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Consider an electron in an electromagnetic field with scalar and vector potentials $\phi, \mathbf{A}$. Suppose for simplicity that $\mathbf{A}$ is time independent. Suppose also that we know the wavefunction $\psi$ of this electron. Then $\psi$ satisfies

$$i \psi_t = \Bigg[ \frac{1}{2m} \left(\hat{\mathbf{p}}- \frac{e \mathbf{A}}{c} \right)^2 + e \phi \Bigg] \psi = \hat{\mathcal{H}} \psi$$

The question concerns showing that if you perform a gauge transformation of the potentials:

$$\mathbf{A} \rightarrow \mathbf{A}'= \mathbf{A} + \nabla \Lambda$$ $$\phi \rightarrow \phi' = \phi - \frac{1}{c} \frac{\partial \Lambda}{\partial t} = \phi$$

for some scalar $\Lambda (t, \mathbf{x})$, the wavefunction transforms as $$\psi \rightarrow \psi' = \mathrm{exp} \left( \frac{i e \Lambda}{\hbar c} \right) \psi$$ i.e. it is multiplied by a phase. It is easy to show that $\psi'$ satisfies the transformed Schroedinger equation: $$i \psi'_t = \hat{\mathcal{H}}' \psi'$$

However, I would like to know if there are other possible solutions to the above equation. If so, what are they? Or, is $\psi'$ the only solution?

I tried to find other solutions by supposing $\psi' = f \psi$, where $f$ is unknown, and then plugging this into the new Schroedinger equation. This gives a new differential equation for $f$. However, so far my attempts at solving this differential equation have failed.

There is probably another way (perhaps via path integrals?) of showing this that I am not aware of. Could you give me a clue, please?

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    $\begingroup$ Please clarify if your question is this : If $\:\psi\:$ and $\:\psi'\:$ satisfy the Schroedinger equations respectively \begin{align} Schr (\psi, \mathbf{A},\phi) & =0 \tag{01}\\ Schr (\psi', \mathbf{A'},\phi') & =0 \tag{01'} \end{align} and $$ \psi'=e^{\tfrac{ie\Lambda}{\hbar}}\;\psi \tag{02} $$ then \begin{align} \mathbf{A'} & = \mathbf{A} + \boldsymbol{\nabla} \Lambda \tag{03a}\\ \phi' & = \phi - \dfrac{\partial \Lambda}{\partial t} \tag{03b} \end{align} $\endgroup$ – Frobenius Jun 17 '16 at 6:21
  • $\begingroup$ Not exactly. My question is: if we start with (01), then perform the transformations (03a) and (03b), what is the function $\psi'$ which satisfies (01'). So my question changes the order of what you are proposing. In other words, I would like to prove directly from the differential equation (01') that $\psi'$ must be (02). Hopefully this will be more clear from my edits. $\endgroup$ – rpf Jun 17 '16 at 13:06
  • $\begingroup$ This is really easy to see in relativistic case, as the equation is linear in momentum. Or in Lagrangian — essentially, the same. I tried a bit to derive this in you non-relativistic case, but the math goes bad. You exactly want the specified Hamiltonian, right? $\endgroup$ – Andrii Magalich Jun 19 '16 at 20:47
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This answer is motivated by the Aharonov-Bohm effect and proves what the OP asks for, but in the special case \begin{equation} \boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A} =\boldsymbol{0}=\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}' \quad \text{that is} \quad \mathbf{B} =\boldsymbol{0} \tag{01} \end{equation}

To simplify the expressions we :

  1. set \begin{equation} \hbar=1, \quad c=1, \quad e=1, \quad m=\dfrac{1}{2} \tag{02} \end{equation}

  2. use a dot for the partial derivative with respect to $\:t$ \begin{equation} \dot{\psi} (\mathbf{x},t ) \equiv \dfrac{\partial \psi (\mathbf{x},t )}{\partial t} \tag{03} \end{equation}

  3. omit the dependence $\:(\mathbf{x},t )\:$ unless otherwise necessary.

