I found in this article a straightforward way to calculate the eigenvalues of the hamiltonian of an electron under the influence of an homogenous magnetic field (p. 5): http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-06.pdf


For this, they express the magnetic potential as $$\vec{A}=\frac{1}{2}B(-y,x,0)$$, which gives rise to the magnetic field using the well-known relation $\vec{\nabla}\times \vec{A}$. Using the Pauli hamiltonian, this leads to the eigenvalue equation:

$$H_0\Psi(r)+\frac{\mu_B B}{\hbar}(L_z+\sigma_z)\Psi(r)=E\Psi(r)$$

which allows a straightforward calculation of the energies.

However, I noticed that one can also calculate the same magnetic field using the potential $$\vec{A}=B(-y,0,0)$$, which is no surprise since the magnetic potential is not unique.

Nevertheless, when trying to perform the same calculations I do not find the same energies as before. In particular, I arrived at the following eigenvalue equation:

$$H_0\Psi(r)+\frac{\mu_B B}{\hbar}(L_z+\sigma_z+i\hbar x\partial_y)\Psi(r)=E\Psi(r)$$

That is, it only differs by an additive term $i\hbar^2x\partial_y$, which is weird since my intuition says I should find the same energies (since the magnetic field is exactly the same in both cases). Why does this happen? Or could I somehow reduce this equation to the previous one?

Additional information: to be more clear on how I arrived at the result, I started by calculating, for $\vec{A}=B(-y,0,0)$ and $e=-e$:

$$\vec{A}\cdot\vec{p}=A_xp_x=B(i\hbar y\partial_x)=B(L_z+i\hbar x\partial_y)$$


The hamiltonian is, in it's expanded form:

$$\hat{H}=H_0+\frac{1}{2m}(-e\vec{A}\cdot\vec{p}+i\hbar\vec{\nabla}\cdot\vec{A}+e^2\vec{A}^2)+e\phi+\mu_b \vec{\sigma}\cdot\vec{B}$$

Substituting, with $\phi=0$, and neglecting second order terms in $\vec{A}$:

$$\hat{H}=H_0+\frac{1}{2m}(eB(L_z+i\hbar x\partial_y))+e\phi+\mu_b \vec{\sigma}\cdot\vec{B}$$

$$\hat{H}=H_0+\frac{\mu_B B}{\hbar}(L_z+\sigma_z+i\hbar x\partial_y)$$

For stationary states, we can use the time-independent Schrodinger-equation:


$$H_0\Psi(r)+\frac{\mu_B B}{\hbar}(L_z+\sigma_z+i\hbar x\partial_y)\Psi(r)=E\Psi(r)$$

  • $\begingroup$ Just FYI you quoted the PDF incorrectly, it's $\vec{A}=\frac{1}{2}B(-y,x,0)$. The curl for $(-y,-x,0)$ is zero. $\endgroup$
    – Cuspy Code
    Oct 2, 2018 at 16:17
  • $\begingroup$ You're right I had a mistake on the sign of the y-component. I'm correcting it right now. $\endgroup$
    – Charlie
    Oct 2, 2018 at 16:25
  • $\begingroup$ Your first equation only contains $B$. Since $B$ does not change how did you arrive at the second one? $\endgroup$
    – my2cts
    Oct 2, 2018 at 19:29
  • $\begingroup$ I guess you refer to the potential (correct me if I'm wrong). The first expression for $\vec{A}=\frac{B}{2}(-y,x,0)$ is proposed in the article, and we can verify by direct calculation that $\vec{\nabla}\times\vec{A}=\frac{B}{2}(\partial_x(x)-\partial_y(-y))\hat{z}=(\frac{B}{2}+\frac{B}{2})\hat{z}=B\hat{z}$. Later, for the second expression of $\vec{A}$, I noticed that we also get the same $\vec{B}=B\hat{z}$ field if we instead use $\vec{A}=B(-y,0,0)$. $\endgroup$
    – Charlie
    Oct 2, 2018 at 19:48
  • 2
    $\begingroup$ Are you sure that the eigenvalues are in fact different? $\endgroup$
    – Javier
    Oct 4, 2018 at 15:20

2 Answers 2


This is a classic case in which gauge invariance plays a large role and one must be careful. So you know that $A'=A+\nabla\Lambda$ is the form of a generic gauge tranformation.

