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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:

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

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.

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.

The answer is no. Below I give a proof for the $n=1$ case that the only operators obeying these constraints are of the form $$B=b P_1+ b^{-1} P_2 \quad , \tag 1$$ 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 $b\in \mathbb R\setminus\{0\}$.

It is easy to see that $B$ is hermitian and that $B=\pm I$ if and only if $b=\pm1$. Also, $B$ is invertible, and $B=B^{-1}$ if and only if $b=\pm 1$.

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

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

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 $BaB=a$ for $B$ in the form of $(1)$ for any $b\in \mathbb R\setminus\{0\}$.

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

We start by recalling the fact that $a^\dagger$ has no eigenvectors. In particular, for every eigenvector $|b\rangle$ of $B$ it holds that $a^\dagger |b\rangle \neq 0$ and thus, by assumption:

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

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

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

which, together with the previous considerations show that $aB|0\rangle=0$ (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 $b$.

Finally, consider the state $|1\rangle=a^\dagger |0\rangle$. From $(1)$ 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 $$ B = b P_1 +b^{-1} P_2\tag 6 \quad ,$$

concluding the proof.

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:

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

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.

The answer is no. Below I give a proof for the $n=1$ case that the only operators obeying these constraints are of the form $$B=b P_1+ b^{-1} P_2 \quad , \tag 1$$ 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 $b\in \mathbb R\setminus\{0\}$.

It is easy to see that $B$ is hermitian and that $B=\pm I$ if and only if $b=\pm1$. Also, $B$ is invertible, and $B=B^{-1}$ if and only if $b=\pm 1$.

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

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

where we've used the fact that $P_1aP_1 = P_2aP_2 =0$. Note that the RHS of $(2)$ is indeed independent of $b$. It remains to show that $a=P_1aP_2 + P_2 aP_1$. To this end, define. Let $A:=P_1aP_2 + P_2 aP_1$ and notice that for even $n$ it holds that

$$A|n\rangle= 0 + P_2aP_1|n\rangle= P_2 a|n\rangle = \sqrt n P_2|n-1\rangle = \sqrt n |n-1\rangle = a|n\rangle \quad , \tag 3$$

and completely analogously for $n$ odd. Thus, $A|n\rangle=a|n\rangle$ for all $n\in \mathbb N_0$, i.e. $A=a$, which finally shows $BaB=a$ with $B$ defined in $(1)$.

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

We start by recalling the fact that $a^\dagger$ has no eigenvectors. In particular, for every eigenvector $|b\rangle$ of $B$ it holds that $a^\dagger |b\rangle \neq 0$ and thus, by assumption:

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

This shows that $b\neq 0$ and $Ba^\dagger |b\rangle\neq 0$, and thus 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

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

which, together with the previous considerations show that $aB|0\rangle=0$ (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 $b$.

Finally, consider the state $|1\rangle=a^\dagger |0\rangle$. From $(1)$ 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 $$ B = b P_1 +b^{-1} P_2\tag 6 \quad ,$$

concluding the proof.

The answer is no. Below I give a proof for the $n=1$ case that the only operators obeying these constraints are of the form $$B=b P_1+ b^{-1} P_2 \quad , \tag 1$$ 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 $b\in \mathbb R\setminus\{0\}$.

It is easy to see that $B$ is hermitian and that $B=\pm I$ if and only if $b=\pm1$. Also, $B$ is invertible, and $B=B^{-1}$ if and only if $b=\pm 1$.

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

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

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 $BaB=a$ for $B$ in the form of $(1)$ for any $b\in \mathbb R\setminus\{0\}$.

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

We start by recalling the fact that $a^\dagger$ has no eigenvectors. In particular, for every eigenvector $|b\rangle$ of $B$ it holds that $a^\dagger |b\rangle \neq 0$ and thus, by assumption:

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

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

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

which, together with the previous considerations show that $aB|0\rangle=0$ (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 $b$.

Finally, consider the state $|1\rangle=a^\dagger |0\rangle$. From $(1)$ 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 $$ B = b P_1 +b^{-1} P_2\tag 6 \quad ,$$

concluding the proof.