# Connection between isometries and projectors in QM

I realize this question is technically a mathematical one but I think it is seen often enough in quantum information so I ask it here. The following is the definition of an isometry in Mark Wilde's book

Let $$\mathcal{H}$$ and $$\mathcal{H}^{\prime}$$ be Hilbert spaces such that $$\operatorname{dim}(\mathcal{H}) \leq$$ $$\operatorname{dim}\left(\mathcal{H}^{\prime}\right)$$ An isometry $$V$$ is a linear map from $$\mathcal{H}$$ to $$\mathcal{H}^{\prime}$$ such that $$V^{\dagger} V=I_{\mathcal{H}}$$. Equivalently, an isometry $$V$$ is a linear, norm-preserving operator, in the sense that $$\||\psi\rangle\left\|_{2}=\right\| V|\psi\rangle \|_{2}$$ for all $$|\psi\rangle \in \mathcal{H}$$.

He also points out that $$V V^{\dagger}=\Pi_{\mathcal{H}^{\prime}}$$ which is a projection onto $$\mathcal{H'}$$.

My questions are about $$V^\dagger$$.

1. By the definition, it is not an isometry but it is a linear map from $$\mathcal{H'}$$ to $$\mathcal{H}$$. Is $$V^\dagger$$ itself a projector from $$\mathcal{H'}$$ to a subspace of $$\mathcal{H'}$$ of dimension $$\text{dim}(\mathcal{H})$$ followed by a unitary from this subspace to $$\mathcal{H}$$?

2. Does every projector have a corresponding isometry? That is, suppose I am given a projector $$\Pi_{\mathcal{H}}$$ onto a subspace of $$\mathcal{H}$$ called $$\mathcal{K}$$. Then does every isometry $$V$$ from $$\mathcal{K}$$ to $$\mathcal{H}$$ satisfy $$VV^\dagger = \Pi_{\mathcal{H}}$$?

• "What is it" is not a well-defined question. Also, what do you mean by "construct an isometry from a projector" -- what condition should it satisfy? Commented May 7, 2020 at 18:27
• @NorbertSchuch, thanks for the feedback. I have edited and hopefully the question is clearer now. Commented May 7, 2020 at 18:56
• I think the best way to think about isometries (as a physicists and in finite dimensions) is as a number of columns taken from a unitary. (So if V are columns of a unitary, then $V^\dagger$ are some rows of a unitary. Commented May 7, 2020 at 19:23

A partial isometry is mapping a sub-vector space $$K$$ of a Hilbert space $$H$$ onto another sub-vector space $$K'$$ of the same dimension isometrically, that is

$$(V\psi, V\phi) = (\psi,\phi)$$

for any two vectors in the initial domain of the isometry, that is $$K=V^*VH$$. The fact that $$V^*VH$$ is the initial domain of $$V$$ can be proved by showing that $$E=V^*V$$ is precisely the projection onto $$K$$. Similarly, one can show that $$F=VV^*$$ is the projection onto $$K'$$, so that $$K' = FH$$. To get an idea of what a partial isometry is, observe that every unitary is a partial isometry, but not every partial isometry is a unitary because $$V^*VH$$ and $$VV^*H$$ are generally not the whole of $$H$$ (although they could be isomorphic to it). Indeed, when $$K$$ is all of $$H$$, one talks about isometries. An important example of an isometry is the adjoint of the shift operator $$S$$ on a separable infinite dimensional Hilbert space $$H$$ with ONB $$\{e_0,e_1,\ldots\}$$,

$$Se_0=0,\qquad Se_k=e_{k-1}.$$

Note how $$S^*$$ maps the whole of $$H$$ onto the orthogonal complement of $$e_0$$ isometrically.

It is also easy to prove the following identities that characterise partial isometries:

$$VV^*V = V\qquad V^*VV^* = V^*.$$

Finally, one sees that projections are (rather trivial) examples of partial isometries.