Observable quantities must be gauge invariant, let's call $\psi$ the wave function when $A$ is involved and $\psi'$ the wave function when $A'$ is chosen. Then the expectation value for an observable $O$ must be the same between the two states: $$\langle \psi \rvert O\rvert \psi\rangle=\langle \psi' \rvert O\rvert \psi'\rangle$$ in particular, in quantum mechanics the position $x$ and cinematic momentum $\Pi=m\dot{x}$ (the one that classically traces the actual trajectory) of a particle must be gauge invariant operators. Notice that $p=\frac{\partial L}{\partial \dot{x}}$ is the canonical momentum and is not a guage invariant quantity. They are connected by the relation $\Pi=p-eA$ (with the speed of light $c=1$).

Then $$\langle \psi \rvert x\rvert \psi\rangle=\langle \psi' \rvert x\rvert \psi'\rangle\\ \langle \psi \rvert p-eA\rvert \psi\rangle=\langle \psi' \rvert p-eA' \rvert \psi'\rangle$$

Furthermore, we assume that since $O$ can be the identity $$\langle \psi \rvert \psi\rangle=\langle \psi' \rvert \psi'\rangle$$ which means that there exists an unitary operator $G$ such that $$\rvert \psi\rangle=G\rvert \psi'\rangle$$ such that if we substitute in the equations above $$G^+xG=x\\ G^+(p-eA-e\nabla\Lambda)G=(p-eA)$$ which implies that $$G=\exp\left(\frac{i e \Lambda(x)}{\hbar}\right)$$ in fact: $$p'=G^+ p\ G=G^+\left[p,G\right]+p=G^+(-i\hbar\nabla G)+p=p+e\nabla \Lambda$$ which if you substitute into the expression of $\Pi$ cancels $\nabla\Lambda$. So the canonical momentum and the vector potential conspire together to keep the cinematic momentum a gauge invariant quantity.

Alternatively the form of $G$ can be seen directly from the Schroedinger equation: $$\left[\frac{(p-eA)^2}{2m}+e\phi\right]\psi=i\hbar \frac{\partial}{\partial t}\psi$$ while in the transformed case $$\left[\frac{(p-eA-e\nabla\Lambda)^2}{2m}+e\phi\right]\psi'=i\hbar \frac{\partial}{\partial t}\psi'$$ Then you just see that if $$\psi'=\exp\left(\frac{ie\Lambda}{\hbar}\right)\psi$$ you get, in accordance with what we have proven before, that if you multiply from the left by $G^+$ the Schroedinger equation comes back to the previous form:

$$\exp\left(\frac{-ie\Lambda}{\hbar}\right)\frac{(p-eA-e\nabla\Lambda)^2}{2m}\exp\left(\frac{ie\Lambda}{\hbar}\right)+e\phi=\frac{(p-eA)^2}{2m}+e\phi$$ and $G^+\psi'=\psi$ ($G$ does not depend on time).

As you can see, we have proven that the Hamiltonian $H$ is a gauge invariant operator $G^+ H G=H$ and then it must have the same eigenvalues in all gauges. The Schoredinger equation to remain gauge invariant needs the states/ eigenvectors to be transformed too! We have shown that they are connected by a unitary transformation which is really a phase factor in our case.

Regarding your particular case: you didn't transorm $p$ in the Hamiltonian and $\psi$ in the eigenvalue problem!

$A=(-\frac{1}{2}B y,\frac{1}{2}B x,0)$ and $A' = (-B y,0,0)$, then it is trivial to choose $\Lambda(x,y)=-\frac{1}{2} B xy$ such that $\nabla \Lambda=(-\frac{1}{2} B y,-\frac{1}{2} B x,0)$ and therefore $$A'=A+\nabla \Lambda=(-\frac{1}{2}B y,\frac{1}{2}B x,0)+(-\frac{1}{2} B y,-\frac{1}{2} B x,0)=(-B y,0,0)$$

But now you know that also $p'=p+e\nabla \Lambda$ and therfore the knietic part of the Hamiltonian doesn't change: $$p'-eA'=p+e(-\frac{1}{2} B y,-\frac{1}{2} B x,0)-e(-B y,0,0)=p-(-\frac{1}{2}B y,\frac{1}{2}B x,0)=p-eA$$ and so when you do the square and the scalar products and all else, nothing changes.


being G an unitary operator we can write it as the exponential of an hermitian operator: $G=\exp(\frac{i}{\hbar} f(x))$, so if you come back to the tranformation equation for the canonical momentum $$p'=G^+ p\ G=G^+\left[p,G\right]+p=G^+(-i\hbar\nabla G)+p=p+\nabla f(x)$$ and so $f(x)=e\Lambda(x)$ if $\Pi$ is to be invariant.

