Expectation value of a wave function

In summary: Try: 1 = \int_{- \infty}^{\infty} \left( \frac{1}{2} \right)^{*} \left( \frac{1}{2} \right) dx This should work because the left hand side is just the complex conjugate of the right hand side.
  • #1
bjogae
20
0

Homework Statement



The wave function of a state is Psi(x)= N*a(x)exp(i*p0*x/h)where a(x) is a quadratically integrable real valued function Show that the expectation value of the function is p0.

Homework Equations





The Attempt at a Solution



The only thing I'm having a problem with is how to integrate the square of the wavefunction so that I could normalize N and put the operator p=-ih(d/dx) in the integral. So I think I know what to do, just not sure how to do it.
 
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  • #2
First off the expectation value of what? Position?

Also what is p0, just a real-valued constant?
 
  • #3
It's the expectation value of the momentum I'm after. Sorry, forgot to put that in. And yes, p0 is a real-valued constant.
 
  • #4
All right so to normalize the wave function the wave function squared (probability) must be one. In this case you need to integrate [itex]\Psi^{*} \Psi[/itex] over all possible values (-inf to inf) and set that equal to 1, then solve for N.
 
  • #5
But that's the thing. I have no idea of how to do the integral.
 
  • #6
Well what have you tried? Start by setting up the integral and working as far as you can.
 
  • #7
well I get stuck when I'm supposed to integrate a(x)^2*exp(i*2*p0*x/h). How do I do that?
 
  • #8
[tex] 1 = \int_{- \infty}^{\infty} \Psi^{*} \Psi dx [/tex]

Psi* is the complex conjugate of Psi. Do you know how to take a complex conjugate?

exp(i*2*p0*x/h) makes it seem like you're trying to do:

[tex] 1 = \int_{- \infty}^{\infty} \Psi \Psi dx [/tex]

When you need to be doing:

[tex] 1 = \int_{- \infty}^{\infty} \Psi^{*} \Psi dx [/tex]
 
  • #9
For instance the complex conjugate of [tex]f(x) = A e^{i x}[/tex] is [tex]f(x)^{*} = A e^{-i x}[/tex] where A is a real constant.
 
  • #10
Thanks, I think I got it right.
 

What is the expectation value of a wave function?

The expectation value of a wave function is the average value of a physical quantity, such as position or energy, calculated over a large number of measurements on a system described by the wave function.

How is the expectation value of a wave function calculated?

The expectation value is calculated by taking the integral of the product of the wave function and the operator representing the physical quantity of interest. This integral gives the average value of the physical quantity over all possible states of the system.

Why is the expectation value of a wave function important?

The expectation value is important because it allows us to make predictions about the behavior of a system described by the wave function. It can be used to calculate the most likely outcome of a measurement and to determine the overall behavior of the system.

What is the relationship between the expectation value and the uncertainty principle?

The uncertainty principle states that it is impossible to know both the position and momentum of a particle with absolute certainty. The expectation value of the position and momentum operators can be used to calculate the uncertainties in these quantities, and the uncertainty principle sets a limit on how small these uncertainties can be.

How does the expectation value change with time?

The expectation value of a wave function can change with time as the system evolves. This change is determined by the time-dependent Schrödinger equation, which describes the time evolution of the wave function. The expectation value can also be used to calculate the average rate of change of a physical quantity over time.

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