# Show that there are unitaries $U_m$ such that $M_m=U_m \sqrt{E_m}$, for any measurement $M_m$ and associated POVM $E_m$

Nielsen and Chuang's QCQI, section 2.2.6, page 92, asks

Suppose a measurement is described by measurement operators $$M_m$$. Show that there exist unitary operators $$U_m$$ such that $$M_m=U_m\sqrt{E_m}$$, where $$E_m$$ is the POVM associated to the measurement.

I can argue:

$$\langle \Phi |M_m^\dagger M_m|\Psi\rangle=\langle\Phi|E_m|\Psi\rangle\quad\text{for all}\quad|\Phi\rangle, |\Psi\rangle\tag1$$

$$\langle\Phi|M_m^\dagger M_M|\Psi\rangle=\langle\Phi|\sqrt{E_m}\sqrt{E_m}|\Psi\rangle.$$

This can only be true if there is a unitary transformation $$U_m$$, such that

$$M_m|\Psi\rangle=U_m\sqrt{E_m}\quad\text{for all}\quad|\Psi\rangle$$

$$M_m=U_m\sqrt{E_m}$$

Why is $$(1)$$ true?

By definition, measurement operators $$M_m$$ are operators such that the probability of outcome $$m$$ is

$$p(m) = \langle \gamma |M_m^\dagger M_m|\gamma\rangle\tag1$$

for any $$|\gamma\rangle$$, see $$(2.92)$$ on page $$85$$. On the other hand, by definition of POVM, we have

$$p(m)=\langle\gamma|E_m|\gamma\rangle\tag2$$

for any $$|\gamma\rangle$$, see discussion following $$(2.117)$$ on page $$90$$. Thus,

$$\langle \gamma |M_m^\dagger M_m|\gamma\rangle = \langle\gamma|E_m|\gamma\rangle.\tag3$$

Now, take $$|\gamma\rangle=(|\Phi\rangle+|\Psi\rangle)/\sqrt2$$. After expanding and eliminating equal terms from both sides of $$(3)$$, we get

$$\langle \Phi |M_m^\dagger M_m|\Psi\rangle = \langle\Phi|E_m|\Psi\rangle\tag4$$

which is the desired equality.

The rest of the argument also remains incomplete. In particular, it is not clear on what basis the claim is made that "this can only be true if there is a unitary transformation...". There could be other ways for the equality to hold without the unitary existing which we are simply failing to imagine. In general, to assert the existence of an object we need construct it or we need to refer to a theorem or axiom that asserts its existence for us. Hint: Note that $$M_m=U_m\sqrt{E_m}$$ looks like the polar decomposition.

Although (1) is true because of Stinespring's dilation theorem. The fact that for any operator $$M$$ there is a unitary $$U$$ and a positive semi-definite operator $$P$$ such that $$M=UP$$ follows from the polar decomposition for arbitrary operators on a Hilbert space.