As we all know, the probability current density in quantum mechanics is defined as: $$\textbf{J}=\dfrac{\hbar}{2mi}(\Psi^* \nabla \Psi-\Psi \nabla \Psi^*)$$ For simplicity let us work in one dimension and let us suppose a wave function $\Psi= A\ \text{cos}\ {kx}$. Applying the above definition and thus using $$J=\dfrac{\hbar}{2mi}\Big(\Psi^* \dfrac{\partial \Psi}{\partial x}-\Psi \dfrac{\partial \Psi^*}{\partial x}\Big)\quad\quad \text{we get:}\quad\quad J=0$$ Using the equation of continuity this means that: $$\dfrac{\partial \rho}{\partial t}=0,$$ which after solving gives us: $\rho=f(x)$. Thus the probability density at any point is independent of time. Now, this result will follow even if we take $\Psi= A\ \text{cos}\ {(kx-\omega t)}$. But here we can clearly see that the probability density i.e. $$|\Psi|^2=|A|^2\ \text{cos}^2\ {(kx-\omega t)}$$ is time dependent. Is it $A$ which carries the time dependence and is responsible for this apparent discrepancy?


3 Answers 3


A solution of the free one-dimensional Schroedinger equation:

$$ i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}\,\,\,\quad \text{(1)} $$


$$\psi = A e^{i(kx -\omega t)} \quad\quad\quad \text{(2)} $$

where $\omega$ fulfills the condition $\hbar \omega = \frac{(\hbar k)^2}{2m}$.

If tentatively one tries to construct a $\cos$-solution one would write

$$\psi = \frac{A}{2} e^{i(kx -\omega t)} + \frac{A}{2} e^{-i(kx -\omega t)} = A \cos (kx -\omega t)$$

Upon checking if $$\psi = A e^{-i(kx -\omega t)}$$ solves the Schroedinger equation one would only find a solution only if the following condition is fulfilled:

$$E = \hbar \omega = -\frac{(\hbar k)^2}{2m}$$

However, negative energy solutions are not allowed in the non-relativistic theory, therefore this solution has to be discarded, consequently also the $\cos$-solution has also to be discarded. This can, of course, be directly checked by inserting the $\cos (kx-\omega t)$ in the free Schroedinger-equation (1); it is not a solution. So one cannot expect it to fulfill the continuity equation.

So the only reasonable solutions in this context are either (2) or

$$\psi(x) = \cos(kx)\quad\quad\quad \text{(3)} $$

for the free time-independent Schroedinger equation

$$ \frac{\partial^2 \psi}{\partial x^2} +\frac{2m}{\hbar^2}E =0$$

with the condition $\frac{(\hbar k)^2}{2m} =E$.

Both solutions (2) and (3) fulfill the continuity equation, even if in the case of (3) it turns out to be quite uninteresting.

Solution (3) can of course be upgraded to a time-dependent solution by choosing

$$\psi(x,t) = e^{-i\omega t} \cos(kx)$$

Of course appropriate superpositions of either (2) or (3) would also be solutions, but using the right sign of $i$ in case of time-dependent solutions.

EDIT In case of the time-dependent solution (2) the probability current $J$ is non-zero, but its gradient is zero, therefore even if $\dot{\rho}=0$

$$ \dot{\rho} + \nabla J =0$$

is fulfilled.


The continuity relation holds for solutions of the Schrodinger equation. $A\cos (\omega t - k x)$ is not a solution.

  • $\begingroup$ I don't think that's true. $\psi = A \cos(kx-\omega t)$ is a superposition of two eigenfunctions of a Hamiltonian $H=v p$ - one with momenta $k$ and the other with $-k$. This can be further formalized by demanding periodic boundary conditions on a finite length $L$, so even the issue of normalization doesn't come into effect here. $\endgroup$
    – user245141
    Commented Jul 13, 2020 at 8:28
  • 1
    $\begingroup$ It is obviously true, just try plugging that function into the time-dependent Schrodinger equation. $\endgroup$
    – AfterShave
    Commented Jul 13, 2020 at 9:48
  • $\begingroup$ @AfterShave if $H = -iv \partial_x$, then $H \psi = iv k A \sin(kx-\omega t)$, $i d\psi/dt = i \omega A \sin(kx-\omega t)$ which are equal as long as $\omega = vk$ so $H\psi = i d\psi/dt $ what am I missing? $\endgroup$
    – user245141
    Commented Jul 13, 2020 at 10:04
  • $\begingroup$ The Hamiltonian in position basis is $$-\frac{\hbar^2}{2m} \partial_x^2$$ $\endgroup$
    – AfterShave
    Commented Jul 13, 2020 at 10:32
  • 1
    $\begingroup$ The expression for the probability current density is only valid for the Hamiltonian that I gave, your Hamiltonian will give a different expression. In fact I wouldn't call it a Hamiltonian at all since it does not have bounded energies. I stress again, the expression for the locally conserved proabability current density is dependent on the Hamiltonian. $\endgroup$
    – AfterShave
    Commented Jul 13, 2020 at 11:11

There are several issues to discuss here.

Since $\nabla\cdot J=\frac{\hbar}{2mi}(\Psi^\ast\nabla^2\Psi-\Psi\nabla^2\Psi^\ast)$, a TDSE solution satisfies$$\nabla^2\Psi=\frac{2m}{\hbar^2}(V\Psi-i\hbar\partial_t\Psi)\implies-\nabla\cdot J=\Psi^\ast\partial_t\Psi+\Psi\partial_t\Psi^\ast=\partial_t\rho,\,\rho:=\Psi^\ast\Psi.$$A choice of $V$ for which $\Psi:=\cos(kx-\omega t)$ solves the TDSE implies $\partial_t\cos^2(kx-\omega t)=0$, which is clearly wrong unless $\omega=0$. If $\omega\ne0$,$$V=\frac{\hbar^2}{2m}\frac{\nabla^2\Psi}{\Psi}+i\hbar\frac{\partial_t\Psi}{\Psi}=-\frac{\hbar^2k^2}{2m}+i\hbar\omega\tan(kx-\omega t)$$is a time-dependent potential with no ground state.

More to the point, this choice of $\rho$ doesn't integrate to $1$ on $\Bbb R$. Even if we try something like a particle in a finite box to get round this, your choice of $\rho$ is dimensionless, so won't integrate to the dimensionless value of $1$ over a region of finite dimensionful length. While we often see $\cos(kx-\omega t)$, $\sin(kx-\omega t)$ or $\exp i(kx-\omega t)$ in physics, in practice there's an overall factor to get the units right.

And in quantum mechanics, we expect $\Psi$ to in general be complex-valued. So let's now consider another option, $\Psi=A\exp i(kx-\omega t)$, where without loss of generality our constant $A$ can be assumed positive rather than of any other phase. So now$$V=-\frac{\hbar^2k^2}{2m}+\hbar\omega,\,\rho=A^2,\,\partial_t\rho=0.$$Again, there's a normalization issue that requires either infinite-potential walls (or $x$ to measure space around a circumference, but let's ignore things like quantum mechanics on a torus). Note that the particle in a box's Hailtonian's eigenfunctions are usually quoted as a sine or cosine in terms of $x$ alone, not $t$; but if we want to flesh out their time-dependence, we multiple by an overall $e^{-i\omega t}$ factor, which gives a behaviour unlike anything discussed above. In particular, this factor is irrelevant to $\rho$, which is time-independent as desired.


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