Now, in agreement with OP, we know that if to the Schroedinger equation of a particle in electromagnetic field $\:[\mathbf{A}(\mathbf{x},t ), \phi (\mathbf{x},t )]\:$

\begin{equation} i\dot{\psi} =\left[\left(-i\boldsymbol{\nabla}-\mathbf{A}\right)^{2}+\phi\right]\psi \tag{04} \end{equation}

we replace the wave function $\:\psi(\mathbf{x},t )\:$ by

\begin{equation} \psi'(\mathbf{x},t )=e^{i \Lambda(\mathbf{x},t )}\psi(\mathbf{x},t ) \quad \text{that is make the substitution} \quad \psi \: \rightarrow \: e^{-i \Lambda}\psi' \tag{05} \end{equation}

then this new wave function obeys the Schroedinger equation of a particle in electromagnetic field $\:[\mathbf{A}'(\mathbf{x},t ), \phi' (\mathbf{x},t )]\:$

\begin{equation} i\dot{\psi'} =\left[\left(-i\boldsymbol{\nabla}-\mathbf{A}'\right)^{2}+\phi'\right]\psi' \tag{06} \end{equation}

where

\begin{align} \mathbf{A}' & = \mathbf{A}+\boldsymbol{\nabla}\Lambda, \quad \text{with} \quad \Lambda(\mathbf{x},t ) \in \mathbb{R} \tag{07a}\\ \phi' & =\phi-\dot{\Lambda} \tag{07b} \end{align}

That is in summary

\begin{equation} \begin{pmatrix} i\dot{\psi} =\left[\left(-i\boldsymbol{\nabla}-\mathbf{A}\right)^{2}+\phi\right]\psi \\ \psi'(\mathbf{x},t )=e^{i \Lambda(\mathbf{x},t )}\psi(\mathbf{x},t ) \end{pmatrix} \Longrightarrow \begin{pmatrix} i\dot{\psi'} =\left[\left(-i\boldsymbol{\nabla}-\mathbf{A}'\right)^{2}+\phi'\right]\psi'\\ \mathbf{A}' = \mathbf{A}+\boldsymbol{\nabla}\Lambda, \quad \phi' = \phi-\dot{\Lambda} \end{pmatrix} \tag{08} \end{equation}

Note : Proof of this statement is found in textbooks and in web : http://www.physicspages.com/2013/02/01/electrodynamics-in-quantum-mechanics-gauge-transformations/

The question, in its 2nd version as in RPF's comment, is the inverse of (08) in the following sense :

\begin{equation} \begin{pmatrix} i\dot{\psi} =\left[\left(-i\boldsymbol{\nabla}-\mathbf{A}\right)^{2}+\phi\right]\psi \\ i\dot{\psi'} =\left[\left(-i\boldsymbol{\nabla}-\mathbf{A}'\right)^{2}+\phi'\right]\psi'\\ \mathbf{A}' = \mathbf{A}+\boldsymbol{\nabla}\Lambda, \quad \phi' = \phi-\dot{\Lambda} \end{pmatrix} \overset{\textbf{???}}{\Longrightarrow} \begin{pmatrix} \\ \psi'(\mathbf{x},t)=e^{i \mathrm{M}(\mathbf{x},t)}\psi(\mathbf{x},t)\\ \mathrm{M}(\mathbf{x},t) \in \mathbb{R} \end{pmatrix} \tag{09} \end{equation}

