How Do You Determine the Outcome and Probability of a Quantum Measurement?

[Lorna]

Homework Statement

I have a wave function, say \Psi(x,t)=Acos(kx) \phi(t).
A quantity is measured (eg momentum, energy ...) and we are asked to find the result of the measurement and its probability.

Homework Equations

The Attempt at a Solution

I'd use the eigenval-eigenfunction equation. For example if we're talking about measureing the momentum:

\widehat{P}|\Psi> = p |\Psi> using definition of \widehat{P} we get:

-i\hbar \partial / \partialx (Acos(kx)) = -i\hbar * k Asin(kx) = p Acos(kx)
now I solve for p and get p = -i\hbar * k tan kx

does that make any sense?

now to find the probability that I get p when I measure the momentum is p^2?

Thanks.

 

[malawi_glenn]

for a continuous wave function, like \Psi (x) in position space.

The probability that particle is bewteen x' and x' + dx :

|\Psi (x')|^2 (a)

That should be in your textbook.

And for momentum, you can do the (inverse)fourier transform to get the wave funtion in momentum space:

\Phi (p) = \dfrac{1}{\sqrt{2\pi \hbar}}\int _0^{\infty}e^{-ipx}\Psi (x) dx

That should ALSO be in your textbook!

Then you use the same definition (a), but momenta as argument.

|\Phi (p')|^2

The probability that particle has momenta bewteen p' and p' + dp

N.B. The prime over x and p, is just a notation, to specify a certain momenta. Very common to have prime as notation for symbols, not just as short hand notation for derivative orders.

 

[Lorna]

so the probability of getting p = 4*hbar is the value of the integral with p replaced by 4*hbar multiplied by its complex conjugate?

 

[malawi_glenn]

from the integral you don't get a value, you get a function. It is very important that you don't see it as a "integral", look at is as the inverse Fourier transform of x, with a normalisation constant infront.

In that function you put p = 4*hbar, then you take the modulus square.

Are these things covered in your course material or not?

 

[Lorna]

We did conversions to momentum space, but I still don't understand how am I going to get the probability if I do not evaluate the integral. Sorry for being slow.

 

[malawi_glenn]

you must perform the integration. But I just wanted to make the point that you don't get a value by doing this integration, you are getting a FUNCTION (the wave function in momentum space). In that function you plug in the value of momenta you are seeking, then you get the value of the wave function in momentum space.