A system of two quantum harmonic oscillators

Question

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 $n=n_1=n_2=1$?

Answer 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.

Answer 2

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}

Answer 3

0. 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.