Now, if $\:\psi(\mathbf{x},t)\:$ obeys (04) under the condition (01) then

\begin{equation} \psi(\mathbf{x},t)=\psi_{0}(\mathbf{x},t) \exp \left[i\int_{\Gamma}\mathbf{A}(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'\right] \tag{10} \end{equation}

where $\:\Gamma(\mathbf{x})\:$ characterizes an arbitrary curve in 3-dimensional space which starts from any constant point $\:\mathbf{x}_{0}\:$ and ends at point $\:\mathbf{x}\:$, as in Figure, and $\:\psi_{0}(\mathbf{x},t) \:$ represents a solution of the Schrodinger equation (04) with $\:\mathbf{A}=\boldsymbol{0} \:$ but otherwise arbitrary $\:\phi(\mathbf{x},t) \:$, that is obeys the reduced Schrodinger equation

\begin{equation} i\dot{\psi}_{0} =\left[\left(-i\boldsymbol{\nabla}\right)^{2}+\phi\right]\psi_{0} \tag{11} \end{equation}

On the same footing after the transformation (07) and since the new wavefunction obeys (06) under the still valid condition (01) then

\begin{equation} \psi'(\mathbf{x},t)=\psi'_{0}(\mathbf{x},t) \exp \left[i\int_{\Gamma'}\mathbf{A}'(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'\right] \tag{12} \end{equation}

where $\:\Gamma'(\mathbf{x})\:$ characterizes an arbitrary curve in 3-dimensional space which starts from any constant point $\:\mathbf{x}'_{0}\:$ and ends at point $\:\mathbf{x}\:$, as in Figure, and $\:\psi'_{0}(\mathbf{x},t) \:$ represents a solution of the Schrodinger equation (06) with $\:\mathbf{A}'=\boldsymbol{0} \:$ but otherwise arbitrary $\:\phi'(\mathbf{x},t)[=\phi(\mathbf{x},t)-\dot{\Lambda}(\mathbf{x},t)]\:$, that is obeys the reduced Schrodinger equation

\begin{equation} i\dot{\psi'}_{0} =\left[\left(-i\boldsymbol{\nabla}\right)^{2}+\phi'\right]\psi'_{0} \tag{13} \end{equation}

Let now the gauge transformation

\begin{equation} \begin{pmatrix} i\dot{\psi}_{0} =\left[\left(-i\boldsymbol{\nabla}-\boldsymbol{0}\right)^{2}+\phi\right]\psi_{0} \\ \xi(\mathbf{x},t )=e^{i \Lambda(\mathbf{x},t )}\psi_{0}(\mathbf{x},t ) \end{pmatrix} \Longrightarrow \begin{pmatrix} i\dot{\xi} =\left[\left(-i\boldsymbol{\nabla}-\mathbf{A}_{\xi}\right)^{2}+\phi_{\xi}\right]\xi\\ \mathbf{A}_{\xi}= \boldsymbol{0}+\boldsymbol{\nabla}\Lambda, \quad \phi_{\xi} = \phi-\dot{\Lambda}=\phi' \end{pmatrix} \tag{14} \end{equation}

that is the wavefunction $\:\xi(\mathbf{x},t )\:$ obeys the Schrodinger equation

\begin{equation} i\dot{\xi} =\left[\left(-i\boldsymbol{\nabla}-\boldsymbol{\nabla}\Lambda\right)^{2}+\phi'\right]\xi \tag{15} \end{equation}

The condition (01) is satisfied for (15) too

\begin{equation} \boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}_{\xi} =\boldsymbol{\nabla}\boldsymbol{\times}\boldsymbol{\nabla}\Lambda=\boldsymbol{0} \tag{16} \end{equation}

so in analogy to the pairs of $\:\psi$-equations (10)-(11) and $\:\psi'$-equations (12)-(13)

\begin{equation} \xi(\mathbf{x},t)=\xi_{0}(\mathbf{x},t) \exp \left[i\int_{\Gamma_{\xi}}\mathbf{A}_{\xi} (\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'\right]=\xi_{0}(\mathbf{x},t) \exp \left[i\int_{\Gamma_{\xi}}\boldsymbol{\nabla}\Lambda(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'\right] \tag{17} \end{equation}

