# Isometric equivalence of purifications of quantum states

Following the notes here (Quantum Information Theory Tips 5 at ETH), we state the following result. For any quantum state $$\rho_A$$ and purifications $$\vert\psi\rangle_{AB}$$ and $$\vert\phi\rangle_{AC}$$, there exists an isometry $$V_{B\rightarrow C}$$ such that $$(I_A\otimes V_{B\rightarrow C})\vert\psi\rangle_{AB} = \vert\phi\rangle_{AC}$$. Consider now $$\rho_{A} = \frac{\mathbb{1}_A}{2}$$, the maximally mixed state, and the following purifications.

$$|\psi\rangle_{A B}=\frac{1}{\sqrt{2}}\left(|0\rangle_{A}|+\rangle_{B}+|1\rangle_{A}|-\rangle_{B}\right) \quad \text{and} \quad|\phi\rangle_{A C}=\frac{1}{\sqrt{2}}\left(|0\rangle_{A}|000\rangle_{C}+|1\rangle_{A}|110\rangle_{C}\right)$$

Is it true that there is an isometry $$V'_{C\rightarrow B}$$ such that $$(I_A\otimes V'_{C\rightarrow B})\vert\phi\rangle_{AC} = \vert\psi\rangle_{AB}$$? Note that here $$\text{dim}(\mathcal{H}_C) > \text{dim}(\mathcal{H}_B)$$. If yes, how is this consistent with the following definition of isometries which state that they go from a smaller Hilbert space to a larger Hilbert space only?

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 $$|\psi\rangle \in \mathcal{H}$$.

This is related to my previous question here but I am still not sure about this dimensional problem.

• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. Commented May 12, 2020 at 14:08
• I'm pretty sure this is just a problem of confusing language. The link states "you have to prove that any two purifications are equivalent up to an isometry on the purifying system", that doesn't always mean there exists an isometry C->B, it might be that there only exists an isometry B->C. Indeed there are isometries both ways if and only if the dimensions of the two purifications are the same. Commented May 12, 2020 at 14:08
• @or1426 I see - so given the purification $\vert\phi\rangle_{AC}$, there is no isometry to obtain $\vert\psi\rangle_{AB}$? Is it then the case that one needs a projection onto a subspace of $C$ (of dimension 2) followed by an isometry from this subspace to $B$ in order to go from $\vert\phi\rangle_{AC}$ to $\vert\psi\rangle_{AB}$? Commented May 12, 2020 at 14:19
• Not only there is no isometry to obtain the state, there is no isometry from $C$ to $B$ at all! An isometry is supposed to map an orthogonal basis to an orthogonal basis, and there just aren't enough orthogonal vectors in $B$ to fit an orthogonal basis mapped from the vectors of $C$. You're going to need to project at some point. There is a statement about partial isometries I think Commented May 12, 2020 at 14:27
• There is no useful way to go from the larger to smaller Hilbert space, I mentioned in my answer that generally we use the partial trace to remove degrees of freedom we don't need but it isn't at all useful here. Just think of isometries as "one way streets". Commented May 12, 2020 at 14:29

The essence is the following: You can write any purification in Schmidt form (note that this is not a transformation, just rewriting the state in a different basis). Then, any two purifications of a given state will be of the form $$|\psi\rangle = \sum \lambda_i |a_i\rangle \otimes |b_i\rangle \in \mathcal H_A\otimes \mathcal H_B$$ and $$|\phi\rangle = \sum \lambda_i |a_i\rangle \otimes |c_i\rangle \in \mathcal H_A\otimes \mathcal H_C\ .$$ To relate the two purifications, you must construct a transformation which maps the orthogonal set of vectors $$\{|b_i\rangle\}$$ to the orthogonal set of vectors $$\{|c_i\rangle\}$$.

Restricted to the span of those vectors, this is a (unique!) unitary transformation. If either $$\mathcal H_A$$ or $$\mathcal H_B$$ is bigger than the span, you can pad this transformation such that it still has orthogonal rows or columns (depending which dimension is bigger), such that one of them is an isometry -- the one from the smaller to the bigger space -- and the converse transformation correspondingly a partial isometry, or the dagger of an isometry.

