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


The wave function is

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



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?

  • $\begingroup$ a unitary matrix has determinant $\pm1$ so that matrix can't be unitary. It is even degenerate $\endgroup$ – Phoenix87 Jan 4 '15 at 15:12
  • $\begingroup$ But that part can easily be fixed by taking the unitary $\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)/\sqrt{2}$. The point is that the unitary depends on the state $\Psi$. $\endgroup$ – Norbert Schuch Jan 4 '15 at 15:21
  • $\begingroup$ As I said in one of the comments, I expect you could manually compute a unitary transformation that gives you the right answer. But you would need to already know the right answer, so it's a bit useless... $\endgroup$ – SuperCiocia Jan 4 '15 at 15:35
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    $\begingroup$ A physical note: unitary => invertable. Physically we know measurements are not always invertable. $\endgroup$ – zzz Jan 4 '15 at 16:36
  • $\begingroup$ @Dims you will find more relevant/useful points on this topic here. $\endgroup$ – Phonon Jan 4 '15 at 17:46


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.

  • $\begingroup$ See my edit please. So, am I correct thinking, that unitarity means that ANY given function will conserve length, while during collapse it is impposible to formulate one? $\endgroup$ – Dims Jan 4 '15 at 14:59
  • $\begingroup$ With regards to your question in the comment: yes, measurement does not obey any unitary transformation. You have to put the measurement by hand, at least in the current framework of QM. I have to say I don't know whether it is impossible to formulate such a matrix, maybe you can formulate one that only works for a specific case, but that'd be a bit useless because, in order to construct it, you'd need to know what the final state is so you'd already know the answer. $\endgroup$ – SuperCiocia Jan 4 '15 at 15:05
  • $\begingroup$ With regards to your edit: When you write out the matrix form of an operator you should always specify what vector representation you are using for your state vectors. I am assuming $|0\rangle = \left( \begin{array}{ c } 1 \\ 0 \end{array} \right)$ and $|1\rangle = \left( \begin{array}{ c } 0 \\ 1 \end{array} \right)$. Well this transformation is not unitary. Remember unitary means transpose and complex conjugate. The transpose of your matrix is not equal to the original one. $\endgroup$ – SuperCiocia Jan 4 '15 at 15:08
  • $\begingroup$ @Dims : which length? Unitarity of a transformation on a wave-function preserves probabilities. The probability of obtaining the value $+\hbar/2$ from an electron polarized with spin up in the direction $z$, but whose spin projection is measured on the axis $x$, remains $1/2$, whatever unitary transformation you do that doesn't change the spin-up polarization of the electron. $\endgroup$ – Sofia Jan 4 '15 at 15:12
  • $\begingroup$ Unitary transformations preserve the norm, i.e. the "length" of a vector in a Hilbert space $\langle \Psi | \Psi \rangle$. The norm can be interpreted as a probability if normalised to 1, in which case there would be a physical reason as to why it needs to be constant through any realistic time evolution. $\endgroup$ – SuperCiocia Jan 4 '15 at 15:18

I assume that you are referring to the outcome of any observable $O$ acting on a state $\psi$, since the act of measuring something is interpreted as "averaging" such operators on some state. General observables are self-adjoint operators which need not be unitary. Perhaps the simplest example of an observable is a projection, i.e. an operator $P$ with the property that $P^*P = P$ (idempotent and self-adjoint). Suppose that, in your case, $P = |0\rangle\langle0|$. The outcome of a measurement of $P$ on your state $\Psi$, when repeated $N$ times, is $|\alpha^2|N$ times YES (and hence $(1-|\alpha|^2)N$ times no. Moreover the result of $P\Psi$ is $\alpha|0\rangle$, which isn't a normalised vector, simply because $P$ is not a unitary.

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    $\begingroup$ You can't say "an observable $O$ acting on a state $\psi$". An observable is a physical quantity, that you measure. But you can use the terminology "operator $\hat O$. Now, a measurement is not averaging. In each single measurement of the operator $\hat O$ you obtain one of its eigenvalues. Measuring $\hat O$ on many particles identically prepared, you obtain statistics, and from it you can calculate different things, one of them being the average value. Another thing: from operators describing observables we don't expect unitary, but self-adjointness. Unitarity we expect from transformations. $\endgroup$ – Sofia Jan 4 '15 at 15:05
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    $\begingroup$ I don't distinguish between observables and self-adjoint operators. For me they are the very same thing. Also a measure is usually performed on a statistical ensemble, and this corresponds to the action of a state on the observable, i.e. $\omega(O)$. I don't see how the single outcome of an experiment can tell anything about the state of a system, unless you know a priori that you are dealing with a pure state. Finally I don't quite get your last remark. $\endgroup$ – Phoenix87 Jan 4 '15 at 15:10
  • $\begingroup$ Idle question while reading this: Since self-adjoint and unitary operators are in one-to-one correspondence, has the exponential of a self-adjoint projection any physical relevant meaning? $\endgroup$ – ACuriousMind Jan 4 '15 at 15:22
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    $\begingroup$ @ACuriousMind That correspondence holds between self-adjoint operators and one-parameter groups of unitaries. For more general unitaries you need more self-adjoint operators and you only hit the connected component of the identity: for every unitary $u$ homotopic to 1 there are self-adjoint operators $h_1,\ldots,h_n$ such that $u = e^{ih_1}\cdots e^{ih_n}$ $\endgroup$ – Phoenix87 Jan 4 '15 at 15:34
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    $\begingroup$ @ACuriousMind Don't know why but I can't edit comments right now. As for the exponential of a projection you simply get $e^{iPt} = 1 + (e^{it}-1)P$. I think this can be interpreted physically if $P$ is some generator of some transformation (e.g. a continuous symmetry). Then its exponential is the generated flow. $\endgroup$ – Phoenix87 Jan 4 '15 at 15:46

Answering the question in the title: a measurement process is intrinsically non-unitary. One way to see this is to realise that the unitarity of a process is equivalent to its being reversible.

A measurement process is intrinsically non-reversible, as some information gets lost. For example, measuring $(|0\rangle+|1\rangle)/\sqrt2$ in the computational basis, you can get either $|0\rangle$ or $|1\rangle$. The same outputs can be obtained measuring a different state, e.g. $(|0\rangle-|1\rangle)/\sqrt2$. This means that, given a measurement result (say $|0\rangle$), there is no way to know from what state it came from. Some information is lost in the process. It follows that the process cannot be described by a unitary matrix.


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