How can one show that $\hat{U}|00\rangle=|00\rangle$ where $\hat{U}=e^{-igt(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)}$ and $|00\rangle$ is the unique joint zero eigenstate of the annihilation operators $\hat{a}_1$ and $\hat{a}_2$, i.e. $\hat{a}_1|00\rangle=\hat{a}_2|00\rangle = 0$.
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Calculate
$$ \begin{align} \hat{U}|00\rangle &= \exp\left(-igt(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)\right)|00\rangle \\ &= \sum_{k=0}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)^k|00\rangle \\ &= |00\rangle + \sum_{k=1}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)^k|00\rangle \\ &= |00\rangle + \sum_{k=1}^\infty 0 \\ &= |00\rangle \end{align} $$
where in the penultimate step we used the fact that
$$ (\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)|00\rangle = \hat{a}^\dagger_2\hat{a}_1|00\rangle+\hat{a}^\dagger_1\hat{a}_2|00\rangle = 0. $$
This implements the first approach suggested by @user1271772.
answered Mar 14, 2021 at 6:26
$\begingroup$
$\endgroup$
1
There's more than one way, and I'll suggest two of them here:
- Expand $\hat{U}$ using the formula for the Taylor series of an exponential ($e^\hat{A}$) centered around $\hat{A}=\hat{0}$, and then you will have a sum of terms where each term no longer involves an exponential operator (i.e. you have just pure creation and annihilation operators and products/powers of them acting on $|00\rangle$. Just as in the answer to your last question (If a Hamiltonian is quadratic in the ladder operator, why is it's time evolution linear in the ladder operator?), you can often find a pattern quite quickly which applies to all of the infinitely many terms in the Taylor Series.
- Use the property that $U$ can be written as $V^\dagger D V$ where $V$ is the matrix built of eigenvectors of $\hat{a}_2^\dagger \hat{a}_1 + \hat{a}_1^\dagger \hat{a}_2$, and $D$ is a diagonal matrix where the entries are just the scalar exponentials of the eigenvalues of $\hat{a}_2^\dagger \hat{a}_1 + \hat{a}_1^\dagger \hat{a}_2$ corresponding to those eigenvectors that were used to build $V$.
answered Mar 14, 2021 at 4:59
-
1$\begingroup$ +1 The first approach turns out to be very simple in this case (see my answer). The simplicity was to be expected because the state $|00\rangle$ survives the action of $\hat{U}$ unchanged suggesting that the action of the first term, i.e. identity, in the exponential expansion is all there is to it. $\endgroup$ Mar 14, 2021 at 6:33
$\begingroup$
$\endgroup$
Let $|\psi\rangle$ be an eigenstate of an operator $A$, $A|\psi\rangle=\lambda|\psi\rangle$. Then $$e^A |\psi\rangle = \sum_{k=0}^\infty \frac{A^k}{k!}|\psi\rangle = \sum_{k=0}^\infty \frac{\lambda^k}{k!}|\psi\rangle = e^\lambda |\psi\rangle.$$ In this particular case, $A=-igt(a_2^\dagger a_1+a_1^\dagger a_2)$, of which $|00\rangle$ is an eigenstate with eigenvalue $\lambda=0$.
