# Matrix Representation of Operators in a Finite Basis

1. Dec 10, 2012

### ggb123

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

I have my quantum mechanics final creeping up on me and I just have a question about something that doesn't appear to be covered in the text.

Let's say you have a wave function of the following form for a linear harmonic oscillator:

$\Psi = c_1 | E_1 \rangle + c_2 | E_2 \rangle$

The basis is just the first two excited energy states. My question is how the Hamiltonian matrix is represented in this case. Is it

$H = \hbar \omega \left( \begin{matrix} 0 & 0 & 0 \\ 0 & \frac{3}{2} & 0 \\ 0 & 0 & \frac{5}{2} \end{matrix} \right)$

Or do you just use the non-zero elements:

$H = \hbar \omega \left( \begin{matrix} \frac{3}{2} & 0 \\ 0 & \frac{5}{2} \end{matrix} \right)$

Any help would be greatly appreciated. Thanks!

Last edited: Dec 10, 2012
2. Dec 10, 2012

### Mute

If your hamiltonian is the usual harmonic oscillator hamiltonian, $\mathcal H = \hbar\omega(a^\dagger a + 1/2)$, with eigenstates $|n\rangle$ for n = 0, 1, 2, etc., then your space is infinite dimensional, and any linear combination of the eigenstates will still leave you with an infinite-dimensional system.

The 3x3 matrix you wrote down is not correct if the elements are $\mathcal H_{mn} = \langle m| \hbar \omega(a^\dagger a + 1/2)|n\rangle$, as the $\mathcal H_{00}$ element is $\hbar\omega/2$, not zero. You would need some sort of additional term in the Hamiltonian to cause this term to vanish, which you presently don't have.

Your 2x2 matrix, however, is a perfectly legitimate finite-dimensional subset of the full infinite-dimensional space. If your initial state is prepared in a superposition of $|1\rangle$ and $|2\rangle$, then the system will not leave those states and you can focus on the 2x2 subset.

Note that the other elements of the matrix $\mathcal H$ are not zero, but since your system was initially prepared in a superposition of the n = 1 and 2 states and there are no off-diagonal terms, meaning no transitions between states of different n, your system will stay in those two states and you can just focus on the 2x2 matrix.

Last edited: Dec 10, 2012
3. Dec 10, 2012

### ggb123

Thanks a lot!

4. Dec 11, 2012

### ggb123

I also would like to know how to represent the ladder operators of the harmonic oscillator in a finite basis of $| E_1 \rangle$ and $| E_2 \rangle$ as matrices.

I get the proper position expectation values using the following representations:

$a = \left( \begin{matrix} 0 & 0 \\ \sqrt{2} & 0 \end{matrix} \right) \hspace{5 mm} a^{\dagger} = \left( \begin{matrix} 0 & \sqrt{2} \\ 0 & 0 \end{matrix} \right)$

Using these, however, I do not have the proper relationship for the Hamiltonian matrix and the ladder operator product:

$H = a a^{\dagger} + \frac{1}{2} = \left( \begin{matrix} \frac{3}{2} & 0 \\ 0 & \frac{5}{2} \end{matrix} \right)$

I get a matrix of the following form for $a a^{\dagger}$:

$a a^{\dagger} = \left( \begin{matrix} 0 & 0 \\ 0 & 2 \end{matrix} \right)$

Am I correct in my initial matrices for the ladder operators and the relationship between H and $a a^{\dagger}$ doesn't hold for finite subsets of $\{ E_n \}$? Or were my initial ladder operator matrices incorrect?

Thanks!

5. Dec 11, 2012

### Mute

First, the Hamiltonian contains $a^\dagger a$, not $a a^\dagger$. This doesn't fix the discrepancy, however.

See what happens if you wrote down 3x3 matrices for $a$ and $a^\dagger$. I suspect that what you'll find is that there are terms from other ($n \neq 1,2$) states that come into the matrix multiplication that you dropped when writing the creation and annihilation operators as 2x2 matrices.

The number/energy wavefunctions are not eigenstates of the annihilation and creation operators, so restricting the matrix representation of the operators to a finite subspace of your basis states will not work out very well because the operators will try to take the input state into a state that is outside the restricted 2x2 representation. You don't have this problem with the full hamiltonian because it is diagonal in the basis you are using.

Also, lastly, use ^\dagger rather than ^\dag to get the daggers in Latex.