Infinite square well expectation value problem

[Fakestreet123]

Homework Statement

A particle in an infinite box is in the first excited state (n=2). Obtain the expectation value 1/2<xp+px>


2. The attempt at a solution

Honestly, I don't even know where to begin.
I assumed V<0, V>L is V=∞ and 0<V<L is V=0

I tried setting up the expectation value formula

1/2∫x|ψ^*Pψ|²+(x|ψ²)^*P x|ψ|²

but what is ψ? is it √(2/L) Sin ((n²πx)/L) because its the solution to the infinite well? are the bounds from 0 to L?

Thanks for viewing my question!

 

[dingo_d]

\psi is the wave function for the given problem, that you find by solving Schrodinger equation for a given potential

 

[vela]

Homework Statement

A particle in an infinite box is in the first excited state (n=2). Obtain the expectation value 1/2<xp+px>


2. The attempt at a solution

Honestly, I don't even know where to begin.
I assumed V<0, V>L is V=∞ and 0<V<L is V=0

You mean V=∞ when x<0 or x>L.

I tried setting up the expectation value formula

1/2∫x|ψ^*Pψ|²+(x|ψ²)^*P x|ψ|²

How did you get this?

but what is ψ? is it √(2/L) Sin ((n²πx)/L) because its the solution to the infinite well? are the bounds from 0 to L?

You do use a solution for the infinite square well, but the function you wrote isn't quite correct.

 

[Fakestreet123]

Thanks for responding guys! I very much appreciate it!

You mean V=∞ when x<0 or
x>L.

Ahh yes, sorry for the crappy notation

How did you get this?

Uhhh Now that I've done some more studying I can see that its wrong but I was, at the time, hoping that the expectation values would be ∫<xp>+<px> = ∫<xp+px> and the expectation values of xp, px will just add together like magic.

Which is wrong! So I've redone the problem in an attempt to simply find the operator first

<xp> = -ihx(dψ/dx) and
<px> = -ih (d(xψ)/dx)
1/2<xp+px> = (-ih)/2 (x(dψ/dx)+(d(xψ)/dx)

Now the expectation value should be <1/2(xp+px)>= ∫ψ*(1/2)<xp+px>ψ dx?

do I just plug in 1/2<xp+px> (assuming its correct) and plug in ψ (the solution to the infinite square well) and take the integral from 0 to L?

You do use a solution for the infinite square well, but the function you wrote isn't quite correct.

Sorry! I was very wrong haha
√(2/L) Sin ((nπx)/L)

 

[DimReg]

One thing that might help you:

Using [x,p] = ihbar, or xp - px = ihbar, or px = -ihbar + xp, you can write

xp + px = xp + xp - ihbar = 2xp - ihbar, which has an easier to calculate expectation value. (you should check my work, I may have made a mistake)

In general an expectation value is:

<Q> = ∫ψ*Qψdx, where Q on the right is the operator for the observable you want to calculate. ψ is the wavefunction of the particle, which should be supplied by your textbook (it looks like you have the right one though).

Removed by moderator[/color]

Where I used the normalization condition ∫ψ*ψdx = 1.

I'll leave it to you do actually do the integral, and to check my work.