In quantum mechanics...

[Eeezee]

I had always been taught last semester that to find the expectation value, one can integrate psistar * operator * psi from negative infinity to infinity (where psi is the wave function)

Suddnly I'm being told that if you divide by the integral of psistar*psi, this is also acceptable and guarantees normalization. This certainly gets rid of any constants out in front, but doesn't it change the value of your answer sometimes?

 

[f95toli]

I am not really sure what you are asking but if you integrate from psistar psi from -inf to inf (in a 1D problem) you will hopefully end up with the answer is 1, simply because what you are calculating is the probability of finding the state *somewhere* in that range.
If you get a different answer it probably means that you have changed the basis of the problem without renormalizing the new state vectors.

There is one exception to this and that is if you use numerical methods to solve a problem, then it is sometimes usefull to renormalize the vectors from time to time simply to keep numerical errors under control.

If you solve the static Schroedinger equation by calculating the eigenvalues of a matrix you need to renormalize the answer simply because the solver calculates a set of eigenvalues (which ususally correspond to energy) and an arbitrary set of corresponding (orthogonal) eigenvectors (meaning as long as you mulitply them all by the same constant you haven't changed anything), in order for the solultion to be meaningfull the wavefunctions (eigenfunctions) should be orthonormal, i.e. the vectors that make up the basis should have the length 1 .

I hope this helps.

 

[silverpig]

Throw 20 red balls and 20 blue balls into a bag. Draw one at random. Without normalization, if you count the ways to draw a red ball, you will find the expectation value of drawing a red ball is 20. This doesn't really make much sense though until you normalize and divide by the total number of balls. You then see the expectation value is 1/2.

It's just a normalization thing.

 

[Eeezee]

Originally posted by: f95toli
I am not really sure what you are asking but if you integrate from psistar psi from -inf to inf (in a 1D problem) you will hopefully end up with the answer is 1, simply because what you are calculating is the probability of finding the state *somewhere* in that range.
If you get a different answer it probably means that you have changed the basis of the problem without renormalizing the new state vectors.

There is one exception to this and that is if you use numerical methods to solve a problem, then it is sometimes usefull to renormalize the vectors from time to time simply to keep numerical errors under control.

If you solve the static Schroedinger equation by calculating the eigenvalues of a matrix you need to renormalize the answer simply because the solver calculates a set of eigenvalues (which ususally correspond to energy) and an arbitrary set of corresponding (orthogonal) eigenvectors (meaning as long as you mulitply them all by the same constant you haven't changed anything), in order for the solultion to be meaningfull the wavefunctions (eigenfunctions) should be orthonormal, i.e. the vectors that make up the basis should have the length 1 .

I hope this helps.

Doh, why didn't I think of that! Yes, of course integrating from negative infinity to infinity will give you 1.

I'm still a little sketchy on this though, since it's effectively known as a way to normalize it. But rather than normalizing, this simply gets rid of any constants that are in front of the integral.

So if the wave function is psi = A*(psi_x)*(psi_t), then the integral from negative infinity to infinity is A^2. I'm being told that if I apply an operator, such as i*hbar*d/dt (the hamiltonian), then I can effectively do this:

(integral from negative infinity to infinity of psistar*i*hbar* d(psi)/dt) / (integral from negative infinity to infinity of psistar*psi)

Yet when I originally learned the integral method with operators, we never divided by that second integral, which should normally go to A^2. So essentially on the top we have the expectation value of energy, and on the bottom we have A^2. Are we really allowed to just divide by A^2 all of a sudden?

 

[silverpig]

No. When you were taught to not do the bottom integral, that was based on the assumption that your wavefunction was already normalized. Dividing by the integral there means that you are doing a normalization step.

Okay maybe this will make it clear. Let's take an un-normalized wavefunction: Psi_1

We can integrate from -inf to inf: Int[Psi_1* Psi_1] = A^2

But we want the probability of finding the particle in space to be 1, not A^2, as that makes more sense, so we divide by A^2: (1/A^2) Int[Psi_1* Psi_1] = 1

We can also find the momentum of the particle: Int[Psi_1* i hbar d/dt Psi_1] / Int[Psi_1* Psi_1] = p

In general, we can find the expectation value of any general operator B: Int[Psi_1* B Psi_1] / Int[Psi_1* Psi_1] = b Where b is an eigenvalue associated with the operator B.

But if we want, we can get rid of having to do that " / Int[Psi_1* Psi_1] " part every time by ensuring our wavefunction is normalized as then that quantity will just be 1. Okay so how do we do that? Well recall that Int[Psi_1* Psi_1] = A^2 right? well let's make a new wavefunction, Psi_2 to be Psi_1/A

Psi_2 = Psi_1 / A

So now we have Int[Psi_2* Psi_2] = Int[Psi_1*/A Psi_1/A] = 1/A^2 Int[Psi_1* Psi_1] = 1/A^2 A^2 = 1 so our new wavefunction Psi_2 is normalized.

Now we can compute expectation values without having to divide by that second integral again:

Int[Psi_2* B Psi_2] = b
Int[Psi_2* i hbar d/dt Psi_2] = p

Note that these give the exact same values for b and p that we got the first way, by integrating Psi_1* B Psi_1 and then dividing by the integral of Psi_1* Psi_1.

It's just easier to normalize your wavefunction first, and then to do all of your calculations using the normalized wavefunction.

 

[Soccerman06]

a^2 + b^2 = c^2
or
y = mx + b

I think these answer your question