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In a lot of quantum mechanics lecture notes I've read the author introduces the notion of a so-called single-particle state when discussing non-interacting (or weakly interacting) particles, but none that I have read so far give an explicit explanation as to what is exactly meant by this term.

Is it meant that, in principle, each individual state constituting a multi-particle system can be occupied by a single particle (contrary to an entangled state, where is impossible to "separate" the particles), such that the state as a whole can be de-constructed into a set of sub-states containing only one particle each, and each being described by its own Hamiltonian?

Sorry to ramble, I'm a bit confused on the subject, in particular as I know that more than one particle can occupy a single-particle state (is the point here that the particles can still be attributed their own individual wave functions, and it just so happens that these individual wave functions describe the same state?).

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Many particle wavefunctions are generally appallingly complicated objects. One way to get a handle on them is to break them down into simpler parts, understand those parts and then put them back together again. We do this by constructing the space of many particle wavefunctions as either a tensor product space or a Fock space.

An obvious way break down a many particle system is to try to consider what each particle is doing individually. Obviously there will be emergent effects in the many body system due to entanglement that were not present when only considering one particle and for strongly interacting systems this breakdown may not be possible, but often it is the only method we have.

So the single particle states are those states which on their own describe a single particle and from which we construct the full space as a tensor product space (i.e. the tensor product of single particle states and linear combinations thereof) or Fock space.

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    $\begingroup$ so when one says that "two (or more) particles occupy the same single-particle state", is it meant that the two (or more) particles have their own wave-functions, but these wave-functions happen to describe the same state?! $\endgroup$ – Will Apr 20 '15 at 12:00
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    $\begingroup$ Essentially yes. When people say "two (or more) particles occupy the same single-particle state" they may or may not be implicitly assuming that the states have have been properly symmetrized (or antisymmeturized of fermions)when taking the tensor product. But asside from that yes. $\endgroup$ – By Symmetry Apr 20 '15 at 12:23
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A single-particle state is a state corresponding to a single particle in isolation. In weakly-interacting translation-invariant systems, for example, a particularly useful set of single-particle states are the plane-wave states $\lvert \mathbf{k}\rangle$, corresponding to a single particle with a plane-wave wavefunction $\langle \mathbf{x} \rvert \mathbf{k}\rangle \propto \mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{x}}$. Now, if one considers two distinguishable particles, it is possible for both to occupy the same single-particle state, in which case one would write the two-particle state as $\lvert \mathbf{k}\rangle \lvert \mathbf{k}\rangle$, which really means the tensor product $\lvert \mathbf{k}\rangle \otimes\lvert \mathbf{k}\rangle$. Such tensor products of single-particle states form a complete basis for the two-particle Hilbert space, so that the most general two-particle state (for distinguishable particles) can be written $$\lvert \psi_2\rangle = \sum_{\mathbf{k}_1,\mathbf{k}_2} a(\mathbf{k}_1,\mathbf{k}_2) \lvert \mathbf{k}_1\rangle \lvert \mathbf{k}_2\rangle,$$ where the coefficients $a(\mathbf{k}_1,\mathbf{k}_2) $ are probability amplitudes for finding particle 1 with wavevector $\mathbf{k}_1$ and particle 2 with wavevector $\mathbf{k}_2$. Note that this general form includes all entangled states, where it is not possible to write down a wave function describing particle 1 or 2 alone.

This generalises straightforwardly to $N$-particle systems. A basis for the $N$-particle Hilbert space is given by an $N$-fold tensor product of single-particle states. These single-particle states do not have to be plane waves, they could be whatever you want. When you consider indistinguishable particles, you also have to take into account the bosonic (fermionic) exchange (anti-)symmetry. This is achievable by including only (anti-)symmetrised combinations of single-particle states in the basis set.

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  • $\begingroup$ Hello. I' d say this is a pretty helpful answer. Thanks. May I ask if there would be any reason to speak of single-particle states in classical statistical mechanics(classical ideal gas, for example)? Thank you. $\endgroup$ – Constantine Black Aug 21 '15 at 10:23
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In a physics of nuclear structure, by the term single particle state is typically understood an excitation, that can be attributed mostly to one proton or one neutron that jumped to a higher orbit. Contrary to collective excitation or collective state, which is an excited level, that many nucleons participate in.

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