# 2 boson harmonic oscillator

1. Dec 13, 2015

### kuy

hy physics forum, im doing an advanced qm course, but i still have some doubts about the correct formalism and so on..
so i have this problem of a one-dimensional harm. oscillator including 2 bosons.
the hamiltonian would thus be $\hat{H}=1/2 (\hat{p_1}^2+\hat{p_2}^2+\omega \hat{x_1}^2+\omega \hat{x_2}^2)$
the ground state would be $|0,0>$ and i have to build first and second excited states.
so my question is, would the first excited state just be $1/\sqrt{2} (|1,0> + |0,1>)$ or do i have to consider the pure states $|1,0>$ and $|0,1>$too? and how would i write that properly with ladder operators?
kuy

2. Dec 13, 2015

### ShayanJ

At first, $\frac{1}{\sqrt 2} (|1.0\rangle+|0,1\rangle)$ is a pure state too!
And yes, this is the first excited state but each of $|1,0\rangle$ and $|0,1\rangle$ can't be because the particles are indistinguishable so you can't know which one is excited.
Each of the oscillators has its own Hilbert space. But when you write things in terms of $|1,0\rangle$ and $|0,1\rangle$, it means you're considering the tensor product of the two Hilbert spaces and so your operators should be product operators too. So if you want to raise $|0,0\rangle$, it means you should use the raising operator associated to the tensor product space which is $\hat a^\dagger=\hat a_1^\dagger \otimes \hat I_2+\hat I_1 \otimes \hat a_2^\dagger$(some may write it simply as $\hat a^\dagger=\hat a_1^\dagger+\hat a_2^\dagger$ but you should be careful each operator acts on which Hilbert space.) , where $\hat I_1$ and $\hat I_2$ are identity operators in the two Hilbert spaces and $\hat a_i^\dagger |0\rangle_i=C^{(i)}_+ |1\rangle_i$ and e.g. $\hat I_1 \otimes \hat a_2^\dagger |0,0\rangle=C_+^{(2)} |0,1\rangle$.

3. Dec 14, 2015

### kuy

wow, this answer was far better than i hoped for :D thanks Shyan, that brought some clarity for me into undistinguishable particles

4. Dec 14, 2015

### kuy

ok i have another question: if i make the second excited state with this tensor product raising operator, i get a linear combination of $|2,0>$, $|0,2>$ and $|1,1>$. Is that THE excited state? i thought $1/\sqrt{2} (|2,0> + |0,2>)$ and $|1,1>$ would be separate states

5. Dec 14, 2015

### ShayanJ

Yes, that's it. Its symmetric so is proper for two bosons. Also its what the raising operator gives us.
But of course you can have a linear combination of $|0,2\rangle \ ,|2,0\rangle \ , |1,1\rangle$ that is orthogonal to the one you got but that would be an antisymmetric linear combination which is used for fermions. Also this is what you get if you raise an antisymmetric first excited state. Of course there may be other linear combinations but they won't be proper base states because we need either symmetric or antisymmetric wave-functions.
Why?

6. Dec 14, 2015

### Staff: Mentor

No. The second excited state is doubly degenerate.

7. Dec 14, 2015

### Staff: Mentor

There is also a linear combination that is orthogonal and symmetric.

8. Dec 15, 2015

### kuy

if you meusure the energy of one particle in the $|1,1>$ state it would be different to one of the $|2,0>$ state, but the total energy is the same.
i was confused because i thought, if a energy level is degenerate, there have to be several states. so is a degenerate state always a linear combination of states with different properties?
or do i need 2 linear combinations for a doubly degenerate state?

Last edited: Dec 15, 2015
9. Dec 15, 2015

### Staff: Mentor

You can use $\frac{1}{\sqrt{2}} \left( \left| 0, 2 \right\rangle + \left| 2, 0 \right\rangle \right)$ and $\left| 1, 1 \right\rangle$ as the two orthogonal degenerate states.