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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 $VV^\dagger = \Pi_{\mathcal{H}}$?

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  • $\begingroup$ "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? $\endgroup$ May 7, 2020 at 18:27
  • $\begingroup$ @NorbertSchuch, thanks for the feedback. I have edited and hopefully the question is clearer now. $\endgroup$ May 7, 2020 at 18:56
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    $\begingroup$ 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. $\endgroup$ May 7, 2020 at 19:23

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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.

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  • $\begingroup$ This is not what Wikipedia says! (There, essentially H=K). $\endgroup$ May 7, 2020 at 18:25
  • $\begingroup$ Then Wikipedia needs fixing $\endgroup$
    – Phoenix87
    May 7, 2020 at 18:26
  • $\begingroup$ 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.) $\endgroup$ May 7, 2020 at 18:28
  • $\begingroup$ @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? $\endgroup$ May 7, 2020 at 18:32
  • $\begingroup$ Edited the comment above (already a few minutes ago). Isometry = (isometric) embedding - of the full $H$. $\endgroup$ May 7, 2020 at 18:33
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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 $|u_k'\rangle$: $$V^\dagger = \sum_{k=1}^d |u_k\rangle\!\langle u_k'| \in \operatorname{Lin}(\mathcal H',\mathcal H).$$

Now to address your questions:

  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 $\mathcal K$).

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  • $\begingroup$ 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. $\endgroup$ May 10, 2020 at 21:02
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    $\begingroup$ @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. $\endgroup$
    – glS
    May 11, 2020 at 7:18

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