The 'wavefunction collapse' upon measurement is usually referred to as being a non-unitary transformation, since it does not preserve the norm of the state vector. Indeed, if a linear superposition like $\psi + \phi$ collapses to let's say just $\phi$, then $||\psi+\phi|| \neq ||\phi||$.

But what if $\psi + \phi$ collapses into $\alpha \phi$ where $\alpha$ is such that $||\alpha \phi|| = ||\psi + \phi||$. Then norm is preserved, and $\alpha \phi$ only differs from $\phi$ by a constant, so it represents the same state as $\phi$. Wouldn't this type of collapse be a unitary transformation, and if so, why can't all types of state collapse be treated like this?

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    $\begingroup$ "as being a non-unitary transformation, since it does not preserve the norm of the state vector" This is not the reason why collapse is non-unitary. The reason is that many different psi functions $\psi$ end up in one of the eigenfunctions of the operator corresponding to measured quantity and the process is thus irreversible; one cannot infer the original state from the resulting state. For example, if spin projection $s_z$ is measured, all possible $|\psi\rangle$ go either to $|z+\rangle$ or $|z-\rangle$. $\endgroup$ Oct 25, 2015 at 10:19

2 Answers 2


The reason is two-fold: 1) A unitary transformation does preserve the norm $\left\|\psi \right\| = \langle \psi | \psi \rangle$, but not only the norm. 2) A quantum measurement must produce a state that is not affected by a repeat identical measurement.

In general, a unitary transformation $U$, $U^\dagger U = U U^\dagger = I$, preserves overlaps: $$ \langle U\phi | U\psi\rangle = \langle \phi |U^\dagger U |\psi \rangle = \langle \phi | \psi \rangle $$ Say a normalized state reads $a\psi + b\phi$ before collapse, for orthogonal and normalized $\psi$ and $\phi$, $\langle \phi | \psi \rangle = 0$, $\langle \psi|\psi \rangle = \langle \phi|\phi \rangle = 1$, and $|a|^2 + |b|^2 = 1$ such that $\left\|a\psi + b\phi \right\| = \langle a\psi + b\phi| a\psi + b\phi \rangle = 1$. Let $a\psi + b\phi$ collapse into $\phi$ upon measurement. By definition, a 2nd identical measurement must leave $\phi$ unchanged. If the collapse were a unitary evolution such that $U(a\psi + b\phi) = \phi$, then the same measurement on $\phi$ would have to result in $U\phi = \phi$. The unitary $U$ would indeed preserve the norm, since $\left\|a\psi + b\phi \right\| = \left\|U(a\psi + b\phi) \right\| = \left\|\phi \right\| = 1$. But $U$ should also preserve the overlap $\langle \phi | a\psi + b \phi \rangle = b$, whereas instead $$ \langle U\phi | U(a\psi + b \phi) \rangle = \langle \phi | \phi \rangle = 1 > b $$ Since we already took care of normalizations, there is no way to remove the above disagreement by a rescaling of $\phi$. So collapse cannot be unitary.

A faster way to arrive at the same conclusion is to consider collapse from a mixed initial state $\rho$, $\rho \neq \rho^2$. The result of the collapse would still be a pure state, so in this case $U$ would have to take a mixed state into a pure state. But unitary transformations always take pure states into pure states, so again this cannot work.

  • $\begingroup$ Wow, thanks, glad to be of help. No books so far, I'm afraid. As for loss of coherence and information, yes, absolutely, although coherence is a relative concept. Information works better and the argument is immediate for mixed states, where info is obviously lost. For pure states, entropy as info doesn't say much, but the overlap does. $\endgroup$
    – udrv
    Nov 14, 2015 at 3:25
  • $\begingroup$ Taking a look right now, but will post tomorrow. As a general idea though, taking the partial trace is equivalent to extracting a partial probability distribution from joint probabilities. $\endgroup$
    – udrv
    Nov 15, 2015 at 4:55
  • $\begingroup$ @user929304 Sorry I didn't manage to post earlier, but have a look at physics.stackexchange.com/questions/204100/… and let me know if you have any further questions. Hope it helps! $\endgroup$
    – udrv
    Nov 16, 2015 at 7:21

A unitary operator doesn't just preserve norms, it is also linear. Therein lies the true problem.

It must send a normalized eigenstate of the operator to itself. So it would end up having to send every state to itself, by linearity.


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