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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?

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    $\begingroup$ 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
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    $\begingroup$ @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

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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 $B = \nabla \times A =0$.

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  • $\begingroup$ 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

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