How Do You Calculate the Expectation Value for a Particle in an Infinite Box?

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SUMMARY

The discussion focuses on calculating the expectation value for a particle in an infinite box in the first excited state (n=2). The general formula for the expectation value is provided as = ∫ψ* Q(x,p) ψ dx, where x and p are the position and momentum operators. The normalized wave function is given by ψ_n(x) = √(2/L) * sin((nπx)/L) for n=2. Participants suggest calculating and separately and applying the respective operators to evaluate the integrals.

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shlomo127
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Homework Statement



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

Homework Equations





The Attempt at a Solution


I'm completely baffled by this problem.
Can anyone just please point me in the right direction and then ill respond with what i get (because i know I am going to get stuck again)

Thanks!
 
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Hey shlomo. As a hint, the general formula for an expectation value is is given by:

&lt;Q(x,p)&gt; = \int_{-\infty}^{+\infty} \psi^* Q(x,p) \psi dx

where x, p are the position and momentum (quantum mechanical) operators.

This should hopefully give you a push in the right direction. As a first suggestion, you will have to find the wave equation \psi for your specific situation.
 
ok, I am going to work on that,
are there any websites you can recommend to help me learn this stuff?
my Prof doesn't explain anything well...
thanks!
 
Well I suppose it's dependent on the situation you're in. What's the outline of the course?

Personally, I find it to be overwhelming if given a resource that goes into far more detail than the course requires.
 
i believe the wave equation is:
\psi(x) = Asin(kx) + Bcos(kx)
and k=sqrt((2mE)/(h-bar)^2)

so then the normalized wave function would be:
\psi n(x)=sqrt(2/L)*sin((n*pi*x)/L)

and the problem said that n=2.

And that's where I get stuck. Am i doing it right thus far? and how do i go further?
 
As far as I know so far you did right. My suggestion for the further is that you can calculate <xp> and <px> separately ans sum them up. To do so, apply the operators with their usual orders and take their integral. I know, it seems a bit tidious but this is the only way that I think.
 
ok, sounds good, can u set up on of the integrals so i can see how to do it?
then ill try to evaluate both integrals and ill post what i get.

Thanks!
 
For the p*x take the derivative of the x*wave function's conjugate and multiply it with the wave function itself and take the integral.
 
Have you figured it out yet? I don't understand 'soul's last comment/step. I understand up to normalizing the wave function. I don't know how to derive px and xp.

Help!
 
  • #10
BUMP,
We need it for tomorrow.
 

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