In his introductory book on quantum mechanics, Griffiths solves the time-independent Schrödinger equation for a delta potential of the form $$V(x) = -V_0\delta(x).$$

Now, while analyzing scattering states he derives the following equations in the various regions,

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Here $\displaystyle k = \frac{\sqrt{2mE}}{\hbar}$. The continuity of $\psi$ and $\frac{d\psi}{dx}$ gives two equations. However, we have four variables ($A$, $B$, $F$, $G$). He now argues:

... Recall that $\exp(ikx)$ gives rise [when coupled with the time-dependent factor $\exp(-iEt/\hbar)$] to a wave function propagating to the right, and $\exp(-ikx)$ leads to a wave propagating to the left. It follows that $A$ is the amplitude of a wave coming in from the left, $B$ is the amplitude of a wave returning to the left, and $G$ is the amplitude of a wave coming in from the right. In a typical scattering experiment particles are fired from one direction-let's say from the left. In that case the amplitude of wave coming in from the right will be zero.

Now the trouble with this argument is where he draws the connection between the direction of firing the particle and the direction of propagation of the wave function. As far as I know, $\psi$ is physically meaningless and these scalar components of $\psi$ (as shown in the figure) are traveling waves only in a purely mathematical sense. How can we draw conclusions about $G$ based on the direction of firing?


1 Answer 1


They're not purely mathematical. This can easily be seen by calculating the Schrödinger probability current, $$ j(x) = \frac{\hbar}{2mi}\left[\psi^*(x)\frac{\partial\psi}{\partial x}(x) - \psi(x)\frac{\partial\psi^*}{\partial x}(x) \right], $$ on the state $\psi(x) = A e^{\pm ikx}$, which will give you $j(x) = |A|^2\cdot(\pm \hbar k)/m$, i.e. if the sign is negative then you have a particle flux to the left.

Similarly, saying "$\psi$ is physically meaningless" is not particularly correct, either - it is a tricky object to interpret, but its physical content goes well beyond just the position-space probability density $\rho_X(x) = |\psi(x)|^2.$ As an example, the phase is crucial in determining the momentum-space density $\rho_P(p) = \left|\tilde \psi(p)\right|^2$, but even that is not enough - even full knowledge of $\rho_X(x)$ and $\rho_P(p)$ is insufficient to completely determine either $\psi$ or the experimental observables that can be obtained from it.


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