# Why does joint ground state not change under action of beam splitting unitary operator?

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. $$\hat{a}_1|00\rangle=\hat{a}_2|00\rangle = 0$$.

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.

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 $$V$$.
• +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. Mar 14 at 6:33

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 $$\lambda=0$$.