# A system of two quantum harmonic oscillators

I have two quantum mechanical harmonic oscillators with the same frequency. The Hamiltonian of the combined system is: $$H= \hbar \omega (2a^\dagger a+b^\dagger b+2)$$ In attempting to find the energy of the combined system, I denoted the energy eigenkets to be $$\lvert n_1\rangle$$ and $$\lvert n_2\rangle$$ for the first and the second oscillator respectively so that the energy eigenket of the combined system is a linear combination of them, i.e. $$\lvert n\rangle=\lvert n_1\rangle+\lvert n_2\rangle$$.

Then I used the definition of the number operator $$N_1 = a^\dagger a$$ and $$N_2 =b^\dagger b$$, where $$N_1\lvert n_1\rangle=n_1\lvert n_1\rangle$$ and $$N_2\lvert n_2\rangle=n_2\lvert n_2\rangle$$.

But when I apply the Hamiltonian onto the eigenkets, I have: $$H_0\lvert n\rangle=H_0(\lvert n_1\rangle+\lvert n_2\rangle)=\hbar \omega((2N_1 + N_2 +2)\lvert n_1\rangle+(2N_1 + N_2 +2)\lvert n_2\rangle)$$

My questions are:
(1)what would $$N_2\lvert n_1\rangle$$ and $$N_1\lvert n_2\rangle$$ be? My intuition tells me they are $$0$$, but I'm not so sure.
(2)I know the ground state is denoted by $$n=0$$, but what about the first excited state? Would the first excited state be denoted by $$n=n_1+n_2=1$$ or $$n=n_1=n_2=1$$?

The combined system eigenstate is not a linear combination. When you combine two quantum systems you take a tensor product of the Hilbert spaces so the combined eigenstates are $$|n_1,n_2\rangle \equiv |n_1\rangle \otimes|n_2\rangle$$ where the first form is the way that a physicist would write it, and the second the way a methematician would write it.

Then $$a^\dagger a |n_1,n_2\rangle= n_1 |n_1,n_2\rangle\\ b^\dagger b |n_1,n_2\rangle=n_2 |n_1,n_2\rangle.$$ In other worrds the $$a$$'s act only on the first factor in the tensor product and the $$b$$'s on the second. A mathematician would probably write $$H= (a^\dagger a +\frac 12)\otimes {\rm identity}+ {\rm identity}\otimes (b^\dagger ab +\frac 12)$$ to make this clear.

• This makes a lot of sense! Thanks a lot! Mar 27, 2021 at 20:12

The energy eigenkets of the combined system are not a linear combination of the two but rather a tensor product: $$\left| n_1 n_2 \right\rangle = \left|n_1 \right\rangle \otimes \left|n_2 \right\rangle$$ This is the proper way to build composite systems in quantum mechanics -- the resulting Hilbert space is a product of the two individual Hilbert spaces and any operator which acts only on one space is implicitly assumed (usually) to act as identity on the other space. Hence: \begin{align} \hat{H} \left| n_1 n_2 \right\rangle &= \hbar \omega \left( a^\dagger a + b^\dagger b + 1\right) \left| n_1 n_2 \right\rangle \\ &= \left(\hbar \omega n_1 + \hbar \omega n_2 + \hbar \omega\right) \left| n_1 n_2 \right\rangle \end{align}

• Thanks a lot! This answers my second question too! Mar 27, 2021 at 20:12
1. Your ket for the system of two uncoupled oscillators will be a direct product, $$|n_1>|n_2>$$. Then it is easy to understand the application of the hamiltonian: $$a^+a$$ sort of acts on $$|n_1>$$ part and $$b^+b$$ on $$|n_2>$$, so you get:

$$(a^+a+b^+b)|n_1>|n_2>=n_1|n_1>|n_2>+n_2|n_1>|n_2>=(n_1+n_2)|n_1>|n_2>$$.

Edit: while I was typing it, another answer was given which I do agree with.

1. In your notation, when you have $$\hat{N}_2|n_1>$$, you apply the creation-elimination operators of oscillator 2 to a ket $$|n_1>$$ which does not know anything about it, it only cares about oscillator 1. The only way out of this dead end is to assume that $$|n_1>$$ effectively means $$|n_1>|0>$$, a state where $$n_2=0$$. Then we are back to the previous point.

2. You will have two "first" excited states, $$|1>|0>$$ and $$|0>|1>$$. If the oscillators have the same frequency, these states are energy-degenerate.

• Thank you! Much appreciated! Mar 27, 2021 at 20:12