Given a Hilbert space of finite dimension $N$ with an orthonormal basis $\mathcal{B}=\{|0\rangle,\ldots,|N-1\rangle \}$ what is the most general unitary operation that commutes with the projector onto one of the basis elements (say $|0\rangle$), \emph{i.e.}, what is the most general $\mathcal{U}$ such that $\mathcal{U}^\dagger \mathcal{U}=\mathcal{U} \mathcal{U}^\dagger=\mathbb{I}$ and $\left[\mathcal{U},|0\rangle \langle 0| \right]\equiv \mathcal{U} |0\rangle \langle 0|- |0\rangle \langle 0|\mathcal{U}=0$. A partial answer to the question is: \begin{eqnarray*} \mathcal{U(\lambda,\theta,\gamma)}&=&\exp\left(i G(\lambda,\theta,\gamma )\right)\\ G(\lambda,\theta,\gamma)&=&\sum_{k=1}^{N-1}\lambda_k |k \rangle \langle k|\\ &&+\sum_{k=1}^{N-1}\sum_{l>k}^{N-1} \theta_{l,k} \left(|k \rangle \langle l|+|l \rangle \langle k|\right)\\ &&+i\sum_{k=1}^{N-1}\sum_{l>k}^{N-1} \gamma_{l,k} \left(|k \rangle \langle l|-|l \rangle \langle k|\right) \end{eqnarray*} For $N=2$ it is indeed the answer to the question, but I am not sure for higher dimensions.

  • $\begingroup$ Are you sure the sums in the definition of $G$ start at $1$ and not at $0$? $\endgroup$
    – Lagerbaer
    May 27 '13 at 14:59
  • $\begingroup$ This question appears to be off-topic because it is about math with absolutely no apparent relevance to physics. While relevance to physics can be inferred, it is not given in the question. My understanding of the site rules indicate that this means the question is off-topic. $\endgroup$
    – DanielSank
    Dec 31 '14 at 6:03

As Lagerbaer pointed out, working with finite dimension can be done in matrix representation. The projector on $|0\rangle$ has only the top left element nonzero, $\hat{\Pi}_0 = \mathrm{diag}(1,0,0,\ldots)$ and we want a unitary $\hat{U}$ that commutes with the projector, $[\hat{U},\hat{\Pi}_0] = 0$.

Let us start with dimension $N = 2$. We find that both off-diagonal elements must be zero (as Lagerbaer argues) but the other diagonal element can be arbitrary, as it is always multiplied by zero. We therefore have $\hat{U} = \mathrm{diag}(a,b)$, where $a, b$ are chosen so that $\hat{U}$ is unitary.

Similar analysis can be done for higher dimensions. The argument of Lagerbaer can be used and you can find out that the first row and first column must be zero (except for the diagonal term). But it says nothing about the rest of the matrix! Once again, we have $\hat{U} = \mathrm{diag}(a,b)$ but this time $b$ is a $N-1$-dimensional matrix and again, $a,b$ have to fulfil unitarity condition.

In conclusion, the unitary factors into two blocks, one for the state at which the projector projects and other for the remaining $N-1$ states. In order to satisfy unitarity, $a = e^{i\phi}$ and $b$ must be unitary.


From geometric viewpoint:

The assertion that $[U,P]=0$, where $P$ is an orthogonal one-dimensional projector is equivalent to $$UP=PU$$. Suppose that $P$ projects on the vector $x$ ($\mathrm{Im}P=\mathrm{Span}\{x\}$). Lets act with the above equation upon $x$. We get: $$ UPx=PUx,\\ y=Py, $$ where $y=Ux$. Now, we see that $y$ lies in the image of $P$ which is spanned by $x$. Thus we have $$ Ux=y=\alpha x. $$ As far as $U$ preserves norms, we deduce that $|\alpha|=1$ thus $\alpha=e^{i\phi}$.

So far we have proven that the image of $P$ is an eigenspace for $U$ : $Ux=e^{i\phi}x$. This is a necessary condition. In fact, this works in an arbitrary vector space (if one allows zero eigenvalues), and $U$ being an isometry is necessary only for the statement $|\alpha|=1$.

We now prove that it is a sufficient condition in the case of a Hilbert space $\mathcal{H}$. The Hilbert space in considiration spilts into an orthogonal direct sum: $$ \mathcal{H}=\mathrm{Im}P\oplus_\perp\mathcal{H_\perp}. $$ It suffices to show that $U\mathcal{H}_\perp\subseteq\mathcal{H}_\perp$. This is obvious since $\mathcal{H}_\perp$ is charactirized as all vectors orthogonal to $\mathrm{Im}P$, while $U\mathrm{Im}P=\mathrm{Im}P$ and $U$ preserves scalar products. Therefore, for arbitrary $y\in\mathcal{H}$ we have $$ y=Py+y_\perp,\\ UPy=e^{i\phi}Py,\\ PUy=e^{i\phi}Py+PUy_\perp=e^{i\phi}Py=UPy, $$ as far as $Uy_\perp\in\mathcal{H}_\perp$ by the above.

So, we have proven the following statements

  • If an operator $U$ acting on a vector space commutes with a one-dimensional projector $P$, it follows that either $\mathrm{Im}P\subseteq\mathrm{Ker}U$ or $\mathrm{Im}P$ is an eigenspace for $U$. If the space is normed and $U$ is an isometry, the the eigenvalue has unit norm.
  • A unitary operator $U$ acting on a Hilbert space commutes with an one-dimensional orthogonal projector $P$ iff the image $\mathrm{Im}P$ is a one-dimensional eigenspace of $U$.

Note that there is no need to require the space to be finite-dimensional.


Since we're talking finite Hilbert spaces, think in matrices. In basis representation, $|0\rangle \langle 0 |$ is a matrix with only zeros, and a $1$ in the upper left corner, if we (without loss of generality) assume that $|0\rangle$ is the "first" vector of the basis.

If you multiply that with the matrix representation of $U$ from the left, you get a matrix that has the first column of $U$ as its first column, and the rest is zero.

If you multiply it with the matrix representation of $U$ from the right, you get a matrix that has the first row of $U$ as its first row, and the rest is zero.

Now these matrices have to be equal, and that means that in the first row and first column only the upper left entry of can be nonzero! The rest of the matrix can be whatever it wants to be as long as it is unitary.

More generally, to commute with the projector onto the $i$-th basis state, the matrix representation of $U$ must have zero non-diagonal elements in the $i$-th row and column.

From physical intuition, that makes sense: To commute with the projector onto a single state $|i\rangle$, it basically means that $U$ should leave the subspace spanned by $|i\rangle$ invariant.

  • 2
    $\begingroup$ Your analysis is correct, but there is an error with the conclusion, the condition is that non-diagonal elements of the first row or coloumn must be zero $\endgroup$
    – Ikiperu
    May 27 '13 at 15:28
  • $\begingroup$ "Non-diagonal elements of the first row or column must be zero" -> Technically, that's what I'm saying, because without loss of generality we can say that $|0\rangle$ is the "first" element of our finite vector space. But I will adjust the answer to be more general and less reliant on the precise matrix structure. $\endgroup$
    – Lagerbaer
    May 27 '13 at 17:20
  • $\begingroup$ @Lagerbaer, I think that "Now these matrices have to be equal, and that means that only the upper left entry of U can be nonzero!" arises the questions. I understand what you mean, but probably you will edit the answer so that more than one element of $U$ is allowed to be non-zero?) $\endgroup$ May 27 '13 at 17:55

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