Assume one has a Hilbert space $\mathcal{H}_K$ of dimension $K=mn$. Associated to it, there exists a set of all possible quantum transformations $\mathcal{E}$. Of those, the most important are, of course, the unitary transformations $\mathcal{U}$ responsible for all the possible ways of evolution, and the set of measurements $\mathcal{M}$.

We also state that $\widetilde{\mathcal{H}}_K$ is a tensor product of two Hilbert spaces $\mathcal{H}_m$ and $\mathcal{H}_n$.

When learning about Hilbert spaces obtained by means of the tensor product, like $\widetilde{\mathcal{H}}_K$, we are always taught about the fundamental difference between the states which are represented by elementary tensors, ${|u\rangle_1\otimes|v\rangle_2}$, and the entangled ones, having more than one term in the Schmidt decomposition. But what is the reflection of this difference in terms of vectors from $\mathcal{H}_K$? In other words, how do we describe the entanglement in terms of a single system $\mathcal{H}_K$?

Do all operations from $\widetilde{\mathcal{E}}$ have their counterparts in $\mathcal{E}$, and vice versa? How to describe measurements within $\mathcal{H}_m$ (but not $\mathcal{H}_n$) in the formalism of $\mathcal{H}_K$? What will be the state of the system after such a measurement in the $\mathcal{H}_K$ language?

Numerous questions arise...


2 Answers 2


Two finite-dimensional Hilbert spaces of the same dimension are isomorphic, and that's it. Sure anything in one has a counterpart in the other. What you are facing here is that some object in $\tilde{\mathcal{H}}_K$ are just more special than their counterparts in $\mathcal{H}_K$ would be.

Take, for example, the state space of two qubits versus a generic 4-dimensional space (so $m = n = 2$), and their bases $\{|0\rangle, |1\rangle\}^{\otimes 2}$ versus $\{|a\rangle,|b\rangle,|c\rangle,|d\rangle\}$. You wouldn't consider $b+d$ any more special than $b+c$, for example, while the difference between $|01\rangle + |11\rangle$ and $|01\rangle + |10\rangle$ is that one we call separated and the other entangled. This semantic difference is lost when we erase the difference between the two spaces. The same goes for quantum operations.

Perhaps the most illustrative example is a change of basis. In the tensor product space, you would likely primarily consider new bases made by tensor multiplying two bases of $\mathbb{C}^2$, to maintain the separation. But you might also just as well use the basis of Bell states, $$\{(|00\rangle + |11\rangle)/\sqrt2, (|00\rangle - |11\rangle)/\sqrt2, (|01\rangle + |10\rangle)/\sqrt2, (|01\rangle - |10\rangle)/\sqrt2\}.$$ Some things are clearer in it, but not e.g. telling which states were separable and which not – that suddenly requires more calculation, basically reducing to transforming back and then doing what you would have done in the original basis. This is because the $\tilde{\mathcal{H}}_4$ was treated like $\mathcal{H}_4$. No information has been lost, it's only a matter of your choice of mathematical description of the same thing.

  • $\begingroup$ I see. What would you say about the measurements? For example, what are the counterparts of measuring the second bit in (b+d) and (b+c) states? (in one case it tells us nothing about the first bit, in another case it tells everything) $\endgroup$
    – mavzolej
    Feb 16, 2018 at 14:16
  • $\begingroup$ Measurements of the first qubit of a two-qubit system are given by (a projector) ⊗ (identity), in the tensor decomposition. These operator are used on the "whole" state space to determine the probabilities and post-measurement states as usual, and give the results you write. But they are just special cases of 4×4 matrices (projectors, Hermitian matrices, measurement operators – depending on the preferred description of quantum measurement) and could be substituted by more general forms. And just like the states they can also easily lose their tensor product form in result of a basis change. $\endgroup$
    – The Vee
    Feb 16, 2018 at 15:37

Assume one has a Hilbert space of dimension ${K= mn}$. It is possible to describe it in (at least) two ways — simply as a $K$-dimensional space, or as a product of $m$- and $n$-dimensional spaces. In the former case we shall call it $\mathcal{H}$, in the latter — $\widetilde{\mathcal{H}}$. All the states in $\mathcal{H}$ and ${\widetilde{\mathcal{H}}=\widetilde{\mathcal{H}}_1\otimes \widetilde{\mathcal{H}}_2}$, as well as the quantum operations, should be in one-to-one correspondence with each other. So, mathematically these two descriptions are equivalent. Let us now explore whether anything is changing from the physical point of view.

