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

[milagros77@ma]

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

Homework Statement

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:

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

Homework Equations

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:
$$\left\langle H \right\rangle = \left\langle {\Psi H\Psi } \right\rangle = \sum\limits_{ij} {\left| {C_{ninj} } \right|} ^2 E_{ninj} = \frac{1} {4}E_1 + \frac{3} {4}(2E_1 + E_2 ) = \frac{2} {4} + \frac{3} {4}(4 + 1) = \frac{{17}} {4}$$

 

[Jhero]

Overall, your approach seems correct. You have correctly identified the Hamiltonian for the system and used the ladder operators to obtain the quantized energy levels. Your calculation for the expected value of the Hamiltonian is also correct.

However, there are a few minor errors in your solution. Firstly, in the expression for the ladder operators, the indices for the position and momentum operators should be swapped. It should be:

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

Also, in the calculation for the expected value of the Hamiltonian, the energy levels should be squared, so the correct expression is:

\left\langle H \right\rangle = \left\langle {\Psi H\Psi } \right\rangle = \sum\limits_{ij} {\left| {C_{ninj} } \right|} ^2 E_{ninj}^2 = \frac{1}
{4}E_1^2 + \frac{3}
{4}(2E_1^2 + E_2^2) = \frac{2}
{4} + \frac{3}
{4}(4^2 + 1) = \frac{{65}}
{4}

Overall, your understanding of the application of the harmonic oscillator to an arbitrary number of free bosons is correct. Keep up the good work!