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The Hamiltonian of a two-level system is given by $$H=E_1|1\rangle\langle 1|+E_2|2\rangle\langle 2|$$ where both the energy eigenstates $|1\rangle$ and $|2\rangle$ are non-degenerate with $E_2>E_1$. Now consider such a 2-level system with $N$ atoms. Let at $t=0$, there is a population inversion so that all the $N$ atoms are in the excited state $|2\rangle$ and zero atoms in the ground state $|1\rangle$.

  1. Does it mean that one implicitly defines a number operator and which has been measured? If yes, I don't know how to define it.

  2. If no, how can I claim that there are $N$ atoms are found in the state $|2\rangle$ and $0$ atoms in $|1\rangle$?

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Your $N$ systems are non-interacting, and thus $$ \hat N=\bigoplus_{i=1}^N \hat N_i $$ where $$ \hat N_i|1_i\rangle=0\ |1_i\rangle\qquad \hat N_i |2_i\rangle=1\ |2_i\rangle $$ for all $i$, that is, $$ \hat N_i=|2_i\rangle\langle 2_i| $$

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  • $\begingroup$ Dear @AFT, thanks for your answer. But two things are unclear. Why do you define the number operator as $\hat N_{i}=|2_i\rangle\langle 2_i|$ for the $i^{th}$ sub-system? The number operator seems to be contrived. I mean why not $|1_i\rangle\langle 1_i|$ or $|2_i\rangle\langle 1_i|$? And a minor point. There should be kets on the RHS of your equations. $\endgroup$
    – SRS
    Commented Jan 8, 2017 at 15:35
  • $\begingroup$ @SRS It's just a convenient definition: if in a system $|\varphi\rangle$ there are $N=100$ atoms, and $32$ of them are in the level $|2\rangle$, then $\hat N|\varphi\rangle=32|\varphi\rangle$. I could have defined $\hat N'_i=|1_i\rangle\langle 1_i|$, in which case $\hat N'|\varphi\rangle=68|\varphi\rangle$. I could even have defined $\hat N''_i=-|1_i\rangle\langle 1_i|+|2_i\rangle\langle 2_i|$, in which case $\hat N''|\varphi\rangle=-36|\varphi\rangle$. Any definition that allows you to count the states works fine 2) fixed, thanks. $\endgroup$
    – AccidentalFourierTransform
    Commented Jan 8, 2017 at 15:38
  • $\begingroup$ I'm not convinced and feeling uneasy about it. If I take $|1_i\rangle\langle 1_i|$ as the number operator, then its action says that there is 1 particle in state $|1_i\rangle$ and no particle in state $|2_i\rangle$. If we take $|2_i\rangle\langle 2_i|$ to be the number operator, it says exactly the opposite. May be I've to give it a little thought and then I'll get back to you. Thanks. $\endgroup$
    – SRS
    Commented Jan 8, 2017 at 16:14
  • $\begingroup$ @SRS they are different number operators, but they all count. They count different things, but if you know the total number of atoms $N$, counting those in the level $2$ is equivalent to counting those in the level $1$. In effect, $\hat N'=N-\hat N$. $\endgroup$
    – AccidentalFourierTransform
    Commented Jan 8, 2017 at 16:15
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    $\begingroup$ @SRS in the notation of my previous comment, $\hat N$ counts the number of particles in $|2 \rangle$ and $\hat N'$ counts the number of particles in $|1\rangle$. Unambiguously. $\endgroup$
    – AccidentalFourierTransform
    Commented Jan 8, 2017 at 16:26
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So what you're kind of orbiting around is a central uncertainty about dealing with multiparticle systems, and that's indeed important because you do indeed have this problem of "now what is my wavefunction, if it does not have a fixed set of coordinates $\psi(\vec r_1, \vec r_2, \dots \vec r_N)$ because I don't yet know what $N$ is..." etc.

Eventually we'll define field operators which will make some sense of "position coordinates" again, but for right now you will just have to let that uncertainty go. Just be willing to do what's algebraically allowed in the bra/ket formalism.

In this case, for example, there is clearly a $|2\rangle\langle 2|$ operator which counts the number of particles in state 2. We know that this is well-defined because we see it in the Hamiltonian and you told us the Hamiltonian was well-defined, so there's no problems here. Furthermore it commutes with the $|1\rangle\langle 1|$ operator which counts the number of particles in state 1, we know this because you are very comfortable saying "there are N particles in state 2 and 0 particles in state 1," which suggests that they are, as single-particle states, disjoint.

