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edit approved Mar 28, 2021 at 15:26

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I have two quantum mechanical harmonic oscillators with the same frequency. The hamiltonianHamiltonian 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

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

But when I apply the hamiltonianHamiltonian 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

My questions are:
(1)what would $N_2\lvert n_1\rangle$ and $N_1\lvert n_2\rangle$ be? My 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 $n=n_1=n_2=1$?

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 $n=n_1=n_2=1$?

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 $n=n_1=n_2=1$?

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edited Mar 28, 2021 at 15:12

Yas0416

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Having trouble finding the energy eigenvalues of a combined A system of two uncoupled quantum harmonic oscillators

I have two quantum mechanical harmonic oscillators with the same frequency. The hamiltonian of the combined system is $$ H_0= \hbar \omega (a^\dagger a+b^\dagger b+1)$$$$ H= \hbar \omega (2a^\dagger a+b^\dagger b+2)$$ 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 $$H_0\lvert n\rangle=H_0(\lvert n_1\rangle+\lvert n_2\rangle)=\hbar \omega((N_1 + N_2 +1)\lvert n_1\rangle+(N_1 + N_2 +1)\lvert n_2\rangle)$$$$H_0\lvert n\rangle=H_0(\lvert n_1\rangle+\lvert n_2\rangle)=\hbar \omega((2N_1 + N_2 +2)\lvert n_1\rangle+(2N_1 + N_2 +2)\lvert n_2\rangle)$$ 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 $n=n_1=n_2=1$?

Having trouble finding the energy eigenvalues of a combined system of two uncoupled quantum harmonic oscillators

I have two quantum mechanical harmonic oscillators with the same frequency. The hamiltonian of the combined system is $$ H_0= \hbar \omega (a^\dagger a+b^\dagger b+1)$$ 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 $$H_0\lvert n\rangle=H_0(\lvert n_1\rangle+\lvert n_2\rangle)=\hbar \omega((N_1 + N_2 +1)\lvert n_1\rangle+(N_1 + N_2 +1)\lvert n_2\rangle)$$ 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 $n=n_1=n_2=1$?

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 $n=n_1=n_2=1$?