# Application of the harmonic oscillator to an arbitrary number of free bosons

1. Jan 18, 2009

### milagros77@ma

Aplication of the harmonic oscillator to an arbitrary number of free bosons

1. The problem statement, all variables and given/known data

I will like to know if my answer to this problem is correct and If not, what I am missing.

Given a system with an arbitrary number of free bosons, where the hamiltonian for one particle with two independent states is given by the equation:

$$H = :(\frac{1}{2}\pi _1^2 + \frac{1}{2}\pi _2^2 + 2\phi _1^2 + \frac{1}{2}\phi _2^2):$$

Calculate the expected value for the Hamiltonian in the state:

$$\left| {\Psi \rangle } \right. = \frac{1}{2}\left| {1,0\rangle + } \right.\frac{{\sqrt 3 }}{2}\left| {2,1\rangle } \right.$$

2. Relevant equations

3. The attempt at a solution

The Hamitonian correspond to two non coupled harmonic oscillators with energies $$E_{1}=2; E_{2}=1$$ in natural units.
Since the Hamiltonian is already diagonalized, we proceed to the quantization introducing the ladders operators:

\eqalign{ & a = \frac{1} {{\sqrt 2 }}\left[ {\sqrt 2 \phi _1^{} + \frac{i} {{\sqrt 2 }}\pi _1 } \right] \cr & b = \frac{1} {{\sqrt 2 }}\left[ {\phi _2 + i\pi _1 } \right] \cr}

we obtain for the Hamiltonian:

\eqalign{ & H = :(aa^ + )\,2(aa^ + )^t \, + (b\,b^ + )(b\,b^ + )^t : \cr & \cr & H = 2\,a^ + a + b^ + b\,\,; \cr & \cr}

Solving the independent Schrödinger equation, we obtain that the energy of the system equals

$$E = 2n_1 + n_2$$, where n1 and n2 indicate the number of particles for the two monoparticular states.

To calculate the value of H in the given state:
$$\left\langle H \right\rangle = \left\langle {\Psi H\Psi } \right\rangle = \sum\limits_{ij} {\left| {C_{ninj} } \right|} ^2 E_{ninj} = \frac{1} {4}E_1 + \frac{3} {4}(2E_1 + E_2 ) = \frac{2} {4} + \frac{3} {4}(4 + 1) = \frac{{17}} {4}$$

Last edited: Jan 18, 2009