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Matrix Representation of Operators in a Finite Basis 
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#1
Dec1012, 06:36 PM

P: 23

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: [itex] \Psi = c_1  E_1 \rangle + c_2  E_2 \rangle [/itex] The basis is just the first two excited energy states. My question is how the Hamiltonian matrix is represented in this case. Is it [itex] H = \hbar \omega \left( \begin{matrix} 0 & 0 & 0 \\ 0 & \frac{3}{2} & 0 \\ 0 & 0 & \frac{5}{2} \end{matrix} \right) [/itex] Or do you just use the nonzero elements: [itex] H = \hbar \omega \left( \begin{matrix} \frac{3}{2} & 0 \\ 0 & \frac{5}{2} \end{matrix} \right) [/itex] Any help would be greatly appreciated. Thanks! 


#2
Dec1012, 10:44 PM

HW Helper
P: 1,391

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 infinitedimensional 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 finitedimensional subset of the full infinitedimensional 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 offdiagonal 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. 


#3
Dec1012, 11:03 PM

P: 23

Thanks a lot!



#4
Dec1112, 03:37 PM

P: 23

Matrix Representation of Operators in a Finite Basis
I also would like to know how to represent the ladder operators of the harmonic oscillator in a finite basis of [itex]  E_1 \rangle [/itex] and [itex]  E_2 \rangle [/itex] as matrices.
I get the proper position expectation values using the following representations: [itex] 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) [/itex] Using these, however, I do not have the proper relationship for the Hamiltonian matrix and the ladder operator product: [itex] H = a a^{\dagger} + \frac{1}{2} = \left( \begin{matrix} \frac{3}{2} & 0 \\ 0 & \frac{5}{2} \end{matrix} \right) [/itex] I get a matrix of the following form for [itex]a a^{\dagger}[/itex]: [itex] a a^{\dagger} = \left( \begin{matrix} 0 & 0 \\ 0 & 2 \end{matrix} \right) [/itex] Am I correct in my initial matrices for the ladder operators and the relationship between H and [itex] a a^{\dagger} [/itex] doesn't hold for finite subsets of [itex] \{ E_n \} [/itex]? Or were my initial ladder operator matrices incorrect? Thanks! 


#5
Dec1112, 03:49 PM

HW Helper
P: 1,391

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. 


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