I'm self-studying Gerry and Knight. To prove Dirac's phase operator is non-existent, the book makes the following argument.
The conventions used are as follows: $\hat{n}$ is the number operator, $\hat{a}$ is the annihilation operator, and $\hat{a}^\dagger$ is the creation operator.
Dirac assumes a Hermitian phase operator $\hat{\phi}$ exists which satisfies the following relation.
$$ \hat{a} = e^{i\hat{\phi}} \sqrt{\hat{n}} $$
$$ \hat{a}^\dagger = \sqrt{\hat{n}} e^{-i\hat{\phi}} $$
The book claims this operator does not exists, for if it did the operator $e^{i\hat{\phi}}$ would have been unitary. This is while:
$$ \left(e^{i\hat{\phi}}\right)\left(e^{i\hat{\phi}}\right)^\dagger = \hat{a}(\hat{n})^{-1/2}(\hat{n})^{-1/2}\hat{a}^\dagger = \hat{a} \hat{n}^{-1}\hat{a}^\dagger \neq I $$
This contradiction proves that the operator $e^{i\hat{\phi}}$ is not unitary; and so the operator $\hat{\phi}$ does not exist.
I fail to understand the last step in the above equation; how is $\hat{a} \hat{n}^{-1} \hat{a}^{\dagger}$ is not the same as the identity operator. Note that for $n \in \mathbb{N} \cup \{0\}$, $|n\rangle$ is defined to be the eigenvector of $\hat{n}$ with an eigenvalue of $n$.
$$ \text{LHS} = \hat{a} \hat{n}^{-1} \hat{a}^{\dagger} |\psi \rangle = \hat{a} \hat{n}^{-1} \hat{a}^\dagger \left(\sum_{n=0}^{\infty} \langle n | \psi \rangle | n \rangle \right) $$
Defining $C_n \equiv \langle n | \psi \rangle$:
$$ \therefore \text{LHS} = \hat{a} \hat{n}^{-1} \left(\sum_{n=0}^{\infty} C_n \sqrt{n+1} |n+1\rangle \right) $$
$$ \therefore \text{LHS} = \hat{a} \left(\sum_{n=0}^{\infty} C_n \dfrac{|n+1\rangle}{\sqrt{n+1}} \right) $$
$$ \therefore \text{LHS} = \sum_{n=0}^{\infty} C_n |n\rangle $$
$$ \therefore \hat{a}\hat{n}^{-1}\hat{a}^\dagger |\psi\rangle = |\psi\rangle $$
So $\hat{a}\hat{n}^{-1}\hat{a}^\dagger = I$. Is the textbook wrong in making this argument?
The only flaw I could think of with the above argument is that $\hat{a}\hat{n}^{-1}\hat{a}^\dagger$ could be undefined for certain vectors in the Hilbert space, in which case it would not be the same as the identity operator. If this is the case, how can we prove this operator is not defined everywhere?
Sorry if this question is trivial.
Best Answer
Yeah, that's a strange thing to say - indeed $an^{-1}a^\dagger$ is defined everywhere and equal to the identity (note that $n^{-1}$ itself is not defined everywhere - but it is defined on the image of $a^\dagger$).
But unitarity has to work both ways - you wrote down $PP^\dagger = I$ for $P = n^{-1/2}a$ (I don't want to write it as an exponential because the exponent doesn't exist), but unitarity is $PP^\dagger = P^\dagger P = I$. The problem here is the value of $P$ at $\lvert 0\rangle$, which is not fixed by the "definition" $a = P \sqrt{n}$ - since $\sqrt{n}\lvert 0\rangle = 0$, this leaves the value of $P\lvert 0\rangle$ arbitary.
Let us write $p_i = \langle i\vert P\vert 0\rangle$ for whatever we might choose. Then we have $$ \langle 0 \vert P^\dagger P \vert0\rangle = \sum_i \langle 0 \vert P^\dagger \vert i\rangle\langle i\vert P \vert0\rangle = \sum_i p^\ast_i p_i $$ but also $$ \langle k \vert P^\dagger P \vert 0\rangle = \sum_i \langle k \vert P^\dagger\vert i\rangle\langle i\vert P \vert 0\rangle = \sum_i \delta_{k,i+1} p_i = p_{k-1}$$
For $P^\dagger P = I$, we would have to therefore have $p_{k-1} = 0$ for all $k$ since the off-diagonals of the identity should vanish, but also $\sum_i \lvert p_i\rvert^2 = 1$. This is a contradiction, so $P^\dagger P \neq I$ for all possible extensions of $P$.
This argument this the same as that from "Quantum mechanical phase and time operator" by Susskind and Glogower, after whom this operator $P$ is sometimes named Susskind-Glogower operator.