[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).

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,

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

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.

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

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

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.