# I What is meant by the terminology "single particle state"?

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1. Apr 19, 2016

### Frank Castle

I'm currently reading through a set of notes on statistical mechanics, and when it comes to deriving the Fermi-Dirac and Bose-Einstein distributions it uses the terminology single-particle state.

By this, is it meant that if the particles can be assumed independent, then each particle can be described by an individual wave function, with the entire system being described by a wave function that is a product of these single-particle wave functions?

If this is the case, is it then true that since each particle has an independent Hamiltonian associated with it (since the particles are independent, there are no interaction cross-terms in the potential), and hence the energy of the system is the sum of the energies of the energies associated with each single-particle wave-function?

If the particles are identical and indistinguishable is it correct to say that one can no longer distinguish which particle is in which state precisely and instead one has to quantify things in terms of the number of particles occupying each single-particle state, and the particular state is identified in by its corresponding energy. One then describes the microstate of the entire system in terms of the occupation numbers of each single-particle state with a given energy.

2. Apr 19, 2016

### Khashishi

There's not much to say about a single particle since there's no interaction and it doesn't matter if it's a fermion or boson. Indistinguishable only matters for more than one particle.

You don't need to use the occupation number to describe identical particles, if you know how many particles there are. You can express the state as a symmetrized wavefunction. E.g. for 2 non-interacting fermions:
$$\Psi(x_1,x_2) = \frac{1}{\sqrt{2}}\left[\psi_1(x_1) \psi_2(x_2) - \psi_1(x_2) \psi_2(x_1)\right]$$

3. Apr 19, 2016

### Frank Castle

By this I mean that the state vector of the system of the form $$\lvert n_{1},\ldots ,n_{k}\rangle =\lvert n_{1}\rangle\lvert n_{2}\rangle\cdots\lvert n_{k}\rangle$$ where $n_{j}$ is the occupation number of the $j^{\text{th}}$ single particle state, and we have neglected spin.

Then we have that $$i\hbar\partial_{t}\lvert n_{1},\ldots ,n_{k}\rangle =i\hbar\partial_{t}(\lvert n_{1}\rangle\lvert n_{2}\rangle\cdots\lvert n_{k}\rangle) \\=i\hbar(\partial_{t}\lvert n_{1}\rangle)\lvert n_{2}\rangle\cdots\lvert n_{k}\rangle +i\hbar\lvert n_{1}\rangle(\partial_{t}\lvert n_{2}\rangle)\cdots\lvert n_{k}\rangle +\cdots +i\hbar\lvert n_{1}\rangle\lvert n_{2}\rangle\cdots(\partial_{t}\lvert n_{k}\rangle) \\ =(\hat{H}\lvert n_{1}\rangle)\lvert n_{2}\rangle\cdots\lvert n_{k}\rangle +\lvert n_{1}\rangle(\hat{H}\lvert n_{2}\rangle)\cdots\lvert n_{k}\rangle) +\cdots +\lvert n_{1}\rangle\lvert n_{2}\rangle\cdots(\hat{H}\lvert n_{k}\rangle) \\= (n_{1}\epsilon_{1}+n_{2}\epsilon_{2}+\cdots n_{k}\epsilon_{k})\lvert n_{1}\rangle\lvert n_{2}\rangle\cdots\lvert n_{k}\rangle\\ =\hat{H}\lvert n_{1},\ldots ,n_{k}\rangle =E\lvert n_{1},\ldots ,n_{k}\rangle$$ and hence $E=\sum_{j}n_{j}\epsilon_{j}$.
I'm unsure how to show that the total number of particles in the state $\lvert n_{1},\ldots ,n_{k}\rangle$ is $N=\sum_{j} n_{j}$? (The number operator doesn't seem to have the same straightforward procedure as was available for the Hamiltonian case).

4. Apr 19, 2016

### Frank Castle

I was referring to describing a system of $N$ independent particles in terms of single-particle states. Is it simply meant by this that each particle can be treated on it's own in terms of a single particle state vector (equivalently, in terms of a single-particle wave function), such that the state vector (equivalently, wave function) of the entire system is described in terms of these single-particle wave vectors, i.e. as a product of single particle state vectors?!