An isometry is a map such that

$$\langle Vx,Vy\rangle=\langle x,y\rangle$$

if the image of $$V$$ has smaller dimension than its domain, then clearly this property cannot hold, as if we have an orthonormal basis

$$\langle x_i,x_j\rangle=\delta_{ij}$$

we cannot have

$$\langle Vx_i,Vx_j\rangle=\delta_{ij}\tag{*}$$

because there aren't enough orthogonal vectors in the image of $$V$$. Instead you can have a partial isometry, i.e. a map $$V$$ such that $$(*)$$ holds for a subset $$\{x_j\}_{j=1}^{d_V}$$ where $$d_V$$ is the dimension of the image of $$V$$, and that sends the other vectors to $$0$$. In practice this means projecting your initial space onto a subspace of the same dimension as the image of $$V$$ and then applying an isometry. More precisely, a partial isometry is a map that is an isometry on the orthogonal complement of its kernel.

what ort1426 says is correct and enough in my opinion, this already shows isometric equivalence, but a more complete statement could be

Let $$|\psi\rangle_{AB}$$ and $$|\psi'\rangle_{AC}$$ be two purifications of $$\rho_A$$. Then there exists a partial isometry $$V_{B\to C}$$ such that $$V|\psi\rangle=|\psi'\rangle$$

You already know how to prove case where $$\mathrm{dim}(B)\leq \mathrm{dim}(C)$$, then $$V$$ is an isometry or a unitary (which are a special case of partial isometry, despite the names), if $$\mathrm{dim}(B)> \mathrm{dim}(C)$$, consider a Schmidt decomposition of $$|\psi\rangle$$ and $$|\psi'\rangle$$

$$|\psi\rangle_{AB}=\sum_{k=1}^{r} s_k |\alpha_k\rangle|\beta_k\rangle\\|\psi'\rangle_{AC}=\sum_{k=1}^{r} s_k |\alpha_k\rangle|\beta_k'\rangle$$

the $$\alpha_k$$ are equal because the states must both partial trace to $$\rho_A$$. We clearly have $$r<\mathrm{dim}(C)$$. Extend the $$|\beta_k\rangle$$ to a basis of $$B$$ arbitrarily and define

$$V_{B\to C}|\beta_k\rangle=\begin{cases} |\beta_k'\rangle \quad &\textrm{if } k\leq r\\ 0 \quad &\textrm{otherwise} \end{cases}$$

$$V$$ is a partial isometry and has the desired property, basically, you didn't need such a big Hilbert space to begin with, as the rank of the Schmidt decomposition is smaller than the dimension of your auxiliary space anyway, and $$V$$ throws away by projection the useless dimensions.

• Could you also comment on whether the correct way to throw away the auxiliary space is by projection or partial tracing? Or are both operations equivalent? Commented May 12, 2020 at 16:54
• @user1936752 you want to relate purifications, and partial tracing isn't guaranteed to yield a pure state. Going from $B$ to and enlarged version of $C$ and then taking a partial trace would be equivalent to applying (the Stinespring representation of) some quantum channel. There is always a quantum channel such that $\mathcal{E}(|\psi\rangle)_{AB}= |\phi\rangle_{AC}$ (you can just define one that does this), so I don't think you can get anything useful from this. Commented May 12, 2020 at 17:00

As far as I can tell the link does not state that

For any quantum state $$\rho_A$$ and purifications $$\vert\psi\rangle_{AB}$$ and $$\vert\phi\rangle_{AC}$$, there exists an isometry $$V_{B\rightarrow C}$$ such that $$(I_A\otimes V_{B\rightarrow C})\vert\psi\rangle_{AB} = \vert\phi\rangle_{AC}$$.

Which is just as well since that claim is incorrect, as your example proves!

It states

any two purifications are equivalent up to an isometry on the purifying system

which is a much more reasonable claim. In particular for the two states to be "equivalent up to an isometry on the purifying system" all that is required is that either there exists an isometry $$V:B\to C$$ or an isometry $$V:C\to B$$. It is not necessary that there are isometries both ways. As I mentioned in my comment there is an isometry both ways if and only if the two isometries are unitaries and both the spaces have the same dimension.

Note that the adjoint of an isometry is not an isometry, let alone an inverse of the isometry you started with. In general an isometry consists of an extension of your Hilbert space (i.e. adding some extra dimensions) followed by doing a unitary. The natural "inverse" operation to adding extra dimensions is the partial trace, but this is certainly not isometric (or useful here).

• Thank you - this was very helpful! Commented May 12, 2020 at 14:45
• How about disentangling the extra dimension, rather than tracing it? Commented May 12, 2020 at 17:37
• @NorbertSchuch You can always disentangle it but the extra space it still there - your state is now $|\xi\rangle | 0 \rangle$ rather than $|\xi\rangle$. You have to do something to get rid of the $|0\rangle$, and then either allow your "something" to be a partial function, or extend it to do something on entangled states. Commented May 13, 2020 at 9:42
• Note that the way the isometric equivalence works is via embedding, not attaching an ancilla (that is, a direct sum, not a tensor product). Commented May 13, 2020 at 11:37