1. Aug 10, 2009

### Johnny010

Hello there. I am a 2nd year uni student studying chemistry. I have a paper in a few weeks on the foundations of physical chemistry.
I am having a problem with some quetions and would appreciate any help. Thank you to any who help.

1.Can the expectation value of the rotational angular momentum of a diatomic molecule
be used to predict what value will be obtained experimentally if the rotational angular
momentum of a single molecule is measure only once? Justify.

The measurement of the observable <L> must be an eigenvalue to L(operator).

L(operator)fn(x)=Lnfn(x) n=1,2,3...

If the system is in an eigenstate of L(operator), the result gained must be of the observable L, that can only be of the particular eigenvalue characteristic of that eigenstate:

say Ψ(x)=f7(x)
then L=L7

Whereas if the system is not an eigenstate of L, the result of one measurement of a single diatomic can be in any one eigenstate of L(operator) but will be completely unpredictable:

Ψ(x)≠fn(x)

The eigenvalue is therefore completely unpredictable.

The expectation value <L> would be given by:
Integration over all space of the wavefunction(in all co-ordinates)*L(operator) d(all co-ordinates).

Therefore <L> and the single measurement are unlikely to be the same.

2. Aug 10, 2009

### jpreed

Your answer is essentially correct, though you may not have a full handle on the topic.

The expectation value is the average value you would expect to obtain if you made the measurement billions of times. A simple example system is one in which you have a linear combination of a spin up and spin down particle wave function

$$\psi = \frac{1}{\sqrt{2}}\big(\,\mid\uparrow\rangle + \mid\downarrow\rangle\,\big)$$

If you make just ONE measurement of the spin you will get either +1/2 or -1/2. But if you make a LOT of measurements and average them, you will get closer and closer to zero, which is also the expectation value.

3. Aug 10, 2009

### Johnny010

Thank you. Can anyone else help a bit more? Using the above question?

4. Aug 11, 2009

### kuruman

Look at it this way: The expectation value of rolling a standard die is 3.5. Can you use this to predict what you will get before you roll it?