# I Probabilities in many body systems...

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1. Dec 14, 2016

### voila

Hi there. Excuse that I write it all in standard text but I don't know how to write it otherwise. I'll get right to the question and try to see if I understand the theory from the answer.

Suppose we have a system of two electrons, one spin up and one spin down in a system with 3 possible states, say 3 atomic levels. Suppose they are in a state

[(up, down, 0)-(0,down,up)]/sqrt(2)

(imagine it in Dirac notation ).

How can I calculate the probability of finding spin down electron in the middle state, or the up electron in any of the other ones?

2. Dec 15, 2016

### blue_leaf77

I don't understand how the state notation is ordered. The first term indicates that the third notation belongs to the atomic level but the second term the atomic level is located in the first notation.

3. Dec 15, 2016

### Staff: Mentor

I assume you mean 3 possible energy eigenstates/energy levels, correct? Spin is part of the state, so a system with 3 energy levels will have 6 possible states (more precisely, 6 basis states) for a single electron, since it could be spin up or spin down in each of the 3 levels.

You should take some time to learn to use the PF LaTeX feature. It makes posts with math in them a lot easier to read. Help on that feature is here:

https://www.physicsforums.com/help/latexhelp/

As I understand it, the state you are trying to write down is

$$\frac{1}{\sqrt{2}} \left( \vert 1 \uparrow \rangle \vert 2 \downarrow \rangle - \vert 3 \uparrow \rangle \vert 2 \downarrow \rangle \right)$$

where each ket gives the energy level and spin of the corresponding electron. However, this state is not a valid state for two electrons, because it is not antisymmetric under electron exchange. That is, if we switch the order of the kets in all the terms, we should get a state that is the same except for a change of sign. For example, the following state is antisymmetric under electron exchange:

$$\frac{1}{\sqrt{2}} \left( \vert 1 \uparrow \rangle \vert 2 \downarrow \rangle - \vert 2 \downarrow \rangle \vert 1 \uparrow \rangle \right)$$

It might help if you would give more details about where you got this problem from and why you are trying to solve it.

4. Dec 15, 2016

### blue_leaf77

What about the "0" in the ket notation?

5. Dec 16, 2016

### Staff: Mentor

The "0" is a state that is unoccupied, so there is no ket corresponding to it. You have two electrons, so each term in the state will have two kets, one for each electron.

Btw, what background in QM do you have? You labeled this thread as "I", which indicates an undergraduate level of knowledge.

6. Dec 18, 2016

### voila

Sorry again for the notation. I'll take some time to learn the PF Latex notation.

PeterDonis, you were right: 3 energy states, plus spin degeneration, so 6 total states, and you got them right. And I proposed a combination of 2 states of 2 electrons each. However, I should have said that they are both already antisymmetrised. So it wouldn't be a tensor product of two kets each, as you wrote, but the antisymmetrised product.

So back to the question. If I want to calculate the probability of finding an electron in a state (say, spin up in the first energy level), should I project the state in each possible state that contains an spin up electron in the first energy level, or is there a more proper way?

PS: I actually just graduated but I take the doubt is "I" level.

7. Dec 18, 2016

### Staff: Mentor

So the actual state would be this?

$$\frac{1}{2} \left( \vert 1 \uparrow \rangle \vert 2 \downarrow \rangle - \vert 2 \downarrow \rangle \vert 1 \uparrow \rangle - \vert 3 \uparrow \rangle \vert 2 \downarrow \rangle + \vert 2 \downarrow \rangle \vert 3 \uparrow \rangle \right)$$

I'm not sure what you mean here, but perhaps writing out the state explicitly as I did above will help. What operation would you perform on this state to get the probability you are looking for?

8. Dec 18, 2016

### blue_leaf77

There are $\frac{6!}{2!3!} = 60$ independent states in your problem, one of which is given by Peter in #3 post (the second expression). If your proposal belongs to this set of states or is a linear combination of them, your state is correct.
To be more precise, you should project the state in each possible antisymmetrized state that contains an spin up electron in the first energy level and then sum them up.

EDIT: sorry the number of independent states should be $\frac{6!}{2!4!}=15$.

Last edited: Dec 19, 2016
9. Dec 19, 2016

### Staff: Mentor

As I understand it, the OP is assuming that the system has already been prepared in one of these states (the one I wrote down in post #7).

Not if the system was already prepared in a particular known state. Then that state is the only one that needs to be projected.

10. Dec 19, 2016

### voila

Yes.

That's what I understood, but I was wondering if there was a simpler way, or at least a more polished formalism to state it. For many body systems that could easily get out of hand, right?

I don't understand what you mean by this, PeterDonis.

11. Dec 19, 2016

### Staff: Mentor

I was saying that, in the scenario as you have described it, we know the particular quantum state that the system is in; it's the one you said "yes" to in your post. In that case, the only state you have to project is that one.

What @blue_leaf77 was describing, when he said "you should project the state in each possible antisymmetrized state", was a scenario in which you don't know what particular quantum state the system is in. In that case, you have to project all possible states that the system could be in and then add up all the probabilities you get (normalizing appropriately) to get the total probability for a particular result.

12. Dec 19, 2016

### blue_leaf77

Well yeah that's what we mean, it's the state at hand (the state of the system) that must be projected to each basis state responsible for the event the OP is trying to measure, namely finding one electron in the first level with spin up.
If we don't know the exact mathematical form of the state (be it pure or mixed state), then projection is not relevant term anymore since it refers to a mathematical operation. In such case, the best we can do is to experimentally measure the system directly.
From the way you formulate the sentence, it seems like you are talking about a mixed state. If yes, I have a hunch that it's the use of the word possible in
that gives you the impression that the OP was referring to a mixed state. May be "possible state" shouldn't have been used there because there are actually infinite number of (antisymmetrized) states that contains one electron in the 1st level with spin up. "Determinantal state" seems a more fitting term there.
As far as I know that's the standard method.

13. Dec 19, 2016

### Staff: Mentor

Yes, I agree. I was simply noting that, in this particular scenario, exactly one of those states was already specified as the state of the system.