Now you do have to decide whether the wavefunction is exactly the same when you interchange two identical particles, or whether the wavefunction is exactly the negative of what it was, or whether the particles are not fundamentally interchangeable and the wavefunction will always know the difference. That last case seems much much harder to treat than the first two.

If the particles are bosonic and saying "all of them are in state 2" literally means "they're all in this single wavefunction state," then you get to invent bosonic annihilators $\hat a,~\hat b$ such that $[\hat a, \hat b] = 0$ but $[\hat a, \hat a^\dagger] = [\hat b, \hat b^\dagger] = 1.$ To model transport between the two you'll have terms like $\hat c~ \hat b^\dagger ~\hat a + \hat c^\dagger~\hat a^\dagger ~\hat b$ in your Hamiltonian, where usually $\hat c$ will describe some other system that the energy goes off into. And this operator we've been calling $|1\rangle\langle1|$ is actually $\hat a^\dagger ~\hat a.$ Easy peasy.

If the particles are fermionic then we generally need to say that there are secretly a collection of states that we're lumping together as "2" and a collection that we're lumping together as "1", so that gets more complicated, we have to peek at the internal states. Invent a fermionic annihilator for each state, so we don't just have $\hat a$ and $\hat b$ but $\hat a_i$ and $\hat b_i$. Now we have $\{\hat a_i, \hat b_j\} = 0$ while $\{\hat a_i, \hat a_j^\dagger\} = \{\hat b_i, \hat b_j^\dagger\} = \delta_{ij}.$ Then this operator that we've been calling $|1\rangle\langle 1|$ is actually $\sum_i \hat a_i^\dagger~\hat a_i.$ Again for transport you probably want to invent a bunch of bosonic modes with annihilator $\hat m_i$ to absorb the energy differences of jumping between the two modes, so that you have a Hamiltonian like (Einstein summation) $$\hat H = 0~ \hat a^\dagger_i ~\hat a_i + \epsilon~\hat b^\dagger_i ~\hat b_i + \epsilon~\hat m^\dagger_i ~\hat m_i + c_{ijk} ~\hat m_i^\dagger ~\hat a_j^\dagger ~\hat b_k + c_{ijk}^* ~\hat m_i ~\hat b_k^\dagger ~\hat a_j.$$So there's no deep mystery in this thing once you abandon this hope of having the explicit wavefunction over all $N$ spatial coordinates written out for you.

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  • $\begingroup$ Dear @CR Drost The Hamiltonian just represents a two-state system. It doesn't contain any information whether the state $|1\rangle$ (or $|2\rangle$) contains 1 particle or 0 particle. When I say $|2\rangle\langle 2|$ counts the number of particles in state $|2\rangle$, it is as if I already know there is 1 particle in state $|2\rangle$. But I can only claim whether a state contains a particle only after measuring the number operator on that state. I don't know apriori, which state contains the particle, what is the unique number operator? $\endgroup$
    – SRS
    Commented Jan 9, 2017 at 12:53
  • $\begingroup$ I think I need to put particles in states before I can define such an operator. Right? $\endgroup$
    – SRS
    Commented Jan 9, 2017 at 12:53
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    $\begingroup$ @SRS Yes, the technical term is "Fock space" and you either have to define the states $|m,n\rangle$ (e.g. $|5,3\rangle$ has 5 bosons in the state you called $|1\rangle$, and 3 bosons in the state you called $|2\rangle$, with their underlying wavefunctions appropriately symmetrized) or else define the states $|a_1,a_2,\dots,a_m/b_1,b_2,\dots,b_n\rangle,$ e.g. $|01011/11000\rangle$ has 5 fermions, with 3 of them in the $a_{2,4,5}$ states and 2 of them in the $b_{1,2}$ states, with their underlying wavefunctions appropriately antisymmetrized. $\endgroup$
    – CR Drost
    Commented Jan 9, 2017 at 16:23
  • $\begingroup$ But in a much more abstract sense, you can basically read the states that you can put the particle in, off of the creation/annihilation operators you're considering. I have a creation operator $\hat a_1^\dagger$ which is fermionic, so I know that there must be an $a_1$ index in my Hilbert space, which can either take on the values 0 or 1. So I just write down a 0 or a 1 there. $\endgroup$
    – CR Drost
    Commented Jan 9, 2017 at 16:26

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