Question about the identity operator and the bosonic ladder operators

Question

Consider a self-adjoint operator $B$, such that for each mode $a_1,...,a_n$ [of a quantum bosonic system with Hilbert space $\cal H$ given by the corresponding Fock space] we have $B a_i B^\dagger = a_i$. Clearly, this works if $B=I$, the identity operator. But does this imply that $B$ is the identity operator?

Answer 1

The answer is no. We will slightly generalize the problem and discuss the task to find normal operators $B$ such that $BaB^\dagger =a$. This includes self-adjoint $B=B^\dagger$ and unitary $B^\dagger=B^{-1}$ operators. Below I give a proof ($n=1$) that the only operators obeying these constraints are of the form $$B=b P_1+ \frac{1}{b^*} P_2 \quad , \tag 1$$ where $P_1=\sum\limits_{n=0}^\infty |2n\rangle\langle 2n|$ and $P_2=\sum\limits_{n=0}^\infty |2n+1\rangle\langle 2n+1|$ for any $b\in \mathbb C\setminus\{0\}$.

It is easy to see that $B$ is hermitian if and only if $b$ is real and that $B=\pm I$ if and only if $b=\pm1$. Also, $B$ is invertible, and $B^{-1}=B^\dagger$ if and only if $b=e^{i\varphi}$ for some $\varphi\in \mathbb R$, which is the result obtained from Schur's lemma in the other answer.

To prove that $B$ in $(1)$ obeys $BaB^\dagger=a$, we compute:

$$BaB^\dagger=\left(bP_1+\frac{1}{b^*}P_2\right) a\left(bP_1+\frac{1}{b^*}P_2\right)^\dagger = P_1 aP_2 + P_2 aP_1 \quad , \tag 2 $$

where we've used the fact that $P_1aP_1 = P_2aP_2 =0$. Note that the RHS of $(2)$ is indeed independent of $b$. Also, by the same argument

$$a=IaI=\left(P_1+P_2\right)a\left(P_1+P_2\right) = P_1aP_2+ P_2aP_1 \quad , \tag 3 $$

which shows that $BaB^\dagger=a$ for $B$ in the form of $(1)$ for any $b\in \mathbb C\setminus\{0\}$.

---

Let us now give a proof that normal operators $B$ obeying $BaB^\dagger=a$ are necessarily of the form $(1)$.

First, we start by recalling the fact that $a^\dagger$ has no eigenvectors. In particular, for every eigenvector $|b\rangle$ of $B$ it holds that $a^\dagger |b\rangle \neq 0$. Second, we also recall that for normal operators it holds that $B|b\rangle=b|b\rangle$ if and only if $B^\dagger |b\rangle =b^* |b\rangle$.

Thus, by assumption:

$$0 \neq a^\dagger |b\rangle = Ba^\dagger B^\dagger|b\rangle = b^* Ba^\dagger |b\rangle \quad . \tag 4$$

This shows $b\neq 0$ and $Ba^\dagger |b\rangle\neq 0$, and hence that $a^\dagger |b\rangle$ is an eigenstate of $B$ with eigenvalue $1/b^*$; in particular $B$ has no vanishing eigenvalue.

Next, the assumptions also imply that

$$a|0\rangle = BaB^\dagger|0\rangle = 0\tag 5 \quad , $$

which, together with the previous considerations show that $aB^\dagger|0\rangle=0$ (because else it would be an eigenstate of $B$ with eigenvalue $0$), and since the vacuum is unique/non-degeneracy of the ground state of the harmonic oscillator, we must have that $B|0\rangle= b|0\rangle$ for some (non-zero) eigenvalue $b$, by the same argument.

Finally, consider the state $|1\rangle=a^\dagger |0\rangle$. From $(4)$ we know that $B|1\rangle=Ba^\dagger |0\rangle=1/b^* |1\rangle$. Similarly, we know $B|2\rangle=\frac{1}{\sqrt 2}Ba^\dagger|1\rangle = b|2\rangle$. The pattern should be clear now: We can re-write $B$ in the occupation-number basis as $$ B = b P_1 +\frac{1}{b^*} P_2\tag 6 \quad ,$$

concluding the proof.

Answer 2

I am a bit confused by your condition that $B$ should be self-adjoint, which would mean $B^{\dagger}=B$. I suspect you meant that $B$ was to be unitary, $B^{\dagger}=B^{-1}$. However, it seems to make no difference in this case, because sandwiching the identity operator between the $B^{\dagger}$ and $B$ gives that $B^{\dagger}B=I$, making $B$ an involution; it is its own inverse. So I will answer for the more general case that $Ba_{i}B^{-1}=a_{i}$ and $Ba_{i}^{\dagger}B^{-1}=a_{i}^{\dagger}$. For a unitary $B$, $Ba_{i}B^{-1}=a_{i}$ and $Ba_{i}^{\dagger}B^{-1}=a_{i}^{\dagger}$ are equivalent, as may be seen from just taking the Hermitian conjugate.

With this condition set, the answer is yes, because this is essentially a statement of Schur's lemma, which states (more or less) that if an operator $M$ commutes with all other operators acting on some subspace $V$, then $M$ is proportional to the identity. The $a$ operators and their adjoints generate all the operators on the bosonic state space, so if $B$ commutes with all of them, it is proportional to the identity.