quantum mechanics - Why aren't transformations caused by measurements unitary?

It is said, that when measured, a quantum system undergoes "wave function collapse", which is a non-unitary transformation.

Why?

The wave function is

$\Psi = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$

where

$\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$

The probabilities sum after measurement is still 1, for example, if system collapsed to $\left|0\right\rangle$, then

$\left|1\right|^2 + \left|0\right|^2 = 1$

For example, if function was

$\Psi = \frac{1}{\sqrt{2}} \left|0\right\rangle + \frac{1}{\sqrt{2}} \left|1\right\rangle$

the transformation was

$ \left[ \begin{array}{ c c } \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & 0 \end{array} \right] $

Isn't this transformation unitary?

Answer

No.

As long as your state is $|\Psi \rangle = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$, then as you said $\alpha$ and $\beta$ need to satisfy $\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$, so say $\alpha = \beta = 1/\sqrt 2$.

If you perform a measurement and find that the system in the $\left|0\right\rangle$ state, then the new wavefunction will be $\Psi =\left|0\right\rangle$. You can write it as $\Psi = \alpha \left|0\right\rangle$ but because of normalisiation $|\alpha|^2$ needs to be 1, so $\alpha$ must be either 1 or a pure phase factor.

You had to change the normalisation by hand (changing $\alpha$ from $1/\sqrt 2$ to $1$). A unitary transformation on $|\Psi \rangle$ would affect only the kets and not the constants. The time evolution of any wavefunction is governed by the Schrodinger equation which, when solved, is effectively a unitary transformation -- unitary transformations leave the norm unchanged. The time evolution due to performing measurements, however, is something entirely different and does not follow the formalism of the Schrodinger equation.

A unitary transformation leaves the norm unchanged, since the norm of $U|\Psi \rangle$ is $\langle \Psi |U^{\dagger}U|\Psi \rangle = \langle \Psi | \Psi \rangle$ if $U$ is unitary. In your specific case this is true, but only because you had to manually change the normalisation of the post-measurement wavefunction. If QM included measurement, then there should be a deterministic way of computing how $\alpha$ or $\beta$ would change. But it doesn't, so you need to re-normalise it.