Dimension of Hilbert spaces for identical particles

[boudreaux]

Homework Statement

What is the Hilbert-space dimension of this two-particle system if the two particles are (a) distinguishable? (b) identical bosons? (c) identical fermions (assuming
their spins are polarized, so the spin-part is completely symmetric)?

Relevant Equations

Two particles occupying N distinct states |1>, |2>, · · · , |N>

My thoughts are:
a) it should just be N^2
b) just N since they're identical
c) due to Pauli exclusion would it be N^2 - N since they have to be different states?

 

[PeroK]

Homework Statement:: What is the Hilbert-space dimension of this two-particle system if the two particles are (a) distinguishable? (b) identical bosons? (c) identical fermions (assuming
their spins are polarized, so the spin-part is completely symmetric)?
Relevant Equations:: Two particles occupying N distinct states |1>, |2>, · · · , |N>

My thoughts are:
a) it should just be N^2
b) just N since they're identical
c) due to Pauli exclusion would it be N^2 - N since they have to be different states?

What do you understand about the total symmetrization requirement in each case?

Can you analyse the cases $N = 2$ and $N = 3$?

 

[boudreaux]

What do you understand about the total symmetrization requirement in each case?

I don't really understand your question - probably not much!

 

[PeroK]

I don't really understand your question - probably not much!

This question is testing your knowldege of symmetrization requirements. First, you should look up the formal definition of the Pauli exclusion principle. I.e. the generalisation, as it applies to all fermions in all cases.

 

[PeroK]

Ps You should look up the symmetrisation requirement for bosons as well!

 

[boudreaux]

Thanks! Still somewhat confused after reading the mathematical definitions. How do I get the answer to my problem from these?

 

[boudreaux]

For fermions they must be antisymmetric so would there just be N states since the second particle's state will have to be antisymmetric to the first?

 

[PeroK]

Thanks! Still somewhat confused after reading the mathematical definitions. How do I get the answer to my problem from these?

I can't do it for you. Try for $N = 2$ first. The answer, once you work it out, ought to be familiar.

Note that for bosons, the entire state/wave-function must be symmetric. You are given a symmetric spin state, which means that the spatial state must also be symmetric.

How many symmetric states are there for each $N$?

For fermions, the entire state must be antisymmetric. In this case, the spatial state must be antisymmetric. How many antisymmetric states are there for each $N$?

Note: if the spin state were antisymmetric, then these answers would be the other way round for bosons and fermions.

 

[boudreaux]

So for fermions they must be antisymmetric so would there just be N states since the second particle's state will have to be antisymmetric to the first?

For bosons would it be 2N?

 

[PeroK]

So for fermions they must be antisymmetric so would there just be N states since the second particle's state will have to be antisymmetric to the first?

For bosons would it be 2N?

No. Do the problem for $N = 2$.

 

[boudreaux]

for N=2 would there be 4 possible states for distinguishable/2 for bosons/1 for fermions?

 

[PeroK]

for N=2 would there be 4 possible states for distinguishable/2 for bosons/1 for fermions?

No. The simplest approach is just to label the states $a$ and $b$. We have four basis states:
$$aa, \ ab, \ ba, \ bb$$
That is the basis for your Hilbert space of two particles with the same two possible states each.

The question is how many linearly independent symmetric states can you construct from those. Hint: two are easy.

And, how many asymmetric states can you construct?

Note: you have to actually do some work here, not just guess an answer!

 

[PeroK]

If you're really stuck, the following should look familiar. There are $3$ symmetric states:
$$aa, \ bb, \ (ab + ba)$$
The normalised version of the third one is, of course, $\frac 1 {\sqrt 2}(ab + ba)$.
And only $1$ antisymmetric state:
$$ab - ba$$
Hence we have a a triplet of symmetric states and a singlet antisymmetric state.

The next step is to look at $N = 3$.