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I have a question about the procedure of generating the matrix elements of the Hamiltonian for a Harmonic oscillator.

I understand how to calculate the matrix for the normal Hamiltonian i.e., $$H=\frac{p^2}{2m}+\frac{1}{2}m\omega x^2$$ as $H \psi_j =E_j \psi_j$ then $$H_{ij}=\langle i|H|j\rangle=E_j\delta_{ij}=(j+1/2)\hbar \omega~ \delta_{ij}$$ which gives us the matrix.

But how would one go about calculating the matrix elements for a more complicated Hamiltonian e.g. an addition of $\cos(x)$ term to the potential $$H=\frac{p^2}{2m}+\frac{1}{2}m\omega x^2+\cos(x).$$

In this case if I remember correctly we can write $H_{mn}=\langle e_m|H|e_n\rangle$ which gives an integral of the new Hamiltonian with the $e_m$ and $e_n$ basis. However my knowledge about this is a bit sketchy and I am unsure what this basis vector is, so any help would be appreciated,

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  • $\begingroup$ In the normal quantum harmonic oscillator case, you should think about what your eigenstates $\psi_j$ are and how you arrived at them. For the more complicated case, what you wrote for $H_{mn}$ is correct. You just need to provide some basis (such as the $\psi_j$ basis in the original case). Also, the matrix element is given by the definition of the inner product on your Hilbert space. For a position space basis, it would look something like $H_{ij} = \psi_i^\dagger(x)H(x)\psi_j(x)$. $\endgroup$ – Aaron Nov 1 '16 at 21:44
  • $\begingroup$ Are you working on a flux qubit or something like that? $\endgroup$ – DanielSank Dec 10 '16 at 23:02
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Since you are just asking about language, I will start the exercise for you but will not solve it to the complete, elegant, punchline. You are probably encouraged to read up on coherent states and their displacement operators.

To start with, dimensionally, your extra interaction term makes no sense, and you need to normalize x in your extra term to yield a dimensionless argument for the cosine. To this end, introduce dimensionless real constants ρ and α, so your hamiltonian is now $$ H= \frac{p^2}{2m} + \frac{m\omega x^2}{2} + 2\rho \hbar \omega \cos \left ( \frac{\alpha x~\sqrt{2m\omega}}{\sqrt{\hbar}}\right ) ~. $$

Dispensing with pesky dimensionful constants, $$ a\equiv x\sqrt{m\omega/2\hbar} +i p/\sqrt{2\hbar m\omega}, \qquad a^\dagger\equiv x\sqrt{m\omega/2\hbar} -i p/\sqrt{2\hbar m\omega}, \qquad [a,a^\dagger ]=1 , $$ so that $$ H=\hbar \omega \left( a^\dagger a + \frac{1}{2} +2\rho \cos(\alpha(a+a^\dagger)) \right). $$

So the “perturbation” has filled up the infinite-dimensional matrix—it is not diagonal anymore, by far !

For example, $$ \langle n| H |0\rangle = \langle n| \hbar \omega \left ( \frac{1}{2}+\rho e^{-\alpha^2/2}(e^{i\alpha a^\dagger } e^{i\alpha a} +e^{-i\alpha a^\dagger }e^{-i\alpha a}) \right ) |0 \rangle \\ =\hbar \omega \left (\frac{1}{2}+\rho e^{-\alpha^2/2} ~(-)^{\lfloor n/2 \rfloor } (1+ (-)^n)~\frac{\alpha^n}{\sqrt{n}} \right )~, $$ etc… where use has been made of the integer floor function $\lfloor x\rfloor $.

Academic aside: Dropping the harmonic potential and just retaining the cosine term would net you the celebrated Sine-Gordon QM problem.

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