where $\:\Gamma_{\xi}(\mathbf{x})\:$ characterizes an arbitrary curve in 3-dimensional space which starts from any constant point $\:\mathbf{x}_{0 \xi}\:$ and ends at point $\:\mathbf{x}\:$, as in Figure, and $\:\xi_{0}(\mathbf{x},t) \:$ represents a solution of the Schrodinger equation (15) with $\:\mathbf{A}_{\xi}=\boldsymbol{0} \:$ but otherwise arbitrary $\:\phi'(\mathbf{x},t) \:$, that is obeys the reduced Schrodinger equation

\begin{equation} i\dot{\xi}_{0} =\left[\left(-i\boldsymbol{\nabla}\right)^{2}+\phi'\right]\xi_{0} \tag{18} \end{equation}

But (18) for $\:\xi_{0}(\mathbf{x},t)\:$ is identical to (13) for $\:\psi'_{0}(\mathbf{x},t)\:$ so we can identify the two functions and so

\begin{equation} \xi_{0}(\mathbf{x},t) \equiv \psi'_{0}(\mathbf{x},t) \tag{19} \end{equation} Combining (12),(19),(17) and the bottom equation in left parentheses in (14), that is $\:\xi=\exp[i\Lambda]\psi'_{0}\:$, we have

\begin{align} \psi'(\mathbf{x},t) & =e^{i \mathrm{M}(\mathbf{x},t)}\psi(\mathbf{x},t) \tag{20}\\ \mathrm{M}(\mathbf{x},t) & = \Lambda(\mathbf{x},t)+\int_{\Gamma'}\mathbf{A}'(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'-\int_{\Gamma}\mathbf{A}(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'-\int_{\Gamma_{\xi}}\boldsymbol{\nabla}\Lambda(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}' \tag{21} \end{align}

If the starting point of any curve is selected then the relative phase integral is independent of the path, since the vector function under the integral has zero curl. The 1rst and the last term of the rhs of (21) give

\begin{equation} \Lambda(\mathbf{x},t)-\int_{\Gamma_{\xi}}\boldsymbol{\nabla}\Lambda(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'=\Lambda(\mathbf{x},t)-\left[ \Lambda(\mathbf{x},t)-\Lambda(\mathbf{x}_{0\xi},t) \right]=\Lambda(\mathbf{x}_{0\xi},t) \tag{22} \end{equation}

If we choose $\:\mathbf{x}'_{0}\equiv \mathbf{x}_{0}\:$ then the 2nd and 3rd terms of the rhs of (21) give \begin{align} \int_{\Gamma'}\mathbf{A}'(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'-\int_{\Gamma}\mathbf{A}(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}' & =\int_{\Gamma'}\boldsymbol{\nabla}\Lambda(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'+\overbrace{\oint_{\Gamma' \cup \Gamma^{-}} \mathbf{A}(\mathbf{x}',t)\boldsymbol{\cdot}\mathrm{d}\mathbf{x}'}^{0} \\ & = \Lambda(\mathbf{x},t)-\Lambda(\mathbf{x}_{0},t) \tag{23} \end{align}

By equations (22) and (23) equation (21) yields

\begin{equation} \mathrm{M}(\mathbf{x},t) = \Lambda(\mathbf{x},t)-\Lambda(\mathbf{x}_{0},t) +\Lambda(\mathbf{x}_{0\xi},t) \tag{24} \end{equation}

Finally if we choose $\:\mathbf{x}_{0\xi}\equiv \mathbf{x}_{0}\:$ then

\begin{equation} \mathrm{M}(\mathbf{x},t) = \Lambda(\mathbf{x},t) \tag{25} \end{equation} enter image description here

Reference : EXAMPLE 1.6 The Aharonov-Bohm effect in "Quantum Mechanics - Special Chapters" by Walter Greiner, 1998 English Edition.