5. Apr 19, 2016

### Khashishi

I guess I don't know what you are asking. Are you talking about a Fock state? https://en.wikipedia.org/wiki/Fock_state . You can write it in terms of single particle states using the Slater determinant?. https://en.wikipedia.org/wiki/Slater_determinant
The number operator for a state just gives the number in that state
$\hat{N}_k\left|n_1,n_2,...n_k,...\right> = n_k\left|n_1,n_2,...n_k,...\right>$
Obviously, the total number of particles is the sum of the number in each state. Not sure what you are asking ??
Maybe you are asking something I'm not qualified to answer.

6. Apr 20, 2016

### Frank Castle

Sorry, I probably haven't worded my questions very well.

My first question relates to the statement that, when one considers a multi-particle system in which there are no mutual interactions between the particles (i.e. they are treated as independent), one can describe the state of the multi-particle system in terms of single-particle states.
By this, is it simply meant that each particle can be treated independently as a single particle system in a given state (i.e. a single-particle state), such that each particle has its own wave function, and the state of the full multi-particle system is described in terms of a tensor product of these single-particle states? (i.e. Would one describe each particle in terms of a single particle state vector $\lvert\psi_{i}\rangle$, such that the state vector of the full multi-particle state is of the form $$\lvert\Psi\rangle =\frac{1}{\sqrt{N}}\left(\lvert\psi_{1}\rangle\lvert\psi_{2}\rangle\cdots\lvert\psi_{N}\rangle \pm\lvert\psi_{2}\rangle\lvert\psi_{1}\rangle\cdots\lvert\psi_{N}\rangle\pm\cdots\right)$$)

The follow-up question to this is, if what I have put about describing the full system in terms of single-particle states is correct, then how does one show that the total energy of the multi-particle state is the sum of the energies of the single-particle states, weighted by the occupation numbers of the respective single-particle states (as more than one particles wave function could correspond to the same state)?
Intuitively I see why it should be of the form $E=\sum_{i}n_{i}\epsilon_{i}$, but I'm unsure as to whether I've understood correctly how one derives this result.

Thirdly, how does one show that the total number of particles of the multi-particle state is the sum of the occupation numbers of the single-particle states comprising the system?
Again, intuitively I can understand why it should be of the form $N=\sum_{i}n_{i}$, but (again) I'm unsure if I've understood how one correctly derives this result.

7. Apr 20, 2016

### A. Neumaier

Your $|\Psi\rangle$ is called a Slater state. These approximate for example ground states of light atoms with about 5% error - although the electrons are highly interacting. This is called the Hartree-Fock approximation. Any book treating it in some detail also derives the corresponding energy formula. Improving the approximation needs superpositions of many Slater states.

8. Apr 20, 2016

### Frank Castle

Have I understood correctly what is meant by the term "single-particle state" though?

9. Apr 20, 2016

### A. Neumaier

Probably not; you described special $N$-particle states. A single-particle state has $N=1$ only and describes only a single particle - hence the name.

10. Apr 20, 2016

### Frank Castle

Yes, I understand that a single particle state is represented by a wave function that describes the state that a single particle can reside in. What I was meaning is that if we consider an $N$-particle system of non-interacting particles then in principle each particle can be described by its own wave function (independently of any other particle), i.e. each particle is in a particular one-particle state. Now, it is possible that more than one particle can be in the same state, and hence their wave functions are identical, but the main idea is that the $N$-particle state can be described in terms of the single-particle states (i.e. in terms of the wave functions associated with each individual particle). This seems to be the point that several notes I've read on statistical mechanics seem to be trying to put across anyway; they all have phrases of the form: " if the particles in the N-particle system can be treated as independent, then one can describe the system in terms of single particle states, in other words, the system can be treated as a collection of N single particle subsystems each with its own associated wave function describing its state".

