5
$\begingroup$

I was recently introduced to the concept of probability as a conserved quantity in the context of QM, and the mathematics behind using the continuity equation on probability, specifically the free-particle equation: $$ \nabla \cdot \mathbf{J} + \frac{\partial |\psi|^2} {\partial t} = 0 $$

Eventually, we came to the well-known result $$ \mathbf{J}(\mathbf{r}, t) = -\frac{i\hbar}{2m}\left[\psi^*\nabla \psi - \psi \nabla \psi^*\right] $$ I would read this as implying that a wavefunction with no imaginary component means $\psi \equiv \psi^*$ and hence $\mathbf{J} = 0 = -{\partial |\psi|^2}/{\partial t}$ and so $|\psi|^2 = \psi^2$ is a function only of position, in other words, a standing wave.

I realise this is probably not the most insightful inference, but I'm interested to know if the reasoning is correct.

$\endgroup$
  • 1
    $\begingroup$ Couldn't you reach this conclusion from the time dependent Schrodinger equation? $\endgroup$ – innisfree Mar 5 '17 at 13:22
  • $\begingroup$ The two formulations are equivalent - I suppose so. $\endgroup$ – catalogue_number Mar 5 '17 at 13:33
5
$\begingroup$

Like any other wavefunction, a real wavefunction evolves according to the Schrödinger equation $$i\hbar \partial_t \psi = \hat{H} \psi.$$ Thus a real wavefunction will in general not stay real, and in the case that $\mathbf J = 0$ holds at $t = 0$, it will in general not hold at $t > 0$.

However, suppose that $\psi$ is an eigenstate of $\hat H$ with eigenvalue $E_\psi$. Then $\psi(t) = e^{-iE_\psi t} \psi(t=0)$ and the current does indeed vanish for all $t$. (You can also perform a unitary transformation $\psi \mapsto \psi' = e^{-iE_\psi t} \psi$ to make the wavefunction $\psi'$ constant in time; this is the same as subtracting off $E_\psi$ from the Hamiltonian.)

What to make of the vanishing current in this case? Well, $|\psi(x)|^2 = \langle \psi |x\rangle\langle x| \psi \rangle $ is the expectation value of the projection operator that projects onto the $x$-eigenstate. In states that are eigenstates of the Hamiltonian, all expectation values and all probabilities are constant in time. So there can be no probability current, because there is no motion of probability mass.

Introductory QM texts emphasize Hamiltonian eigenstates and the time-independent Schrödinger equation, but all the interesting dynamics is in linear combinations.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Suppose you are in eigenstate of the Hamiltonian with non-zero angular momentum component $L_z$. For example the Hydrogen atom with $n=2$ and $l=-1$. If I understand correctly in this case the wave function should be intrinsically complex, is this true? Does it contradict the above argument? $\endgroup$ – Rudi_Birnbaum Oct 11 at 17:40
3
$\begingroup$

Your manipulations are correct, but not particularly meaningful.

As you've discovered, a purely real-valued wavefunction (i.e. one which obeys $\psi^*=\psi$) does not have any current associated with it. However, this does not mean that the wavefunction is time-invariant, as you can see by considering the evolution of a free particle that starts in a gaussian bump $\psi(x,0) = N e^{-x^2/2\sigma^2}$: the current density starts off equal to zero but then grows as the wavepacket starts to expand.

More genearlly, what your math should be telling you is that restricting yourself to real-valued wavefunctions is much too restrictive, because you are missing out on all wavefunctions with nonzero particle current. The current in quantum mechanics is intrinsically tied to the spatial gradient of the phase, and you can see this directly: write your wavefunction as the Ansatz $$ \psi(\mathbf r,t) = p(\mathbf r,t)e^{iS(\mathbf r,t)/\hbar}, $$ for $p$ and $S$ real-valued and differentiable, calculate the current density, and you will obtain $$ \mathbf J(\mathbf r,t) = \frac{1}{m}\nabla S(\mathbf r,t). $$ That is, the particle flow is encoded in spatial changes in the phase. If there is no such spatial change (e.g. if the wavefunction is purely real-valued) then there is obviously no current, but that's only because you chose a boring example.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

It is possible that the wave function is real, i.e., $\psi(\textbf{r}, t) = \psi^{\ast}(\textbf{r},t)$, at one particular moment $t$. Then, as you have correctly observed, $\textbf{J}(\textbf{r}, t) = 0$, and $\partial\psi/\partial t = 0$, but only at that moment. At later (or earlier) times, $\text{J}$ will become nonzero, and $\psi$ will evolve to develop an imaginary part.

The reason is quite simple. Suppose that $\psi(\textbf{r}, t) = \psi^{\ast}(\textbf{r},t)$ for all $\textbf{r}$ and $t$. Comparing the free-particle Schrödinger equation $i\hbar \partial \psi/\partial t = -\hbar^2/2m \,\nabla^2\psi$ and its complex conjugate shows that $\partial \psi/\partial t = 0$ and $\nabla^2\psi=0$. This means $\psi(\textbf{r}, t) = 0$ for all $\textbf{r}$ and $t$, so in any nontrivial case, $\psi(\textbf{r}, t)$ can't stay real.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Your expression for the current is only valid if there is no external field (https://en.wikipedia.org/wiki/Probability_current#Spin-0_particle_in_an_electromagnetic_field). For example, for a particle in electromagnetic field, there is gauge invariance, and you can make any solution of the Schroedinger equation real by a gauge transformation, at least locally (this was noted by Schroedinger in 1952).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It is important to note that if you do the manipulation in the second sentence of this answer, then the expression for the current will change and the OP's conclusions will no longer be valid. $\endgroup$ – Emilio Pisanty Mar 5 '17 at 16:30
  • $\begingroup$ @EmilioPisanty: There is a general expression for the current (I gave a link to it; one has to use it if there is an external field, say, external potential), and it remains valid under gauge transformation. The OP's expression for the current is only correct for a free particle. $\endgroup$ – akhmeteli Mar 5 '17 at 16:36

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.