# Why is a post-measurement state a valid quantum state?

Postulate 3 of the postulates of quantum mechanics in Nielsen & Chuang states that

Quantum measurements are described by a collection $${M_m}$$ of measurement operators... and the state of the system after the measurement is $$\frac{M_m |\psi \rangle}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}}$$

I have two related questions:

1. Why is this a valid quantum state? I am looking for a proof-technique that handles an arbitrary choice of $$M_m$$, but I know generally speaking we want to show that $$\langle \phi | \phi \rangle = 1$$ to show that an arbitrary vector $$|\phi \rangle$$ is a valid quantum state.
2. Additionally, $$M_m$$ is a measurement operator and measurement operators are generally not unitary operators. Can we see the the expression $$\frac{M_m}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}}$$ as a unitary operator acting upon $$|\psi \rangle$$? If so, how can we show that? If not, how should we view the said expression?
• – glS
Commented Apr 25 at 6:55

1. Define $$|\phi\rangle = \frac{M_m |\psi \rangle}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}}$$, then you want to show that $$\langle \phi |\phi\rangle = 1$$. Note that taking the the inner product $$\langle \phi | \phi \rangle$$ is the same as multiplying $$\langle \phi |$$ with $$|\phi\rangle$$, where $$\langle \phi |$$ is just the conjugate transpose $$|\phi\rangle^*$$. So, we have $$\langle \phi | = \left(\frac{M_m |\psi \rangle}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}}\right)^* = \frac{(M_m |\psi\rangle)^*}{\left(\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}\right)^*}$$ $$= \frac{(|\psi\rangle)^* M_m^*}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}} = \frac{\langle\psi|M_m^*}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}},$$ where we've used the fact that the denominator is a real number. Now, if you take the product of $$\langle\phi|$$ and $$|\phi\rangle$$, can you see why $$\langle \phi | \phi \rangle = 1$$?

2. No, the measurement operators are not generally unitary. Instead, they form a complete set of projectors onto the Hilbert space in which our quantum mechanical objects live. For example, qubits are just vectors $$\begin{pmatrix}a \\ b \end{pmatrix} = a|0\rangle + b|1\rangle \in \mathbb{C}^2,$$ where we've defined $$|0\rangle = \begin{pmatrix}1 \\ 0 \end{pmatrix}, \quad |1\rangle = \begin{pmatrix}0 \\ 1 \end{pmatrix}.$$ When we measure a qubit with respect to the basis $$\{|0\rangle, |1\rangle\}$$, we obtain some classical outcome $$i=0$$ or $$1$$ and the quantum state is projected into the subspace of $$\mathbb{C}^2$$ spanned by $$|i\rangle$$. E.g. if we measure $$0$$, the post-measurement state is $$M_0 (a|0\rangle + b|1\rangle) = a |0\rangle$$ where $$M_0 = |0\rangle \langle 0| = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$ is the projector. But, the state a$$|0\rangle$$ isn't normalised, so we normalise this state by dividing it by the square root of its own inner product, as we saw in part 1, to just get $$|0\rangle$$ (up to some complex phase which is irrelevant -- see section 2.2.7 of the book).

Let me know if you require any further details!

• So if $\frac{M_m}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}}$ is not a unitary, how can we explain the fact that $\frac{M_m |\psi \rangle}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}}$ is a valid quantum state and the transformation was not unitary? Do these types of transformation (non-unitary transformations that take a quantum state to another quantum state) have a special name or properties? I was under the impression that only unitary transformations can output a valid quantum state.
– Josh
Commented Apr 25 at 14:46
• @Josh Note that $\frac{M_m}{\sqrt{\langle \psi | M^\ast_mM_m |\psi \rangle}}$ depends on $|\psi\rangle$ so it does not define a linear transformation.
– JiK
Commented Apr 25 at 14:50
• If I remember correctly, Nielsen and Chuang allow the measurement operators to be more general than just projections. The only requirement is that $\sum_mM_m^*M_m$ is the identity operator. Commented Apr 25 at 19:35
• @Josh just to be clear, you could have a linear transformation that takes some state $|\psi\rangle$ to some other state $|\phi\rangle$ (e.g., act on your state with $A=|\phi\rangle\langle\psi|$). If your transformation does not change the norm of any state and it is linear, then it must be unitary (e.g., my $A$ acting on some other state will only give a constant times a valid state, and your $M_m/\sqrt{\cdots}$ operation is not linear due to the denominator depending on $|\psi\rangle$) Commented Apr 25 at 21:55
• Why does this dependency on $| \psi \rangle$ make it non-linear?
– Josh
Commented Apr 26 at 1:04

Question 1 is just basic linear algebra: Any nonzero vector (with finite norm) divided by its norm is a unit vector. Or in other words, given an arbitrary nonzero $$|\phi \rangle$$, we know that $$\frac{1}{\sqrt{\langle \phi | \phi \rangle}} | \phi \rangle$$ has norm $$1$$. In this case $$| \phi \rangle = M_m | \psi \rangle$$.