# Small psi in the time-independent Schrödinger equation

I'm a total beginner in quantum mechanics, and I am learning about time-independent Schrödinger equation. we separate the wave function into two functions $$\Psi(x, t) = \psi(x)\phi(t).$$ Does the probability density $$p = \psi(x)\psi(x)^*~ ?$$ Shouldn't it be $$p = (\psi(x)\phi(t))(\psi(x)\phi(t))^*~?$$ my instructor's slides are very misleading about this point.

The two expressions you give are probabilities for different things.

You have some wavefunction $$\Psi(x,y,z,t)$$ say (it need not be a product wavefunction like in your post, but if it is we'll get some nice properties). In general, the probability density to observe the particle at some point in space-time is:

$$p(x,y,z,t)=|\Psi|^2$$

If we only want to know the probability density to be at some point along the $$x$$ axis and don't care about the other variables then we need to integrate them out:

$$p(x)=\int dt \int dz \int dy \ p(x,y,z,t)$$

In the special case that $$\Psi(x,t)=\psi(x)\phi(t)$$ (I'm going back to 2D to make the typing easier, but the extension is obvious) we have:

$$p(x)=\int dt \psi^*(x)\psi(x)\phi^*(t)\phi(t) = \psi^*(x)\psi(x) \int dt \phi^*(x)\phi(x)$$

But the exact way we wrote our product $$\Psi=\psi\phi$$ we could have moved constants arbitrarily around. Thus we are free to declare that each factor separately has norm 1 and that the last integral is 1. This is a normalization choice and we could choose to do things differently (e.g. when working in spherical polars one might choose to have the integral over angles come out to $$4\pi$$, but then the radial integral must come out to $$\frac{1}{4\pi}$$).

• $p(x)=\int dt \psi^*(x)\psi(x)\phi^*(t)\phi(t)$ is not correct. I mean you can evaluate such a quantity but it wouldn't be meaningful, the probability (density) that is meaningful is the one evaluated at a given moment in time.
– ACat
Mar 22 at 2:32