Two-particle fermionic basis states

[Niles]

Hi guys

Say we are looking at a two-particle system consisting of two electrons (fermions). In my book it says that the basis states are given by

$$\left| {\psi _{\alpha ,i} (r_m )} \right\rangle \left| {\psi _{\beta ,j} (r_n )} \right\rangle$$

where r_m and r_n denote the two particles. My question is, how do we know that the basis states are merely the product of the single-particle states, and not some obscure linear combination?

 

[naima]

Try f1(rm)f2(rn)-f1(rn)f2(rm)

 

[Niles]

Try f1(rm)f2(rn)-f1(rn)f2(rm)

That was not really my question. In my book they specifically state that single particle states are given by
$<br /> \left| {\psi _{\alpha ,i} (r_m )} \right\rangle \left| {\psi _{\beta ,j} (r_n )} \right\rangle <br />$. My question is why.

 

[strangerep]

In my book they specifically state that single particle states are given by
$<br /> \left| {\psi _{\alpha ,i} (r_m )} \right\rangle \left| {\psi _{\beta ,j} (r_n )} \right\rangle <br />$. My question is why.

A linear combination of two states S1, S2 means it's either one or the other (with probabilities
given by the weightings of the two terms in the superposition). I.e., it's a state of 1
particle which could either be in state S1 or S2. But here we want a state representing a
composite system that's definitely got two particles...

The product expression corresponds to a state in a tensor product of two single-particle
Hilbert spaces. Check out Wiki for the mathematical detail on tensor product spaces.
We use them in QM because the tensor product of two Hilbert spaces is itself a Hilbert
space. Ballentine's text covers this topic fairly well, iirc.

 

[arkajad]

The intuitive idea is simple: every matrix $$a(i,j)$$ can be represented as a linear combination of simple matrices $$a(i)a(j)^T$$ where $$a(i)$$ is a vector that has i-th component 1 and all other zero. Every matrix, thought as a square table is a linear combination of matrices with one entry =1 and all other entries =0.

That's the idea. Then come tensor products and mathematical theorems stating that every function f(x,y) of two variables can be approximated by sums of products of functions of one variable.

 

[naima]

My question is, how do we know that the basis states are merely the product of the single-particle states, and not some obscure linear combination?

This comes from the choice of a z axis upon which the spin of an electron is measured. We have a up state and a down state relative to this axis. these states form a basis in a 2 dimensional space.
For 2 electrons we use the tensor product of two such basis:
(I note it ¤)
up ¤ up, up ¤ down, down ¤ up, down ¤ down
We can choose another basis:
(up¤down - down¤up)/sqrt(2) , up¤up, (up¤down + down¤up)/sqrt(2) , down¤down
The 4 dimensional space is the sum of a singlet (maximum spin = 0) and a triplet (maximum spin = 1)

 

[arkajad]

Adding to the above: of course you could take "obscure linear combinations" as your basis. But what for if a mathematical theorem states that "(tensor) products of the basis vectors form a basis for the (tensor) product space"? (In fact for indistinguishable fermions you should antisymmetrize the products as mentioned by naima.)

 

[Niles]

Adding to the above: of course you could take "obscure linear combinations" as your basis. But what for if a mathematical theorem states that "(tensor) products of the basis vectors form a basis for the (tensor) product space"? (In fact for indistinguishable fermions you should antisymmetrize the products as mentioned by naima.)

Does such a theorem exist? If yes, can I ask what it is called (I'll google it afterwards)?