Let us consider a classical Hamiltonian of a many body system \begin{equation*} H = \sum_{j=1}^N\frac{p_j^2}{2m}+V(\mathbf q) \end{equation*} and let us pass to quantum dynamics by promoting the positions and momenta to operators: \begin{equation*} q_j\rightarrow \hat q_j \;\;\; p_j\rightarrow \hat p_j \end{equation*} where the index label still refers to THE particle $j$. Additionally, we could define a bosonic Fock space via the creation and destruction operators (by taking $\hbar\equiv 1$) \begin{equation*} a^\dagger_j=\frac{\hat q_j+\hat p_j}{\sqrt 2} \;\;\; a_j = \frac {\hat q_j+i\hat p_j}{\sqrt 2}\hspace{8mm} j=1\cdots N \end{equation*} creating/destroyting A particle in A STATE $j$. To me the fact that particles are seen as indistinguishable once a Fock space is build appears natural. In particular the label $j$ does not refer anymore to a particle but rather to a state. Instead, if we stick to the quantum dynamical system $\hat H(\mathbf{\hat q}, \mathbf{\hat p})$, is it true that particles are still distinguishable? In particular, I think that in the calculation of the canonical partition function \begin{equation*} Z(\beta) = \int_{\mathbb R^N} \mathrm d \mathbf q \langle \mathbf q|e^{-\beta \hat H(\mathbf {\hat q}, \mathbf {\hat p})}|\mathbf q\rangle \end{equation*} the states in the braket doesn't need to be symmetrized (as we would do for a trace in the Fock space) and that we can stick to the tensor product of the single particle Hilbert spaces: \begin{equation*} |\mathbf q \rangle = |q_1\rangle \otimes\cdots \otimes |q_N\rangle \end{equation*} Is my line of arguments correct?

  • $\begingroup$ Careful with your definition of particle and state. For example if $V(q)=\sum_\limits{j=1}^{N}V_j(q_j)$. Then an excitation by $a^\dagger_j$ can still be distinguished from one by $a^\dagger_{i\neq j}$ $\endgroup$ – E. Bellec Jan 9 '19 at 13:04
  • $\begingroup$ I agree, so I guess in case of a one body potential as the one you mentioned we would have to rewrite the interaction in Fock space as $\hat V = \sum_{rs} c^\dagger_r<r|v|s>c_s$ (by assuming $V_j=v \;\;\forall j$) which is indistinguishible. However I think in the coordinates representation $\mathbf q$ no such procedure is needed. Is it true? $\endgroup$ – Graz Jan 9 '19 at 13:09
  • $\begingroup$ I think I see what you mean. The excitation $a^\dagger_j$ for a given j are indistinguishable. As for the $q_j$ between themselves, that's up to you. You have to choose either $[q_j,p_k] = i\hbar \text{ or } 0$ for $k\neq j$. $\endgroup$ – E. Bellec Jan 9 '19 at 13:44

This seems to be a rather confused question. Given a general interaction potential $V(\mathbf{q}_i)$, your operators $a_j^\dagger$ and $a_j$ simply aren't creation or annihilation operators, in the naive sense. Their commutators with the Hamiltonian are much more complicated.

Assuming that $V(\mathbf{q}_i)$ has the very special form $$V(\mathbf{q}_i) = \frac12 \sum_i k_i q_i^2$$ then you indeed do get a set of $N$ independent harmonic oscillators. In this case, your conclusions are still confused because you're describing the same state in two incompatible ways.

For simplicity, let's restrict to $N = 3$ and consider the state $$|2, 1, 0 \rangle = a_1^\dagger a_1^\dagger a_2^\dagger |0 \rangle.$$ We can describe this state in two ways: either "there are three harmonic oscillators; the first harmonic oscillator is in the second excited state, and..." or "there are three units of excitation; two are in the first harmonic oscillator, and...".

The source of your confusion is that you're calling the harmonic oscillators themselves the particles in the first case, but then switching to calling the units of excitation the particles in the second case. Of course, "particle" is just a word which you can freely attach to either concept, with one or the other being more useful depending on the situation, but it shouldn't surprise you that the properties of the particles change if you change what the word "particle" means halfway through.

To hopefully clear some of the confusion, let's do an explicit example. Let $N = 3$ and consider the state $a_1^\dagger a_3^\dagger |0 \rangle$.

  • In the original picture, the Hilbert space is $\mathcal{H}_1^{\otimes 3}$ where the tensor product is not symmetrized. We describe this state in words as "there are three particles; the first particle is in the first excited state, the second is in the ground state, and the third is in the first excited state".
  • Next, we may switch to a completely different description of the states. We rename "particles" to "modes" and rename "particle in $n^{\text{th}}$ excited state" to "$n$ particles in the mode". Hence to describe the state we say "there are two particles; the first particle is in the first mode, and the second is in the third mode". We can write this state as $|1, 3 \rangle$ or as $|3, 1 \rangle$.
  • Since the "particles" are identical, the states $|1, 3 \rangle$ and $|3, 1 \rangle$ are both the same state, namely $a_1^\dagger a_3^\dagger |0 \rangle$. So we must work with a symmetrized Fock space.

To say the distinction yet another way: suppose I own $3$ sheep among a large herd, and to pick them out I give them identical bells. In the first langauge, the particles are the sheep, and I specify the state by saying where they are. In the second language, the particles are the bells, and I specify the state by saying what sheep each bell is on. Since the bells are identical, I then have to quotient out by permutations of the bells. But that doesn't mean the sheep have become identical.

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  • $\begingroup$ I think to the labels assigned to $a_j$ and $a_j^\dagger$ as states because a Fock state naturally constructed via them $|n_1...n_N\rangle\propto(a_1)^{n_1}\cdots(a_N)^{n_N}$ reads as the state where $n_j$ bosons are in the state $n_j$. Instead, starting from the classical Hamiltonian H, $\hat q_j$'s and $\hat p_j$'s are the direct quantum correspondent of physical particles. $\endgroup$ – Graz Jan 9 '19 at 14:45
  • $\begingroup$ I can rephrase my question asking whether I need to fix bosonic indistinguishability already at the level of the operators $\hat q_j$'s and $\hat p_j$'s or I can 'postpone' the creation of the symmetrized Fock space once I define $a_j^\dagger$ and $a_j$. $\endgroup$ – Graz Jan 9 '19 at 14:45
  • $\begingroup$ @Graz Again, you're using two totally different definitions of the word "particle", which is leading to all the confusion. For example, consider the state $a_1^\dagger |0 \rangle$ for $N = 3$ harmonic oscillators. In your original language, there are $3$ particles. In your new language, there is $1$ particle. It's the same state, you've just completely changed the words you use to describe it. $\endgroup$ – knzhou Jan 9 '19 at 15:46
  • $\begingroup$ I see what you mean. Though is it right that the Hilbert spaces for the two set of coordinates are interpreted differently? In particular that for the canonical coordinates $\mathcal H = \mathcal H_1^{\otimes N}$ (where $\mathcal H_1$ is a one-particle Hilbert space), while for the creation destruction operators we have to work in a Fock space? This would imply that to take traces in the case of $(\mathbf q,\mathbf p)$ we can simply consider a factorized basis, while for creation/destruction operators a complete set of orthonormal vectors is symmetrized $\endgroup$ – Graz Jan 9 '19 at 16:11
  • $\begingroup$ @Graz I edited with some more explicit explanation, does that help? $\endgroup$ – knzhou Jan 9 '19 at 16:37

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