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  • $\begingroup$ Isn't this an overkill? $\endgroup$ – Andrii Magalich Jun 19 '16 at 23:08
  • $\begingroup$ @Frobenius: Can you please answer my question at: physics.stackexchange.com/questions/264916/… $\endgroup$ – N.G.Tyson Jun 27 '16 at 17:24
  • $\begingroup$ OK.Then don't answer. Please just tell (through comments) where I am wrong in my question. $\endgroup$ – N.G.Tyson Jun 28 '16 at 11:10
  • $\begingroup$ @Frobenius: I already finished digging up Maxwell's treatise on "Ampere Force Law" . Now I want to know something which is not in detail in Maxwell's treatise; that is about Newton's Third Law in weak form (equal action reaction) in the four force equations. I might be somewhere wrong in my question. That is why I end up getting the wrong conclusion,i.e.there would be an equal and opposite force in all four cases. I need the answerer to point out where I am wrong. $\endgroup$ – N.G.Tyson Jun 28 '16 at 13:36
  • $\begingroup$ @Frobenius: I plan to put the derivation of Ampere's Force Law on a website in an easy to understand language. I searched the internet for over two months for a derivation of Ampere's Force Law and I could find it only in Maxwell's Treatise. It might be useful for those people who wish to know a derivation of Ampere's Force Law. These forgotten theories of electromagnetism must be revived. $\endgroup$ – N.G.Tyson Jun 28 '16 at 13:45
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I can show this backwards and then explain the motivation. The uniqueness of the solution follows from the constraint on the transformation to be unphysical.

Say, we take your equation and transform $\psi$:

$$ i \partial_t \psi = \left( \frac{\Pi^2}{2m} + e \phi \right)\psi $$

becomes

$$ i \partial_t \left(e^{i\frac{e}{c} \Lambda}\psi\right) = \left( \frac{\Pi^2}{2m} + e \phi \right)e^{i\frac{e}{c} \Lambda} \psi $$

Lets evaluate this in stages.

$$ i \partial_t \left(e^{i\frac{e}{c} \Lambda}\psi\right) = i \partial_t \left(e^{i\frac{e}{c} \Lambda}\right) \psi + i e^{i\frac{e}{c} \Lambda} \partial_t \psi = e^{i\frac{e}{c} \Lambda} \left(-\frac{e}{c}\partial_t \Lambda \, \psi + i \Lambda \partial_t \psi \right) $$

Now, consider effect of $\hat{\Pi} = \hat{p} - \frac{e}{c} \vec{A}$ on the exponent.

$$ \hat{p} e^{i\frac{e}{c} \Lambda} = -i \vec{\nabla} e^{i\frac{e}{c} \Lambda} = e^{i\frac{e}{c} \Lambda} (\vec{\nabla}\Lambda + \hat{p}) $$

(due to Leibnitz rule of differentiation)

By fiddling with this for some time, you can show that

$$ \hat{\Pi}^2 e^{i\frac{e}{c} \Lambda} = e^{i\frac{e}{c} \Lambda} \left(\hat{\Pi} + \frac{e}{c} \vec{\nabla} \Lambda\right)^2 $$

Here you can already see how $e^{i\frac{e}{c} \Lambda} $ cancels out of both sides of the equation and how the additions are absorbed into $\phi$ and $A$. Basically, we commuted the exponent with differential operators to achieve this.


The rationale

Gauge symmetry was known from classical electrodynamics, so it had to be incorporated into quantum mechanics. However, it is a priori unclear how the electromagnetic potentials should enter the equation. The inspiration in the equation above comes from classical mechanics where the correct equations of motion can be achieved with the form of the Hamiltonian that we see (e.g., here; the gauge invariance is used implicitly).

But if we apply gauge transformation directly, we get a mess in our Schrodinger's equation. But we expect that this transformation has no physical consequence — so all observables have to stay the same. The state of the system is defined by it's wavefunction, but only the amplitude is physical, while it's phase is unobservable.

So, we want to absorb excessive terms in the equation into the coordinate-dependent phase-shift of the wavefunction that gives us no choice other then the transformation above.

So, this is the only transformation that does not influence the amplitude.

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