# The quantum state just after a position measurement

The wave function of a free particle is given as, $$\psi(x) ~=~ e^{-{ x }^{ 2 }/{ a }^{ 2 }}.$$

Then a position measurement is made and the position of the particle is found to be at $x=a$.

My question is: What is the state of the particle just after this measurement? Is it equal to $\psi (x=a)$ ?

• Ján Lalinský, Why is a delta function not normalizable? As far as I know, $\int\limits_{-\infty}^{+\infty} \delta(x-a)dx=1$. Isn't it? – SRS Jan 8 '14 at 17:36
• Normalisation involves the square of the wavefunction. – gj255 Jan 8 '14 at 17:44

In a true measurement procedure of position, the outcome is an interval $(a-\delta,a+\delta)$, $\delta>0$ being the precision of the instrument. In view of Luders-von Neumann's postulate on the reduction of the state, if the state before the measurement was described by the normalized vector $\psi \in L^2(R)$, immediately after the measurement the state is represented by the vector: $$\psi'(x) = \frac{\chi_{(a-\delta,a+\delta)}(x) \psi(x)}{\sqrt{\int_{a-\delta}^{a+\delta}|\psi(s)|^2 ds} }$$ where $$\chi_{(a-\delta,a+\delta)}(x) =1 \quad \mbox{if x \in (a-\delta,a+\delta)}\qquad \chi_{(a-\delta,a+\delta)}(x) =0 \quad \mbox{if x \not\in (a-\delta,a+\delta)}$$ This is coherent with the fact that: $$\int_{a-\delta}^{a+\delta}|\psi(s)|^2 ds$$ is the probability to find the particle in $(a-\delta,a+\delta)$. The new wavefunction belongs to $L^2(R)$ again, as it is due.

• +1. However, when $\delta \to 0$, may we morally use the awful notation $\psi'(x) = \sqrt{\delta(x-a)}$ ? – Trimok Jan 8 '14 at 20:04
• Actually, it does not make much sense, rigorously speaking, there is no such thing as $\delta^{1/2}$, as a distribution I mean. However all this issue is very difficult, because in real measurement procedures for observables with continuous spectrum it is very difficult to give a general recipe for the real state after the measurement, it mostly depends on the instrument (even for the same observable). A better approach is to try to picture the procedure in terms of a "quantum operation". – Valter Moretti Jan 8 '14 at 20:11
• I am trying to fuse qm theory with the real world in my mind. What is the instrument with which we measure position? The only instrument we do in theory is the Stern-Gerlach for spin measurements. – user1584421 Apr 18 '19 at 17:58

When the position is measured on this state, the wavefunction collapses to a position eigenstate which is a Dirac-delta function $\delta(x-a)$ spiked about $x=a$. The measurement of position on any wavefunction collapses the state a position eigenstate. This is the case immediately after the measurement. This is not a stationary state and with time it will evolve in a non-trivial fashion which can easily be worked out if the governing Hamiltonian is known.

• This is a theoretical idealization which is experimentally false. At the LHC, to measure momentum, 3 consecutive position measurements are made. These define the ~circle along which the particle is moving, which in the presence of a uniform magnetic field yields the momentum. However, if the wave function were really collapsed to a $\delta$ after the first position measurement, the momentum $\psi(p)$ would be a plane wave ~uniformly distributed about all real values of momentum. This is not the case. Walter Moretti's answer, though I think also not perfect, is closer to reality. – doublefelix Nov 23 '19 at 15:51

Every measurement of a continuous variable, although it may give one definite result $a$, is loaded with uncertainty (error) which means that the actual value $x^*$ could be different than the result of the measurement $a$. This means that after the measurement, the state of the particle is such that it is somewhere around $a$. The uncertainty in knowledge of how far from $a$ is the actual $x^*$ can be quantified by variance $\langle (x^* - a)^2 \rangle$. This is the lower the higher the accuracy of the original measurement.

So $\psi(x = a)$ is not a state, because it isn't a function --- it's just a number. In fact, evaluating this number, you've just asked "is the state of the particle 0.368?" Hopefully you see that this is a nonsensical statement!

