I am having trouble understanding Niemark's theorem from books (e.g watrous). The wikipedia page is clearer but most calculations are not justified. I want to contruct the PVM associated with the state discrimination POVM. Can someone explain to me where does the expression for ${\displaystyle {\sqrt {F_{?}}}={\sqrt {\frac {2|\langle \varphi |\psi \rangle |}{1+|\langle \varphi |\psi \rangle |}}}|\gamma \rangle \langle \gamma |}$ come from where ${\displaystyle |\gamma \rangle ={\frac {1}{\sqrt {2(1+|\langle \varphi |\psi \rangle |)}}}(|\psi \rangle +e^{i\arg(\langle \varphi |\psi \rangle )}|\varphi \rangle ).}$? and why $U_{UQSD}$ outputs what it does ?

  • 1
    $\begingroup$ it's help to make the post self-contained by including what $\sqrt F$ and $U_{UQSD}$ are. Also, what exactly do you find unclear in the cited explanations? $\endgroup$
    – glS
    Commented Dec 30, 2023 at 14:11
  • $\begingroup$ In the cited explanations, the details of the calculations are omitted and only the results are shown. What I don't understand is how do they get the formulas above. And If possible I would like to see an explicit application of niemark theorem (by constructing a PVM from an explicit POVM) $\endgroup$
    – yosh
    Commented Jan 1 at 11:10
  • $\begingroup$ To get this expression for $\sqrt{F_?}$ you just diagonalize $F_?$ and take the square root of the eigenvalues. This is the canonical square root of a positive semidefinite operator. $|\gamma\rangle$ is then the eigenvector with nonzero eigenvalue. $U_\text{UQSD}$ is defined as in the third equation in the subsection "Post-measurement state", just replace the appropriate operators. $\endgroup$ Commented Jan 2 at 17:53

1 Answer 1


The tl;dr of how to go from a given POVM to a PVM is the following:

  1. Take your state on which you do the measurement to a larger Hilbert space using a linear isometry $A$.

  2. Do a projective measurement in the ancilla registers of the larger Hilbert space.

For a POVM with elements $\{E_i\}$ where $\sum_i E_i = I$, consider the operator

$$A = \sum_i U\sqrt{E_i}\otimes \vert i\rangle,$$

where $U$ is an arbitrary unitary. Note that you have a second unitary freedom in defining the basis of the ancilla system i.e. you can replace all the states $\vert i\rangle$ with $U'\vert i\rangle$ for some other unitary $U'$ and everything that follows still works.

You can work out that $A^\dagger A = I$ i.e. $A$ is an isometry.

Having taken your quantum state to a larger Hilbert space using $A$, you can now do a projective measurement. The PVM elements $\{\Pi_i\}$ are given by $\Pi_i = I\otimes \vert i\rangle\langle i\vert$, which obviously satisfy the requirements of a PVM. Since $E_i = A^\dagger (I\otimes \vert i\rangle\langle i\vert)A$, we have that the post-measurement probabilities are correct i.e.

$$\mathrm{Tr}(E_i\rho) =\mathrm{Tr}(\Pi_i (A\rho A^\dagger)).$$

The right hand side is manifestly a composition of the isometry followed by a projective measurement.

Following the notation from the Wikipedia page, for state discrimination between two states, you have a POVM with elements $\{F_\psi, F_\phi, F_?\}$.

You first find these three POVM elements (see here for how). Next, take the square roots of those elements and append an ancilla register in order to construct the required isometry $A$.

It is notationally cumbersome to write $A$ out so I will omit the explicit answer you desire but hopefully you understand how to get there.

Once you have $A$, you do the projective measurement on the ancilla registers and this implements the POVM.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.