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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?

glS
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Dims
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  • a unitary matrix has determinant $\pm1$ so that matrix can't be unitary. It is even degenerate – Phoenix87 Jan 04 '15 at 15:12
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    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 $\Psi$. – Norbert Schuch Jan 04 '15 at 15:21
  • 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... – SuperCiocia Jan 04 '15 at 15:35
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    A physical note: unitary => invertable. Physically we know measurements are not always invertable. – zzz Jan 04 '15 at 16:36
  • @Dims you will find more relevant/useful points on this topic here. – Ellie Jan 04 '15 at 17:46

4 Answers4

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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.

SuperCiocia
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  • 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? – Dims Jan 04 '15 at 14:59
  • 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. – SuperCiocia Jan 04 '15 at 15:05
  • 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.

     – SuperCiocia Jan 04 '15 at 15:08
  • @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. – Sofia Jan 04 '15 at 15:12
  • 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. – SuperCiocia Jan 04 '15 at 15:18
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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.

glS
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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.

Phoenix87
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    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. – Sofia Jan 04 '15 at 15:05
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    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. – Phoenix87 Jan 04 '15 at 15:10
  • 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? – ACuriousMind Jan 04 '15 at 15:22
  • @Phoenix87 : did you tell the fellow who asked, that you don't distinguish? Nobody can read others' minds. The people that ask such questions as the one here, are beginners. Many of our explanations are very detailed, so as to be clear for them. Next, single outcomes don't tell, but many single measurements give us the statistics, as I said. This statistics is not only for calculating the average. I just refer to your words. You don't answer here to people having your level of knowledge. Next: he asked about unitarity of the measurement process, not of $\hat O$ - read what you wrote. – Sofia Jan 04 '15 at 15:24
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    @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 $u = e^{ih_1}\cdots e^{ih_n}$ – Phoenix87 Jan 04 '15 at 15:34
  • @Phoenix87 and the Curious Mind : what you talk about, all of you? Do you indeed want to help the OP? Please see how simple and understandable answered the SuperCiocia. – Sofia Jan 04 '15 at 15:37
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    @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. – Phoenix87 Jan 04 '15 at 15:46
  • @Sofia: My question was just triggered by the observation of this answer that the projectors are self-adjoint, it has indeed nothing to do with the OP's question. Phoenix87: Alright, so there's nothing special about the projector in this respect. Thanks, anyway. – ACuriousMind Jan 04 '15 at 16:22
  • @ACuriousMind I was also a bit surprised when I learnt about this, that any unitary operator $U$ can be realized in the form $U = exp(iK)$ for some Hermitian operator $K$. It turns out this is extremely useful in most quantum computation schemes, which allows one to build all sorts of register operators (quantum gate). This link covers some of them if you're interested. – Ellie Jan 04 '15 at 16:50
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Measurement is not unitary, because it does not implement a unitary. It selects a unitary.

We can certainly say that a preparation $|\psi_i\rangle$ and (renormalized) measurement outcome $|\psi_f\rangle$ are related by a unitary transformation $$|\psi_f\rangle = U_{fi} |\psi_i\rangle,$$ where $U_{fi} \in U(2)$. However, we can only say this once a measurement has occurred.

Before measurement, the possible outcomes are $|{+}\psi_f\rangle$ and $|{-}\psi_f\rangle$, so all we can say is that $|\psi_i\rangle$ and the not-yet-measured outcome $|\psi_f\rangle$ are related by either $U_{fi}^+$ or $U_{fi}^-$, where $$|{+}\psi_f\rangle = U_{fi}^+ |\psi_i\rangle \hspace{1em} \text{and} \hspace{1em} |{-}\psi_f\rangle = U_{fi}^- |\psi_i \rangle.$$

Formally, we can construct an equivalence class of unitaries $U_{fi}^+ \sim U_{fi}^-$ over the possible outcomes of the measurement process, so that $[U_{fi}^+] = [U_{fi}^-] \in U(2)/\mathbb Z_2$ represents the part of the relation between the initial and final state that is fixed by our experimental setup.

The point is that the measurement process implements a transition $$ U(2)/\mathbb Z_2 \to U(2) \\ [U_{fi}^\pm] \mapsto U_{fi}^\pm $$ that breaks the $\mathbb Z_2$ symmetry and selects a particular representative of $[U_{fi}^\pm]$. In other words, the unitary relation $U_{fi}^+$ or $U_{fi}^-$ between an initial state and measurement outcome is not a description of the measurement process. It is the result of the measurement process.

Luke Burns
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