11. Apr 20, 2016

### A. Neumaier

In a Slater state, the N-particle state is described by N different 1-particle wave functions, one for each of the N particles. It still remains a N-particle state, just one of the special Slater form. If you assume this, you effectively assume the Hartree-Fock approximation, and get (in general, if you also allow for quasiparticles to enhance independence) what is called a mean field theory.

12. Apr 20, 2016

### Frank Castle

So is this what they are referring to then, that one can approximate the N-particle state in terms of N single-particle states, and in the ideal case, where all of the N particles are truly independent, then such an approximation would be exact?

13. Apr 20, 2016

### A. Neumaier

Yes, with the qualification that the final statement only holds if the initial state is already of this form.

14. Apr 20, 2016

### Frank Castle

Ok, I think I understand it now.
With this in mind, does one then show that the total energy and total number of particles of the N-particle system are $E=\sum_{i}n_{i}\epsilon_{i}$ and $N=\sum_{i}n_{i}$, respectively (where $n_{i}$ is the occupation number of the $i$-th single-particle state, by noting that the Hamiltonian can be written as $$\hat{H}=\sum_{i}\hat{H}_{i}$$ where $\hat{H}_{i}$ is the Hamiltonian of the $i$-th single-particle sub-system (assuming the particles are independent). As such $$\hat{H}\lvert\Psi\rangle =\sum_{i}\hat{H}_{i}\lvert\Psi\rangle = \sum_{i}n_{i}\epsilon_{i}\lvert\Psi\rangle$$ where $\epsilon_{i}$ is the energy eigenvalue of the $i$-th single-particle state (i.e. $\hat{H}_{i}\lvert\psi_{i}\rangle =\epsilon_{i}\lvert\psi_{i}\rangle$).
Similarly, the number operator can be written as $$\hat{N}=\sum_{i}\hat{N}_{i}$$ where $\hat{N}_{i}$ is the number operator of the $i$-th single-particle sub-system. As such, $$\hat{N}\lvert\Psi\rangle =\sum_{i}\hat{N}_{i}\lvert\Psi\rangle = \sum_{i}n_{i}\lvert\Psi\rangle$$ where $n_{i}$ is the number eigenvalue of the $i$-th single-particle state (i.e. $\hat{N}_{i}\lvert\psi_{i}\rangle =n_{i}\lvert\psi_{i}\rangle$).

15. Apr 20, 2016

### A. Neumaier

Please read something about the Hartree-Fock method (e.g., Chapter 18 in the second volume of Messiah's quantum mechanics).

16. Apr 20, 2016

### Frank Castle

I don't have access to Messiah's book unfortunately, however, I have had a look at another set of notes in which the author states that "if the N particles in a system are not interacting, then the Hamiltonian of the system can be written as a sum of the single-particle Hamiltonians corresponding to the N single particle sub-systems". This seems to agree with what I wrote in my previous post in the derivation of the total energy for a system of N independent particles. I appreciate that I was lazy and should have written the state vector as something like $$\lvert\Psi_{i_{1},\ldots ,i_{N}}(1,\ldots ,N)\rangle =\frac{1}{N!}\lvert\left(\begin{matrix}\lvert\psi_{i_{1}}(1)\rangle &\cdots &\lvert\psi_{i_{1}}(N)\rangle\\ \vdots &\ddots &\vdots\\ \lvert\psi_{i_{N}}(1)\rangle &\cdots &\lvert\psi_{i_{N}}(N)\rangle\end{matrix}\right)$$ (where the subscripts denote the possible single-particle states the N particles in the system), but otherwise, I can't see what I've done wrong?!

17. Apr 21, 2016

### A. Neumaier

Nothing wrong, just irrelevant - particles of interest are never independent. The Slater determinant approximaltion is always used in the interacting case. There you still have the energy relation but its meaning is quite different and much more interesting, since it is relevant for the interpretation of the periodic system of elements. Any book discussing the latter also discusses the Hartree-Fock approximation.

18. Apr 21, 2016

### Frank Castle

Ah ok. Would there be any other sets of notes or textbooks that you would recommend for studying this?

19. Apr 21, 2016

### A. Neumaier

20. Apr 21, 2016

### Frank Castle

Ok thanks, I'll take a look.