  • $\begingroup$ Thanks, I was just thinking about gauge invariance but I wasn't sure how to proof the Hamiltonian is gauge invariant after finding the corresponding gauge transformation between both vectorial fields. Just one last question, when you deduced the form of $G$, I suppose you used the general expression for an unitary operator $U=(-\frac{i}{\hbar}O(x))$, am I correct? $\endgroup$
    – Charlie
    Oct 4, 2018 at 19:35
  • $\begingroup$ Precisely, I have edited the answer just for completeness. $\endgroup$
    – Fra
    Oct 4, 2018 at 19:44
  • $\begingroup$ Thanks a lot, I understood better the ideas, I'm going to reproduce the results to practice a bit more. I'll mark this post as answered then (I'll award the bounty as soon as the option is available). $\endgroup$
    – Charlie
    Oct 4, 2018 at 19:46
  • $\begingroup$ Why when you calculated $G$ you used the gauge invariance of $p-qA'$, but when you substituted in the hamiltonian you used $p'-qA'$? If $G$ transforms such that $G^\dagger (p-qA')G=p-qA$, shouldn't you be using $(p-qA')$ to preserve the hamiltonian gauge invariance, since $G^\dagger H'G=H$? $\endgroup$
    – Charlie
    Oct 5, 2018 at 1:04
  • $\begingroup$ If $H_{g}$ is the Hamiltonian in a particular gauge, you have that the transformation rule is $H_{g_1}=G^+ H_{g_2} G$, since $H$ is gauge invariant you discover that $H_{g_1}=H_{g_2}$. The Schroedinger equation gives you $H_{g_2}G\psi=E\ G \psi$, so if you multiply by $G^+$, you get $G^+H_{g_2}G\psi= E\psi$ and so you get the transformation rule because the momentum acts on $G$. You can do the opposite but then you have to change $ G\rightarrow G^+$. I used $p'-qA'$ becuase $G^+(p-qA')G=p'-qA'$. $\endgroup$
    – Fra
    Oct 5, 2018 at 11:34

my intuition says I should find the same energies (since the magnetic field is exactly the same in both cases). Why does this happen?

In general, gauge transformation of external 4-potential may change the energy eigenvalues. The EM field is not affected by the transformation, but energy eigenvalues may be, because energy is a quantity more like the electric potential - defined by physical conditions up to an additive constant, which can be chosen arbitrarily, and the same is true for the energy operator and thus also for energy eigenvalues. For example, consider the gauge transformation

$$ \phi \rightarrow \phi + C $$

$$ \mathbf A \rightarrow \mathbf A $$

where $C$ is a number independent of position and time. With Hamiltonian $H=\mathbf{p}^2/2m + q\phi$, the above transformation implies transformation in the energy eigenvalues

$$ E_n\rightarrow E_n + qC. $$

So, in general, your intuition is not correct. Gauge transformations may change energy eigenvalues.

However, in your case of particle in pure magnetic field, if the change of vector potential is done in the usual way with some differentiable function $\chi$ : $$ \mathbf A \rightarrow \mathbf A + \nabla \chi $$ then your intuition is correct: when only the vector potential is so transformed, and the electric potential is unchanged, the eigenvalues remain unchanged. It is true that the eigenvalue equation

$$ \frac{(\mathbf p - q\mathbf A)^2}{2m}\psi = E_n\psi $$

changes into $$ \frac{(\mathbf p - q\mathbf A -q\nabla \chi)^2}{2m}\psi' = E'_n\psi' $$ so it looks like the transformation will have effect on the eigenvalues because the operators changed. However, actually this has no effect on the eigenvalues, because $\psi$ transforms as

$$ \psi \rightarrow e^{\frac{q}{i\hbar}\chi} \psi $$

and with this it follows that the new equation is just multiple of the original one: $$ e^{\frac{q}{i\hbar}\chi}\frac{(\mathbf p - q\mathbf A )^2}{2m}\psi = e^{\frac{q}{i\hbar}\chi}E'_n\psi. $$ The exponential factor depends on coordinates, but is everywhere non-zero, so can be written off and we conclude $E_n' = E_n$.


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