Griffiths 1.5: Normalization of a wave function

[CAF123]

When solving for expectation values, is t set to zero as well?

In general no, you will have to keep the t dependence. (Your system may not be in a stationary state of the Hamiltonian for example.)
Also, in your example, if you had to compute say, $\langle p \rangle$ then by differentiating once, you pull down another factor before the exponents cancel.

 

[Matterwave]

Can I just inject a little physics into this discussion. I see a lot of math, and that's all great, but some physical intuition would be something good to have. Let's examine the given wave function.

1) First notice its time dependence. The time dependence of this wave function is merely an overall phase factor. There can be no room for interference effects since the same phase factor multiplies the entire expression. Therefore, we know that this state is, in fact, a stationary state wherein all the expectation values of all operators will be time independent. So, looking for the expectation values, I should expect no dependence on t. This is NOT true for general states. A general state may have expectation values dependent on t, so that you may have to find something like $\left<x(t)\right>$.

2) Notice that the wave function is an even function of x. In other words, x is symmetric about x=0. This tells you that I have as much chance of finding the particle to the left of x=0 as to the right of x=0. From this consideration alone, without doing ANY calculations, I can immediately conclude $\left<x\right>=0$.

3) Notice that the wave function is "spread out" (i.e. it's not a delta function at some x). Therefore, we can conclude that the statistical properties of this wave function are such that if I measured x multiple times on identical copies of this wave function, that I should get some different values of x. The standard deviation of my statistical ensemble of measurement should be $\sigma_x=\sqrt{\left<x^2\right>-\left<x\right>^2}\neq 0$ . Thus, from this argument alone, I can see that $\left<x^2\right>\neq 0$.

 

[micromass]

2) Notice that the wave function is an even function of x. In other words, x is symmetric about x=0. This tells you that I have as much chance of finding the particle to the left of x=0 as to the right of x=0. From this consideration alone, without doing ANY calculations, I can immediately conclude $\left<x\right>=0$.

I'm going to be nitpicky and not agree with this. For example the Cauchy distribution http://en.wikipedia.org/wiki/Cauchy_distribution is symmetric around $0$. But the mean does not exist, since the integral does not exist.

 

[dauto]

I'm going to be nitpicky and not agree with this. For example the Cauchy distribution http://en.wikipedia.org/wiki/Cauchy_distribution is symmetric around $0$. But the mean does not exist, since the integral does not exist.

Well, you could use Cauchy's principal value to redefine that integral in which case the mean turns out to be zero as expected. At any rate, this is quite a pathological distribution. The main use of that distribution is to showcase the possibility of pathological distributions...