The Schrödinger equation of electron in a magnetic field is

$$ \frac{1}{2m} \left(-\mathrm{i}\hbar\nabla+\frac{e}{c}\mathbf{A}\right)^2 \psi + V\psi = E\psi $$

where $V=-e\phi$ and $\phi$ is the scalar potential.

And the solution is

$$ \psi(\mathbf{r}) = \psi_0(\mathbf{r}) \exp\left[-\frac{\mathrm{i}e}{\hbar c} \int^{\mathbf{r}}_{\mathbf{r}_0} \mathbf{A}(\mathbf{r}') \cdot \mathrm{d}\mathbf{r}'\right]. $$

It's easy to check this solution by putting into the previous equation. But how can I achieve the solution directly from Schrödinger equation?

  • 1
    $\begingroup$ Are you sure that this does not involve the potential $V$? Or your vector-potential is purely "gauge" one? $\endgroup$ Jun 15, 2018 at 15:44
  • 1
    $\begingroup$ @Stone-Zeng Why doesn't the solution involve the potential $V$? $\endgroup$ Jun 15, 2018 at 16:48
  • $\begingroup$ @probably_someone $\psi_0$ does involve the potential $V$. It's actually the solution of the equation without $\frac{e\mathbf{A}}{c}$ term. $\endgroup$
    – stone-zeng
    Jun 15, 2018 at 17:16
  • $\begingroup$ If $\mathbf{A}$ is a purely gauge function $\mathbf{A}=\nabla f(\mathbf{r})$, then your "solution" may be correct, but in general case I am not sure we can factorize the solution like that. $\endgroup$ Jun 15, 2018 at 18:01
  • $\begingroup$ The first term of your original equation $\propto (...)^2$ contains derivatives of $\mathbf{A}$, so it is not reduced in general case to the factor with $\mathbf{A}$ in the exponential. $\endgroup$ Jun 15, 2018 at 18:19

1 Answer 1


If $A$ is not a pure gauge, or more weakly if $\nabla \times A \neq 0$, then it is false that $$\nabla_r \int_{r_0}^r A(r')\cdot dr' = A(r)\tag{0}$$ Without this result, by direct inspection you see that your statement is false. Otherwise it is true. The proof is easy, it is the direct generalization of this elementary identity $$\left(-i\frac{d}{dx} + f(x)\right) \psi(x) =-ie^{+i \int^x_0 f(y) dy} \frac{d}{dx} e^{-i \int^x_0 f(y) dy}\psi(x) \tag{1}\:.$$ From (1), implementing once more the identity you have $$\left(-i\frac{d}{dx} + f(x)\right) \left(-i\frac{d}{dx} + f(x)\right) \psi(x) =(-i)^2e^{+i \int^x_0 f(y) dy} \frac{d}{dx} e^{-i \int^x_0 f(y) dy} e^{+i \int^x_0 f(y) dy} \frac{d}{dx} e^{-i \int^x_0 f(y) dy}\psi(x)\:,$$ that is $$\left(-i\frac{d}{dx} + f(x)\right)^2 \psi(x)= e^{+i \int^x_0 f(y) dy} \frac{d^2}{dx^2} e^{-i \int^x_0 f(y) dy}\psi(x)$$ and finally $$ e^{-i \int^x_0 f(y) dy} \left[\left(-i\frac{d}{dx} + f(x)\right)^2 + U(x)\right]\psi= \left(-\frac{d^2}{dx^2} +U(x) \right)e^{-i \int^x_0 f(y) dy}\psi(x)\:.$$ The crucial point in the computations above is that $$\frac{d}{dx}\int_0^x f(y) dy = f(x).\tag{2}$$ Unfortunately, the procedure does not work in dimension $>1$, since the generalization of $\int^x_0 f(y) dy$ depends on the integration path unless the vector field $A$ which replaces $f$ has zero curl. Also fixing arbitrarily a path, the generalisation (0) of (2) generally does not hold in dimension greater than $1$. Validity of (2) for dimension greater than $1$ is equivalent to saying that $A$ is a pure gauge at least locally, and however $B = \nabla \times A =0$.

  • $\begingroup$ If A has nonzero curl, is the solution even well defined? A path should be specified for the integral. $\endgroup$
    – Javier
    Jun 15, 2018 at 22:10
  • $\begingroup$ In this case, it is however false that $\nabla_r \int^r_\gamma A(r') dr' = A(r)$ no matter the path$\gamma$ you fix. The procedure does not work if $A$ is not a pure gauge field. $\endgroup$ Jun 16, 2018 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.