91影视

The expression for hamilontonia of electrodynamics is as follows,

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

The expression for Gauge transformation is as follows,

${\mathit{A}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{A}\mathbf{+}\mathbf{鈭�}\mathit{蚂}$

The expression for gauge wave function is as follows,

${\mathit{蠄}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{e}}^{\mathbf{i}\mathbf{q}\mathbf{蚂}\mathbf{/}\mathbf{鈩�}}\mathit{蠄}$

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

$$\begin{gathered} \mathrm{B}\text{'}=鈭�脳\mathrm{A}\text{'} \\ =鈭�\left(\mathrm{A}+鈭�鈭�\right) \\ =鈭�脳\mathrm{A}+鈭�脳鈭�鈭� \\ =鈭�脳\mathrm{A} \\ =\mathrm{B} \end{gathered}$$

It is known that$鈭�脳鈭�蚂=0$

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

$$\begin{gathered} \mathrm{E}\text{'}=-鈭�\mathrm{蠁}\text{'}-\frac{鈭�\mathrm{A}\text{'}}{鈭�\mathrm{t}} \\ =-鈭�\left(\mathrm{蠁}-\frac{鈭�鈭�}{鈭�\mathrm{t}}\right)-\frac{鈭�\left(\mathrm{A}+鈭�鈭�\right)}{鈭�\mathrm{t}} \\ =-鈭�\mathrm{蠁}+鈭�\left(\frac{鈭�\mathrm{A}}{鈭�\mathrm{t}}\right)-\frac{鈭�\mathrm{A}}{鈭�\mathrm{t}}-\frac{鈭�}{鈭�\mathrm{t}}\left(鈭�鈭�\right) \\ =-鈭�\mathrm{蠁}+鈭�\frac{鈭�\mathrm{X}}{鈭�\mathrm{t}}-\frac{鈭�\mathrm{A}}{鈭�\mathrm{t}}-鈭�\left(\frac{鈭�\mathrm{X}}{鈭�\mathrm{t}}\right) \\ =-鈭�\mathrm{蠁}-\frac{鈭�\mathrm{A}}{鈭�\mathrm{t}} \\ =\mathrm{E} \end{gathered}$$

So, $\frac{鈭�}{鈭�t}(鈭�蚂)=鈭�\left(\frac{鈭�蚂}{鈭�t}\right)$.

Due to the continuity of$鈭�$ , 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.

$$\begin{gathered} \mathrm{H}\text{'}\mathrm{蠄}\text{'}=\left\{\frac{1}{2\mathrm{m}}{\left(\mathrm{p}\text{'}-\mathrm{qA}\text{'}\right)}^2+\mathrm{辩蠁}\text{'}\right\}\mathrm{蠄}\text{'} \\ =\left\{\frac{1}{2\mathrm{m}}\left[\frac{鈩�}{\mathrm{i}}鈭�-\mathrm{q}\left(\mathrm{A}+鈭�鈭�\right)\right]+\mathrm{q}\left(\mathrm{蠁}-\frac{鈭�鈭�}{鈭�\mathrm{t}}\right)\right\}{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄} \\ =\left\{\frac{1}{2\mathrm{m}}\left[\frac{鈩�}{\mathrm{i}}鈭�-\mathrm{q}\left(\mathrm{A}+鈭�鈭�\right)\right].\left[\frac{鈩�}{\mathrm{i}}鈭�-\mathrm{q}(\mathrm{A}+鈭�鈭�)\right]+\mathrm{q}\left(\mathrm{蠁}-\frac{鈭�鈭�}{鈭�\mathrm{t}}\right)\right\}{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄} \end{gathered}$$......(1)

Calculate the equation.

$$\begin{gathered} \left[\frac{\sout{\mathrm{h}}}{\mathrm{i}}鈭�-\mathrm{q}\left(\mathrm{A}+鈭�鈭�\right)\right]{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}=\frac{\sout{\mathrm{h}}}{\mathrm{i}}.\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{qe}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}鈭�鈭�+\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}-{\mathrm{qe}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄础}-{\mathrm{qe}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}鈭�鈭� \\ ={\mathrm{qe}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}鈭�鈭�+\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}鈭�\mathrm{蠄}-{\mathrm{qe}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄础}-{\mathrm{ge}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}鈭�鈭� \\ =\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}鈭�\mathrm{蠄}-{\mathrm{qe}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄础} \end{gathered}$$

So, this can be concluded that $\mathrm{蠄}={\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}$satisfy the Schr枚dinger equation

Consider equation (1) and (2).

role="math" localid="1658223145289" $$\begin{gathered} \mathrm{H}\text{'}\mathrm{蠄}\text{'}=\left\{\frac{1}{2\mathrm{m}}{\left[\frac{鈩�}{\mathrm{i}}鈭�-\mathrm{q}\left(\mathrm{A}+鈭�鈭�\right)\right]}^2+\mathrm{q}\left(\mathrm{蠁}-\frac{鈭�鈭�}{鈭�\mathrm{t}}\right)\right\}{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄} \\ =\frac{1}{2\mathrm{m}}{\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}\left\{{\left(\frac{鈩�}{\mathrm{i}}鈭�-\mathrm{qA}\right)}^2\mathrm{蠄}\right\}+\mathrm{q}\left(\mathrm{蠁}-\frac{鈭�鈭�}{鈭�\mathrm{t}}\right){\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄} \\ ={\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}\left\{\frac{1}{2\mathrm{m}}{\left(\frac{鈩�}{\mathrm{i}}鈭�-\mathrm{qA}\right)}^2+\mathrm{辩蠁}-\mathrm{q}-\frac{鈭�鈭�}{鈭�\mathrm{t}}\right\}\mathrm{蠄} \\ ={\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}\left\{\mathrm{H}-\mathrm{q}\frac{鈭�鈭�}{鈭�\mathrm{t}}\right\}\mathrm{蠄} \end{gathered}$$

$$\begin{gathered} \mathrm{H}\text{'}\mathrm{蠄}\text{'}={\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\left\{\mathrm{i}鈩�\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{t}}-\mathrm{q}\frac{鈭�鈭�}{鈭�\mathrm{t}}\mathrm{蠄}\right\} \\ =\mathrm{i}鈩�\frac{鈭�}{鈭�\mathrm{t}}\left({\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}\right) \\ =\mathrm{i}鈩�\frac{鈭�\mathrm{蠄}\text{'}}{鈭�\mathrm{t}} \end{gathered}$$

With gauge-transformed potentials are $蠁\text{'}$and$A\text{'}$, it is infered thatlocalid="1658218210175" $\mathrm{蠄}\text{'}={\mathrm{e}}^{\mathrm{iq蚂}/鈩�}\mathrm{蠄}$satisfies the time-dependent Schrodinger's equation.