Proof of Momentum Operator as Hermitian: Step-by-Step Guide

[aaaa202]

Now, I know this has been discussed a few times, but I still don't get the proof of how p is a hermitian operator, so I thought someone might be able to help me. So i got this to start off with:
$$\int_{-\infty}^\infty \Psi^\ast(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi(x,t) dx$$
Which when using the integration by parts for me yields:
$$[\frac{\hbar}{i}\Psi^\ast(x,t)\Psi(x,t)]^\infty_{-\infty}- \int_{-\infty}^\infty \Psi(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi^\ast(x,t) dx$$
Now everyone else gets this in front of the integral:
$$\frac{\hbar}{i} [\Psi^\ast(x,t)\Psi(x,t)]^\infty_{-\infty}$$
And i don't understand how. Do the bars with the limits represent an integral, or what do they represent? I'm not that good at maths so could someone step for step show how they get to there?
And if the bars represent and integral, why does it disappear as the wave functions tends to plus or minus inf? I thought a wave function integrated over the entire space had to yield 1.

 

[bapowell]

There is a typo here. The bar with the limits should appear also in the first term of the second line -- this is shorthand for saying, "I've solved the indefinite integral, it remains to evaluate it at these limits."

 

[Fredrik]

Now everyone else gets this in front of the integral:
$$\frac{\hbar}{i} [\Psi^\ast(x,t)\Psi(x,t)]^\infty_{-\infty}$$
And i don't understand how. Do the bars with the limits represent an integral, or what do they represent? I'm not that good at maths so could someone step for step show how they get to there?

They're using the fundamental theorem of calculus: If F'=f, then

$$\int_a^b f(x)dx=[F(x)]_a^b=F(b)-F(a)$$

The last equality shows how the bracket notation is defined.

The expression you're integrating is of the form f(x)g'(x), which can be rewritten using the product rule (for derivatives): f(x)g'(x)=(fg)'(x)-f'(x)g(x). The fundamental theorem of calculus is easy to apply to the first term on the right.

And if the bars represent and integral, why does it disappear as the wave functions tends to plus or minus inf? I thought a wave function integrated over the entire space had to yield 1.

You're supposed to use that |ψ(x)|→0 as x→±∞. (This actually isn't true for all ψ in L²(ℝ), but your teacher may not even be aware of that. I think it's true for all wavefunctions of physical interest though).

 

[aaaa202]

Okay, call me stupid, but I just can't get it to work:
Using product rule we obtain (as you said):
∫f(x)g’(x) dx = ∫(f(x)g(x))’ – f ’(x)g(x) dx = ∫(f(x)g(x))’ - ∫f ’(x)g(x) = f(x)g(x) – f ’(x)G(x) - ∫f ’’(x)G(x) dx
I'm assuming that ∫(f(x)g(x)'dx = f(x)f(x) right?
Tell me where I'm wrong please : (

 

[Fredrik]

You're supposed to start out like this:

$$\int_a^b f(x)g'(x)dx=\int_a^b((fg)'(x)-f'(x)g(x))dx=[f(x)g(x)]_a^b-\int_a^b f'(x)g(x)$$

$$=f(b)g(b)-f(a)g(a)-\int_a^b f'(x)g(x)$$

...and then conclude that since your functions satisfy f(a)=g(a)=f(b)=g(b)=0, you have

$$\int_a^b f(x)g'(x)dx=-\int_a^b f'(x)g(x)$$

At this point, you're supposed to stop trying to integrate stuff and instead think about what your result means.