$$H=\hbar\omega \left(a^\dagger a+\frac{1}{2}\right)+\hbar \omega_0\left(a^\dagger+a\right)$$ How to diagonalize $H$ and find its eigenenergies?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ What are your attempts, what have you tried? $\endgroup$– Tobias FünkeCommented Jul 5, 2023 at 18:55
-
$\begingroup$ I am very clueless could you give me a hint $\endgroup$– SophieCommented Jul 5, 2023 at 19:03
-
$\begingroup$ you can take a look at this at the section "Matrix representation": en.wikipedia.org/wiki/Creation_and_annihilation_operators $\endgroup$– MatteoCommented Jul 5, 2023 at 19:44
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Using creation and annihilation operators is a bit misleading here, since the problem is much easier to solve in ordinary notation. Note that $$ x=\sqrt{\frac{\hbar}{2m\omega}}\left(a+a^{\dagger}\right) \ , $$ so your Hamiltonian can be rewritten as $$ H=\frac{1}{2m}p^2+\frac{m\omega^2}{2}x^2+\eta x $$ for $\eta=\hbar\omega_0\sqrt{2m\omega/\hbar}$. Now you only need to complete the square to obtain a harmonic oscillator Hamiltonian with a shifted center and a constant term: $$ H=\frac{1}{2m}p^2+\frac{m\omega^2}{2}\left(x+\frac{\eta}{m\omega^2}\right)^2 -\frac{\eta^2}{2m\omega^2}. $$ This Hamiltonian has the usual eigenvalues, with a constant subtracted: $$ E_n=\hbar\omega\left(n+\frac{1}{2}\right)-\frac{\eta^2}{2m\omega^2} =\hbar\omega\left(n+\frac{1}{2}-\frac{\omega_0^2}{\omega^2}\right) \ , $$ for $n=0,1,2,...$.
answered Jul 5, 2023 at 19:41
-
$\begingroup$ Nice solution! Would it be possible to solve this by writing the matrix representation of $H$ on the Fock basis? $\endgroup$– MatteoCommented Jul 5, 2023 at 20:28
-
1$\begingroup$ @Matteo I'm not sure. Due to the properties of $a$ and $a^{\dagger}$, we will have a tridiagonal matrix of the form $H_{kl}=k\delta_{kl}+q(\sqrt{k}\delta_{k,l+1}+\sqrt{l}\delta_{l,k+1})$ (I subtracted the $1/2$ term of the ZPE, and turned to $\hbar\omega$ units; $q=\omega_0/\omega$). However, I don't see any straightforward way to diagonalize it. I'll have to think about this... (Of course, it is easy to check numerically that the eigenvalues do tend to $k-q^2$ as the basis size is increased.) $\endgroup$ Commented Jul 6, 2023 at 9:26
-
1$\begingroup$ You can also complete the square with the ladder operators. More generally, note that the Hamiltonian is up to a constant shift unitarily equivalent to the harmonic oscillator Hamiltonian, where the unitary operator is just the translation operator. $\endgroup$ Commented Jul 6, 2023 at 10:45
-
1$\begingroup$ @dennismoore94 I asked about this on mathematics stack exchange, if you are interested: math.stackexchange.com/questions/4731225/… $\endgroup$– MatteoCommented Jul 6, 2023 at 11:27