expected value of Q(x,p)
To find the expected value of Q(x,p) we evaluate [itex]<\psiQ(x,i\hbar \frac{\partial }{\partial x})\psi>[/itex]. But what do you do if you want to find, say, <p^(3/2)>. How do you raise the derivative operator to the threehalves?

Re: <p^(3/2)>
Quote:

Re: <p^(3/2)>
You would make like you do for angular momentum. Since you can't take square roots of operator, you have to study the square of the quantity you are interested in. So in your case, you would have to settle for [tex]\sqrt{ \langle p^3\rangle}[/tex], but you would have to put absolute values somewhere.
...edit: go with count iblis' suggestion instead 
Re: <p^(3/2)>
Fourier transform the wave function to the momentum representation, getting
[tex]\phi(p)[/tex]. Then p is just like a c number. 
Re: <p^(3/2)>
Okay, thanks for the replies.

Re: <p^(3/2)>
Next problem:
How would you handle: <x^(1/2)p^(3/2)> 
Re: <p^(3/2)>
Quote:
[tex] \frac{d^nf[x]}{dx^n}=\frac{1}{2\pi}\int_{\infty}^{\infty}dk(ik)^ne^{ikx} \int_{\infty}^{\infty}dxf[x]e^{ikx} [/tex] also when [tex]n=\sqrt{pi}/i^5.8[/tex], for example. So the <x^(1/2)p^(3/2)> involves a tripple integral. 
Re: <p^(3/2)>
Quote:

Re: <p^(3/2)>
Ok, that was too easy for you two. :smile: Let me think of a more difficult problem. Well, why not just consider <f(x,p)> where f is some arbitrary function, like e.g.:
[tex]f(p,x)=\sqrt{x^2 + p^2}[/tex] Now, you can just modify the Fourier transform method and write this too as a triple integral. However, that may not be the simplest way, particularly not in the case when f is given as above. :smile: 
Re: <p^(3/2)>
For that you may need a Taylor expansion.

Re: <p^(3/2)>
Diagonalizing the operator x^2 + p^2 is easier. The eigenstates are the harmonic oscillator eigenstates, let's denote them by n>. You can thus compute the average as:
<psisqrt(x^2 + P^2)psi> = sum over n of <psisqrt(x^2 + p^2)n><npsi> sqrt(x^2 + p^2)n> = sqrt[(n+1/2)C] were C follows from the usual H.O. algebra (i'm too lazy to compute it right now) So, the average is: sum over n of sqrt[(n+1/2)C] <npsi>^2 
Re: <p^(3/2)>
Expanding in SHO states may or may not be easier than Taylor expansion.
This would depend on the original wave function and operator. SHO works for the particular operator x^2+p^2, but TE works for most operators. 
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