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I'm interested in applying the adiabatic theorem to the forced harmonic oscillator with time dependent hamiltonian of the form:

$$H(t) = \hbar \omega(a^{\dagger}a + \frac{1}{2}) - f(t)a - f^{*}(t)a^{\dagger}$$

where $f(t)$ is an arbitrary function of time and $f^{*}(t)$ is its complex conjugate. I've solved the problem exactly for the system state $|\Psi (t) \rangle$ which is a coherent state. In order to apply the adiabatic theorem I need to solve for the instantaneous eigenstates of the Hamiltonian $|E^{r}(t)\rangle$, which are not the same as the system state $|\Psi (t)\rangle$. $|E^{r}(t')\rangle$ is an eigenstate of $H(t')$ only at time $t = t'$

I'm not sure where to begin, I tried expanding the eigenstates as a linear combination of the excited states of the simple harmonic oscillator, just like a coherent state. But have gotten stuck. Can anyone point me in the right direction?

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  • $\begingroup$ The adiabatic theorem refers to an energy gap between states. As far as I understand, your Hamiltonian is about a single isolated state since there is no an index at the creation/annihilation operators. $\endgroup$ – freude Apr 6 '16 at 6:09
  • $\begingroup$ This post should help : physics.stackexchange.com/questions/129664/… $\endgroup$ – Adam Apr 6 '16 at 6:42
  • $\begingroup$ @Adam the post your referenced to has a hamiltonian where the only constant is $\omega$ so they are able to factor their hamiltonian. I'm not sure I'm able to get mine in a form like theirs. $\endgroup$ – CStarAlgebra Apr 6 '16 at 12:32
  • $\begingroup$ @CStarAlgebra: have a look at the second answer. There is the general case. $\endgroup$ – Adam Apr 6 '16 at 13:51
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To find the instantaneous energy eigenstates, you need to treat $t$ as a parameter and solve the problem for a time independent Hamiltonian depending on the extra parameter $t$.

The best way to do that is to complete the square and write the Hamiltonian as:

$$H = \hbar \omega (A^{\dagger} A + \frac{1}{2}) - \frac{ f(t)^2}{\hbar \omega}$$

where

$$A = a - \frac{f(t)}{\hbar \omega}$$ $$A^{\dagger} = a^{\dagger}-\frac{f(t)}{\hbar \omega}$$

Since, the commutation relations do not change: $$[A, A^{\dagger}] = [ a, a^{\dagger}] = 1$$ This Hamiltonian is just a shifted Harmonic oscillator Hamiltonian, whose (instantaneous )eigenvalues are: $$E_n = \hbar \omega (n+\frac{1}{2}) - \frac{ f(t)^2}{\hbar \omega}$$
Now, the following caution must be exercised. In order to compare the exact and the instantaneous solutions and verify the adiabatic theorem they must be expressed in terms of the same coordinates. In the instantaneous case, the shift in the raising and lowering operators will be translated to the position operator: $$X = A + A^{\dagger} = a + a^{\dagger} = x-2\frac{f(t)}{\hbar \omega}$$ The dependence of the instantaneous eigenfunctions will be on the shifted position coordinate $\Psi_n(X)$.

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