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.

  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/292633 Especially the last answer mentioning Suskind-Glogower operator. $\endgroup$
    – Meng Cheng
    May 3, 2022 at 16:36
  • 3
    $\begingroup$ Tip: Speaking of 'the last answer' in SE's dynamical Q&A format is ambiguous. $\endgroup$
    – Qmechanic
    May 4, 2022 at 8:18
  • 1
    $\begingroup$ I suggest you look up polar decomposition in mathematical books on operator theory. $\endgroup$
    – Blazej
    May 4, 2022 at 18:03

4 Answers 4


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.

  • 1
    $\begingroup$ Ah this make sense, so $\hat{a} \hat{n}^{-1} \hat{a}^\dagger$ is indeed the same as the identity operator. Doing it the other way around ($P^\dagger P \neq I$) seems like the better approach, especially since $P$ is not even defined over the entire Hilbert space anyway (e.g. $|1\rangle$ is not in its domain). Thanks. $\endgroup$ May 3, 2022 at 16:58

I'd do it the other way round: $$ (e^{i\phi})^\dagger e^{i\phi} $$ should also be the identity, but $$ (e^{i\phi})^\dagger e^{i\phi}= n^{-1/2} a^\dagger a n^{-1/2} $$ is not defined on $|0\rangle$ as the rightmost $n^{-1/2}$ it would give $1/0$.

Basically we want $e^{-i\phi}|n\rangle = |n+1\rangle$ and $e^{i\phi}|n\rangle =|n-1\rangle$. Then stepping up and stepping down gives the identity, but stepping down and then up gives the zero vector when when applied to $|0\rangle$.

  • 1
    $\begingroup$ Yeah this is a more reasonable approach; doing it the other way around should be much harder if not impossible. $\endgroup$ May 3, 2022 at 16:56

We have in fact to check whether for $U:=\exp i\hat \phi$ it holds that $$UU^\dagger = \mathbb I = U^\dagger U \quad . $$

In Phase and Angle Variables in Quantum Mechanics (where in section 5 it is also noted that $UU^\dagger = \mathbb I$, as you've found) it is shown that

$$ U^\dagger U \neq \mathbb I \quad .$$ Indeed, the paper shows that if $U^\dagger U =\mathbb I$, then $$\langle 0|U^\dagger U |0\rangle = 0 \quad , $$ which is a contradiction.

Consequently, $U$ is not unitary.


There is an even shorter way than the one pointed out by @ACuriousMind. Suppose that there is a unitary $V$ such that $$a= Vn^{1/2}\:, \quad a^* = n^{1/2}V^*$$ on the corresponding domains. As a consequence $aa^* = VnV^*$ that is $n + I = VnV^*$ because $aa^*-a^*a=I$ and $n=a^*a$. From $V^*V=I$ and $n + I = VnV^*$, we have $$nV +V = Vn\:.$$ This is impossible because it would imply $$nV|0\rangle = -V|0\rangle$$ but $\sigma(n) \not \ni -1$ so that $V|0\rangle=0$ which is in contradiction with $Ker V = \{0\}$ which arises form the fact that $V$ is unitary.

Actually all that (and the theorem of polar decomposition of closed operators) proves that $V$ exists but it is only a partial isometry with kernel made of the span of $|0\rangle$. In particular it cannot be of the form $V=e^{iA}$ for some selfadjoint operator $A$, because $V$ would be unitary.


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