# When one can reduce the Hilbert space in Jaynes-Cummings Model?

I found that : $$[N,H] = 0$$ , With $$N=|e> is the number of excitations and $$H$$ is the Hamiltonian of Jaynes-Cummings.

and i want to describe the dynamics of the system(atom of two levels and one mode of quantized field).

so why one can reduce the whole Hilbert space to the subspace spanned by the two states {|e,n>,|g,n+1>}?

The Jaynes-Cummings model describes coupling between two-level system and single boson mode.

The model Hamiltonian is : $$\hat{H}_{JC}^{}=\hbar \omega_{a}\hat\sigma_{}^{\dagger}\hat\sigma_{}^{}+\hbar\omega_{f}^{}\hat a_{}^{\dagger}\hat a_{}^{}+\frac{\hbar \Omega_{af}^{}}{2}\left[\hat a_{}^{\dagger}\hat\sigma_{}^{}+\hat\sigma_{}^{\dagger}\hat a_{}^{}\right],$$ where $$\hat a_{}^{\dagger}$$, $$\hat a_{}^{}$$ are bosonic operators and $$\hat\sigma_{}^{\dagger}=|1\rangle\langle 0|$$ and $$\hat\sigma_{}^{}=|0\rangle\langle 1|$$. ($$\mathbf{\text{Notice few differences from the standard conventions.}}$$)

As already pointed out by OP, $$\left[\hat{H}_{JC}^{},\hat N_{ex}^{}\right]=0$$ where $$\hat N_{ex}^{}=\hat\sigma_{}^{\dagger}\hat\sigma_{}^{}+\hat a_{}^{\dagger}\hat a_{}^{}$$ is the excitation number operator. Hence number of excitations in the composite system doesn't change during unitary evolution ($$e_{}^{-\frac{i}{\hbar}t \hat{H}_{JC}^{} }$$).

Arbitrary state of the composite system at any time lives in the tensor product space, spanned by $$\{|2 n + \alpha\rangle:= |\alpha\rangle \otimes |n\rangle : \alpha=0,1 \hspace{3pt}\&\hspace{3pt} n \in \mathbb{W}\}$$ ($$\mathbf{\text{Notice the labeling of basis states.}}$$): $$|\Psi(t)\rangle = \sum_{n=0}^{\infty}\psi_{n}^{}(t)|n\rangle .$$ Evolution of any given initial state is given by : $$i\hbar\frac{d}{dt}|\Psi(t)\rangle = \hat{H}_{JC}^{}|\Psi(t)\rangle\Rightarrow \sum_{n=0}^{\infty}i\hbar\frac{d}{dt}\psi_{n}^{}(t)|n\rangle =\sum_{n=0}^{\infty}\psi_{n}^{}(t)\hat{H}_{JC}^{}|n\rangle .$$ Taking inner product of $$|\Psi(t)\rangle$$ with $$|m\rangle$$ : $$i\hbar\frac{d}{dt}\psi_{m}^{}=\sum_{n=0}^{\infty}\langle m|\hat{H}_{JC}^{}|n\rangle \psi_{n}^{}(t) \equiv i\hbar\frac{d}{dt} \begin{bmatrix}\psi_{0}^{}(t)\\ \vdots \\ \psi_{n}^{}(t) \\ \vdots \end{bmatrix}= \begin{bmatrix} \ddots & \vdots & \ddots\\ \ddots & \langle m|\hat{H}_{JC}^{}|n\rangle & \ddots \\ \ddots & \vdots & \ddots \end{bmatrix} \begin{bmatrix} \psi_{0}^{}(t)\\\vdots \\ \psi_{n}^{}(t) \\ \vdots \end{bmatrix}.$$ Now note $$\hat N_{ex}^{}|n\rangle=\epsilon(n)|n\rangle$$, where $$\epsilon(n)=\begin{cases} \frac{n}{2} & \text{n is even} \\ \frac{n+1}{2} & \text{n is odd}\end{cases}$$ $$\hspace{5pt}$$ and use $$\left[\hat{H}_{JC}^{},\hat N_{ex}^{}\right]=0$$ : $$\langle m|\left[\hat{H}_{JC}^{},\hat N_{ex}^{}\right]|n\rangle=0 \Rightarrow \left[\epsilon(m)-\epsilon(n)\right]\langle m|\hat{H}_{JC}^{}|n\rangle =0 .$$ Hence $$\left[\epsilon(m)-\epsilon(n)\right] \neq 0 \Rightarrow \langle m|\hat{H}_{JC}^{}|n\rangle=0$$.

Further $$\langle m|\hat{H}_{JC}^{}|n\rangle$$ can only be non-zero for ordered pairs : $$\left(m,n\right) \in \mathbb{S} :=\{\left(n,n+1\right) :\text{n is odd}\} \cup\{\left(0,0\right)\}$$ i.e., only states $$|0\rangle \otimes |n+1\rangle$$ & $$|1\rangle \otimes |n\rangle$$ are coupled by $$\hat{H}_{JC}^{}$$.

Hence during temporal evolution, wave amplitudes $$\psi_{m}^{}(t)$$ & $$\psi_{n}^{}(t)$$ and are coupled only if $$\left(m,n\right) \in \mathbb{S}$$.

This analysis reveals that apart from state $$|0\rangle \otimes |0\rangle$$, all other states are coupled pair wise i.e, $$|0\rangle \otimes |n+1\rangle$$ & $$|1\rangle \otimes |n\rangle$$ are only coupled. Hence evolution of wave amplitude pairs $$\{\psi_{m}^{}(t), \psi_{n}^{}(t)\}$$ for $$\left(m,n\right) \in \mathbb{S}$$ is independent for different $$\left(m,n\right)$$.

You always work the Hilbert space over the states that you are interested in. All the other stuff that may be going on is irrelevant to the interaction.

You could think of the states as $$\vert e, n\rangle\otimes\vert\psi\rangle$$ and $$\vert g, n + 1\rangle\otimes\vert\psi\rangle$$. Whatever happens to the rest of the state, $$\vert\psi\rangle$$, which lives in a large Hilbert space still goes on, but you don't care about it.

• exactly , but what's the relation between that and the fact of N is a constant of motion ? – Qninja Jan 22 at 22:01
• I would say there isn't any link between focusing on a reduced Hilbert space and $N$ being a constant of the motion. You always work in a reduced space in QM, in nearly all situations. – user1936752 Jan 22 at 22:09