For example, in the case of $\widetilde{\mathcal{H}}$ we can measure only a `part' of the state corresponding to one of the Hilbert spaces $\widetilde{\mathcal{H}}_1$ and $\widetilde{\mathcal{H}}_2$. This will cause the partial collapse of the wave function. By performing measurements in $\widetilde{\mathcal{H}}_1$, one gets (here and in what follows for simplicity we assume ${\dim \widetilde{\mathcal{H}}_1 = \dim \widetilde{\mathcal{H}}_2 = 2}$, and also that the measurements in $\widetilde{\mathcal{H}}_{1,2}$ are performed in the computational basis):

\begin{alignat}{9} \label{pr1} \dfrac{1}{2}(|0\rangle+|1\rangle) \otimes (|0\rangle+|1\rangle) &\longrightarrow \left[\begin{alignedat}{9} &|0\rangle\otimes \dfrac{1}{\sqrt{2}} (|0\rangle+|1\rangle)\\ &|1\rangle\otimes \dfrac{1}{\sqrt{2}} (|0\rangle+|1\rangle) \end{alignedat}\right.\quad&&,\qquad\qquad(1)\\ \label{pr2} \dfrac{1}{\sqrt{2}} (|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle) &\longrightarrow \left[\begin{alignedat}{9} &|0\rangle\otimes |0\rangle \\ &|1\rangle\otimes |1\rangle \end{alignedat}\right.\quad&&.\qquad\qquad(2) \end{alignat} In the former case, the measurement in $\widetilde{\mathcal{H}}_1$ does not affect the part of the state belonging to $\widetilde{\mathcal{H}}_2$. In the latter case the situation is opposite — due to entanglement, the outcome of the $\widetilde{\mathcal{H}}_1$ measurement ensues the collapse of $\widetilde{\mathcal{H}}_2$. It natural to ask what are the reflections of these effects in the case of $\mathcal{H}$.

Denoting the four states of $\mathcal{H}$ with letters, we assign the states as: \begin{alignat}{99} |00\rangle & = |a\rangle \quad&&,\qquad |01\rangle & = |b\rangle \quad&&,\qquad |10\rangle & = |c\rangle \quad&&,\qquad |11\rangle & = |d\rangle \quad&&.\qquad\qquad(3) \end{alignat}

We now want to recast (1) and (2) in terms of $\mathcal{H}$: \begin{alignat}{9} \label{pr3} \dfrac{1}{2}(|a\rangle+|b\rangle+|c\rangle+|d\rangle) &\longrightarrow \left[\begin{alignedat}{9} &\dfrac{1}{\sqrt{2}} (|a\rangle+|b\rangle) \\ &\dfrac{1}{\sqrt{2}} (|c\rangle+|d\rangle) \end{alignedat}\right.\quad&&,\qquad\qquad(4)\\ \label{pr4} \dfrac{1}{\sqrt{2}} (|a\rangle+|d\rangle) &\longrightarrow \left[\begin{alignedat}{9} &|a\rangle \\ &|d\rangle \end{alignedat}\right.\quad&&.\qquad\qquad(5) \end{alignat}

It's tempting to call the operations above `projecting onto the states on the RHS of (1)'. To convince ourselves, let us take a look at the quantum channels corresponding to these measurements. The $\widetilde{\mathcal{H}}_1$ measurement operators of $\widetilde{\mathcal{H}}$ are: \begin{equation} \widetilde{M}_1 = \dfrac{1}{\sqrt{2}} (|0\rangle\langle0|) \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}} \qquad,\quad \widetilde{M}_2 = \dfrac{1}{\sqrt{2}} (|1\rangle\langle1|) \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}} \quad.\qquad\qquad(6) \end{equation} They represent a complete set of measurement operators in $\widetilde{\mathcal{H}}_1$, and, of course, an incomplete set in $\widetilde{\mathcal{H}}$. Rewriting those in the basis of $\mathcal{H}$ renders: \begin{equation}\begin{alignedat}{9} M_1 &= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\\end{pmatrix} = |a\rangle\langle a| + |b\rangle\langle b| \quad&&,\qquad\qquad(7)\\ M_2 &= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\\end{pmatrix} = |c\rangle\langle c| + |d\rangle\langle d| \quad&&.\qquad\qquad(8) \end{alignedat}\end{equation} As expected, these are the projectors onto $\operatorname{span}\{|a\rangle,|b\rangle\}$ and $\operatorname{span}\{|c\rangle,|d\rangle\}$.

Now, speaking of physics. What would one say about the state ${\dfrac{1}{2}(|0\rangle+|1\rangle) \otimes (|0\rangle+|1\rangle)}$ when measuring in the computational basis? — 'Not sure about either of two qubits'. And about the post-measurement state ${|0\rangle\otimes \dfrac{1}{\sqrt{2}} (|0\rangle+|1\rangle)}$? — 'Not sure about the second qubit but the first one is definitely $|0\rangle$'. Clearly, (4) permits for the similar interpretation.

The celebrated feature of the maximally entangled Bell state is the observer's ability to completely identify the post-measurement state of a composite system by performing a measurement on its part. By looking at (5), we see how this mechanism works in the language of $\mathcal{H}$: the identification of a general post-measurement state after projecting the input on a two-dimensional subspace is, of course, impossible. However, for certain special states such a `soft' measurement may give the full information about the post-measurement state.


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