Is it possible to prepare experimentally a quantum state which is a superposition of two states with the different number of particles? For example $|\psi\rangle=|N=1\rangle+|N=2\rangle$

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    $\begingroup$ I guess you can form $|N=1 \rangle \otimes | \mathrm{vacuum} \rangle_2 $ for $|N=1 \rangle$ for consistency $\endgroup$ – user26143 Aug 24 '13 at 11:31
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    $\begingroup$ Because QM is linear, nothing bad happens if you form a superposition of states chosen completely freely. But such a superposition may not be very interesting if it's not possible for its different parts to interfere. $\endgroup$ – user4552 Aug 24 '13 at 14:20
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    $\begingroup$ @richard In absence of any mechanism for production or annihilation of particles, time evolution will not be able to mix up the states with different particles numbers and hence you can study them separately too. $\endgroup$ – user10001 Aug 24 '13 at 16:14
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    $\begingroup$ @jjcale: I would stress that the ground state of a superconductor doesn't actually consist of a superposition of different particle numbers. This may be clearer in the case of nuclear physics, where the particle number fluctuations in the BCS ground state are usually quite large, and obviously can't represent anything real. This is an example where the states with different particle numbers can't physically interact in reality, but we introduce such an interaction purely as an approximation scheme in order to make the many-body problem more tractable. $\endgroup$ – user4552 Aug 24 '13 at 20:06
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    $\begingroup$ @jjcale: The BCS state is not just an approximation but the true ground state in the thermodynamic limit. Yes, but the thermodynamic limit is obtained with $N\rightarrow\infty$. In that same limit the fluctuations in the particle number become negligible in the sense that $\sigma_N/<N>\rightarrow0$. That's why I used the example of nuclear physics, where we only get $N\sim100$, and it is very clear and uncontroversial that the BCS ground state is an approximation. Condensed matter physicists don't usually have to worry about this because $N$ is on the order of Avogadro's number. $\endgroup$ – user4552 Aug 25 '13 at 21:26

The comments already give the hint answer, each number state would evolve independent if there are no mechanism of creation or annihilation. One situation is that the number operator commutes with the Hamiltonian $\mathcal{\hat{H}}$, so the number state in the superposition would preserve over time with extra phase on each of them:

$|\psi(t)\rangle = e^{-i\mathcal{\hat{H}}t/\hbar}|\psi(0)\rangle = \sum_N e^{-i\mathcal{\hat{H}}t/\hbar}c_N|N\rangle = \sum_N e^{-iE_Nt/\hbar}c_N|N\rangle \tag{1}$

The state with different number will not mix in this case.

For the preparation, suppose the system can absorb photons and create excitation. If the photon state is something like $|\psi\rangle = a|1\rangle+b|2\rangle$, then the electron excitation in the system could get the same state and evolve like equation (1) when there is no creation and annihilation. (I am not sure what systems have such properties.)


Sure, that comes up all the time in quantum optics. Search for "coherent state".


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