbf{\text{'}}}$and${\mathit{A}}^{\mathbf{\text{'}}}$, Since${\mathit{\Psi }}^{\mathbf{\text{'}}}$differs from$\mathit{\psi }$only by a phase factor, it represents the same physical state, 48and the theory is gauge invariant (see Section 10.2.3for further discussion).

Short answer

(a)The potentials results in the same fields.

(b)The equation satisfies the Schrödinger equation withgauge invariant.

Step-by-step solution

Expression for the hamilontonia of electrodynamics and gauge transformation

The expression for hamilontonia of electrodynamics is as follows,

$\mathbf{H}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{m}}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{qA}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{q\phi }$

The expression for Gauge transformation is as follows,

${\mathit{A}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{A}\mathbf{+}\mathbf{\nabla }\mathit{\Lambda }$

The expression for gauge wave function is as follows,

${\mathit{\psi }}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{e}}^{\mathbf{i}\mathbf{q}\mathbf{\Lambda }\mathbf{/}\mathbf{\hslash }}\mathit{\psi }$

(a) Verification of the given expression

Define a new magnetic field by taking the curl of the new vector potential.

$$\begin{gathered} \mathrm{B}\text{'}=\nabla \times \mathrm{A}\text{'} \\ =\nabla \left(\mathrm{A}+\nabla \wedge \right) \\ =\nabla \times \mathrm{A}+\nabla \times \nabla \wedge \\ =\nabla \times \mathrm{A} \\ =\mathrm{B} \end{gathered}$$

It is known that$\nabla \times \nabla \Lambda =0$

Determine the new electric field by using the gradient of the new potential.

$$\begin{gathered} \mathrm{E}\text{'}=-\nabla \mathrm{\phi }\text{'}-\frac{\partial \mathrm{A}\text{'}}{\partial \mathrm{t}} \\ =-\nabla \left(\mathrm{\phi }-\frac{\partial \wedge }{\partial \mathrm{t}}\right)-\frac{\partial \left(\mathrm{A}+\nabla \wedge \right)}{\partial \mathrm{t}} \\ =-\nabla \mathrm{\phi }+\nabla \left(\frac{\partial \mathrm{A}}{\partial \mathrm{t}}\right)-\frac{\partial \mathrm{A}}{\partial \mathrm{t}}-\frac{\partial }{\partial \mathrm{t}}\left(\nabla \wedge \right) \\ =-\nabla \mathrm{\phi }+\nabla \frac{\partial \mathrm{X}}{\partial \mathrm{t}}-\frac{\partial \mathrm{A}}{\partial \mathrm{t}}-\nabla \left(\frac{\partial \mathrm{X}}{\partial \mathrm{t}}\right) \\ =-\nabla \mathrm{\phi }-\frac{\partial \mathrm{A}}{\partial \mathrm{t}} \\ =\mathrm{E} \end{gathered}$$

So, $\frac{\partial }{\partial t}(\nabla \Lambda )=\nabla \left(\frac{\partial \Lambda }{\partial t}\right)$.

Due to the continuity of$\wedge$ , shift between the time-derivative and the gradient can be done.

Under gauge transformations, the electric and magnetic fields, E and B, remain invariant.

Thus, both the potentials give the same field.

(b) Verification of the given equation that it satisfies Schrodinger’s equation

Write the expression for the gauge transformation.

$$\begin{gathered} \mathrm{H}\text{'}\mathrm{\psi }\text{'}=\left\{\frac{1}{2\mathrm{m}}{\left(\mathrm{p}\text{'}-\mathrm{qA}\text{'}\right)}^2+\mathrm{q\phi }\text{'}\right\}\mathrm{\psi }\text{'} \\ =\left\{\frac{1}{2\mathrm{m}}\left[\frac{\hslash }{\mathrm{i}}\nabla -\mathrm{q}\left(\mathrm{A}+\nabla \wedge \right)\right]+\mathrm{q}\left(\mathrm{\phi }-\frac{\partial \wedge }{\partial \mathrm{t}}\right)\right\}{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi } \\ =\left\{\frac{1}{2\mathrm{m}}\left[\frac{\hslash }{\mathrm{i}}\nabla -\mathrm{q}\left(\mathrm{A}+\nabla \wedge \right)\right].\left[\frac{\hslash }{\mathrm{i}}\nabla -\mathrm{q}(\mathrm{A}+\nabla \wedge )\right]+\mathrm{q}\left(\mathrm{\phi }-\frac{\partial \wedge }{\partial \mathrm{t}}\right)\right\}{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi } \end{gathered}$$......(1)

Calculate the equation.

$$\begin{gathered} \left[\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{q}\left(\mathrm{A}+\nabla \wedge \right)\right]{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }=\frac{\sout{\mathrm{h}}}{\mathrm{i}}.\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{qe}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\nabla \wedge +\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }-{\mathrm{qe}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi A}-{\mathrm{qe}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\nabla \wedge \\ ={\mathrm{qe}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\nabla \wedge +\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\nabla \mathrm{\psi }-{\mathrm{qe}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi A}-{\mathrm{ge}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\nabla \wedge \\ =\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\nabla \mathrm{\psi }-{\mathrm{qe}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi A} \end{gathered}$$

So, this can be concluded that $\mathrm{\psi }={\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }$satisfy the Schrödinger equation

Consider equation (1) and (2).

role="math" localid="1658223145289" $$\begin{gathered} \mathrm{H}\text{'}\mathrm{\psi }\text{'}=\left\{\frac{1}{2\mathrm{m}}{\left[\frac{\hslash }{\mathrm{i}}\nabla -\mathrm{q}\left(\mathrm{A}+\nabla \wedge \right)\right]}^2+\mathrm{q}\left(\mathrm{\phi }-\frac{\partial \wedge }{\partial \mathrm{t}}\right)\right\}{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi } \\ =\frac{1}{2\mathrm{m}}{\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\left\{{\left(\frac{\hslash }{\mathrm{i}}\nabla -\mathrm{qA}\right)}^2\mathrm{\psi }\right\}+\mathrm{q}\left(\mathrm{\phi }-\frac{\partial \wedge }{\partial \mathrm{t}}\right){\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi } \\ ={\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\left\{\frac{1}{2\mathrm{m}}{\left(\frac{\hslash }{\mathrm{i}}\nabla -\mathrm{qA}\right)}^2+\mathrm{q\phi }-\mathrm{q}-\frac{\partial \wedge }{\partial \mathrm{t}}\right\}\mathrm{\psi } \\ ={\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\left\{\mathrm{H}-\mathrm{q}\frac{\partial \wedge }{\partial \mathrm{t}}\right\}\mathrm{\psi } \end{gathered}$$

Evaluate the above expression further.

$$\begin{gathered} \mathrm{H}\text{'}\mathrm{\psi }\text{'}={\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\left\{\mathrm{i}\hslash \frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}-\mathrm{q}\frac{\partial \wedge }{\partial \mathrm{t}}\mathrm{\psi }\right\} \\ =\mathrm{i}\hslash \frac{\partial }{\partial \mathrm{t}}\left({\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }\right) \\ =\mathrm{i}\hslash \frac{\partial \mathrm{\psi }\text{'}}{\partial \mathrm{t}} \end{gathered}$$

With gauge-transformed potentials are $\phi \text{'}$and$A\text{'}$, it is infered thatlocalid="1658218210175" $\mathrm{\psi }\text{'}={\mathrm{e}}^{\mathrm{iq\Lambda }/\hslash }\mathrm{\psi }$satisfies the time-dependent Schrodinger's equation.