• This is not what Wikipedia says! (There, essentially H=K). Commented May 7, 2020 at 18:25
• Then Wikipedia needs fixing Commented May 7, 2020 at 18:26
• To me, an isometry $V:H\to H'$ maps a vector space $H$ to a subspace of $H'$ (preserving the scalar product), and not only a subspace of $H$. Differently speaking, it is a number of columns taken from a unitary. (Note that the OP cites the same definition.) Commented May 7, 2020 at 18:28
• @NorbertSchuch in the definition of Wilde, it says the isometry maps $\mathcal{H}$ to a larger or equally large space $\mathcal{H'}$. But you say it maps to a smaller space? Commented May 7, 2020 at 18:32
• Edited the comment above (already a few minutes ago). Isometry = (isometric) embedding - of the full $H$. Commented May 7, 2020 at 18:33

You can characterise isometries as those linear maps that can be written in the form $$V = \sum_{k=1}^d |u_k'\rangle\!\langle u_k| \in \operatorname{Lin}(\mathcal H,\mathcal H'),$$ where $$\{|u_k\rangle\}_k$$ is an orthonormal basis for $$\mathcal H$$, $$\{|u_k'\rangle\}_k$$ is an orthonormal set in $$\mathcal H'$$ (but not a basis if $$\operatorname{dim}(\mathcal H)<\operatorname{dim}(\mathcal H')$$), and $$d\equiv\operatorname{dim}(\mathcal H)$$.

In this notation, $$V^\dagger$$ is obtained by simply switching $$|u_k\rangle$$ and $$|u_k'\rangle$$: $$V^\dagger = \sum_{k=1}^d |u_k\rangle\!\langle u_k'| \in \operatorname{Lin}(\mathcal H',\mathcal H).$$

1. Indeed, $$V^\dagger$$ is not an isometry if $$\operatorname{dim}\mathcal H<\operatorname{dim}\mathcal H'$$. You can write it as $$V^\dagger = \left( \sum_{j=1}^d |u_j\rangle\!\langle u_j'| \right) \left( \sum_{k=1}^d |u_k'\rangle\!\langle u_k'| \right)=V^\dagger \left( \sum_{k=1}^d |u_k'\rangle\!\langle u_k'| \right).$$ This amounts to simply multiplying to the right with a projector onto the support of $$V^\dagger$$, which you can always do freely. This clearly is not a very insightful statement. However, you could think of $$V^\dagger$$ as a unitary operation when restricting its domain to its support. In other words, $$V^\dagger|_{\operatorname{supp}(V^\dagger)}$$ is unitary. That's probably how close you can get to your statement.
2. Let $$W:\mathcal K\to\mathcal H$$ be an isometry, with $$d'\equiv \operatorname{dim}\mathcal K\le d$$. Then we can write it as $$W = \sum_{k=1}^{d'} |v_k\rangle\!\langle v_k'|,$$ with $$|v_k'\rangle$$ orthonormal basis for $$\mathcal K$$ and $$|v_k\rangle$$ orthonormal set in $$\mathcal H$$. Then, $$W W^\dagger = \sum_{k=1}^{d'} |v_k\rangle\!\langle v_k|.$$ This is therefore a projector onto a subset of $$\mathcal H$$ of dimension $$d'$$ (though not necessarily a projector onto $$\mathcal K$$).
• Thank you for answering! I'm sorry if I keep repeating the same question but the aspect that has me confused is that you have also called $V^\dagger$ an isometry. But according to the definition (in my original question), if $\text{dim}(\mathcal{H})<\text{dim}(\mathcal{H'})$, then $V^\dagger$ is not an isometry. Is this correct? According to Wilde's definition, an isometry cannot be a map from a larger to a smaller space. Commented May 10, 2020 at 21:02
• @user1936752 good point, that is correct. An explicit example probably makes it easier to understand. Suppose $d=2$. Then $V= |3\rangle\!\langle 0|+|2\rangle\!\langle 1|$ is an isometry. But $V^\dagger=|0\rangle\!\langle3|+|1\rangle\!\langle 2|$ is not an isometry. E.g. it doesn't preserve the norm of $|1\rangle\in\mathcal H'$. Although I think one could argue that it is an isometry (and unitary) as a map between its support and its range.