One of the central rules of quantum mechanics is that, when a measurement of an observable $O$ is made, the outcome is one of the eigenvalues of $\hat{O}$, and the system collapses into the corresponding eigenstate of $\hat{O}$.

So if position is measured, and found to be $x=a$, then the system is going to collapse into the eigenstate of position with eigenvalue $a$. Informally, this state is the delta function.

$$\psi(x) = \delta(x-a) \qquad \mathrm{after\ measurement}$$

To see this, note that we're looking for a function $\psi$ that, when multiplied by $x$, returns the same function scaled by a factor $a$. At first glance it might appear that there is no such function: $\sin x$ is certainly not proportional to $x \sin x$, and $1/x^2$ looks rather different to $1/x$ etc. The delta function is the only thing that works, because it's only non-zero at one specific point. Thus, multiplying it by $x$ has the effect of changing the height of the spike by a certain amount $a$ corresponding to where the spike is, leaving everything else zero. This is equivalent to multiplying the whole function by $a$. Hence it is an eigenstate.

• gj255, your application of the projection postulate is understandable, but in case of continuous spectra it has to be reformulated at least. It is not possible to measure position with infinite accuracy, and even if it was, the wave function cannot be set to $\delta(x-a)$, as this is not normalizable function. This is just a situation in which the original projection postulate draws shorter straw and has to be replaced by some more realistic description (for example, instead of delta peak, we may use Gaussian peak, which is also usual in the theory of errors.) – Ján Lalinský Jan 8 '14 at 17:26
• You're right of course. I used the word 'informally' to try to cover my back here. Perhaps my answer is further from the truth than is reasonable, but given the nature of the question asked I felt that the projection postulate was what user37000 ought to grasp, since (s)he appeared to have a misunderstanding about the whole essence of collapse of the wavefunction. Though it may not be quite applicable here, with the delta function not even a true function, let alone normalisable, that's a subtlety that should be dealt with later, no? – gj255 Jan 8 '14 at 17:31
• I do not think so. When people begin to learn quantum theory, they already know about measurements and their uncertainties. Why not use that to their advantage? Even if we disregard that, it makes no sense to teach projection postulate on an example for which it is not valid. The projection postulate is much better explained and applied to measurements of discrete variables, like spin projections in Stern-Gerlach type experiments. – Ján Lalinský Jan 8 '14 at 17:50
• We shall have to disagree here. The Born rule is at the heart of QM; all sources discussing measurement I've encountered revolve solely around it. To say that it is not valid for position measurements is unfair; better would be to say that it needs some modification due to subtleties with the mathematics. I agree that it is better explained in the context of e.g. spin, but user37000 is not asking about spin. Reading again the final line of her/his question, it is clear to me that this idea of wavefunction collapse needs to be emphasised, even if this is not the most 'convenient' location. – gj255 Jan 8 '14 at 18:10
• The first Born rule says that $|\psi(x)|^2$ gives density of probability that the particle is at $x$. This is valid for position measurements, even without the projection postulate and collapse to $\delta$. It is the collapse to $\delta$ that is invalid in this situation, not the Born rule. – Ján Lalinský Jan 8 '14 at 18:19

Thanks. So, I understand that after this measurement the wave function can be written as $\psi (x)=\delta (x-a)$. What if a second measurement is made after some time t, to find an eigenvalue of another observable? What's the form of the wave function that's going to evolve with time? Is it $\psi (x)={ e }^{ -\frac { { x }^{ 2 } }{ { a }^{ 2 } } }$ or $\psi (x)=\delta (x-a)$ ?. In other words, what the wave function after some time t? $\psi (x,t)={e}^{-iHt/\hslash}{ e }^{ -\frac { { x }^{ 2 } }{ { a }^{ 2 } } }$ or $\psi (x,t)= {e}^{-iHt/\hslash}\delta (x-a)$. Here, H is the Hamiltonian of the above free particle. I think the latter is true.

• I'm no veteran of this site, but it would be better I think to ask this as a new question, rather than an answer to your current one. You'd probably get more views that way. In any case, the latter is true, as you say. As soon as you've made the measurement, all information about the previous state is forgotten. The system has a new state, given by this delta function, and this evolves with time (according to the time-dependent Schrodinger equation) precisely as you've written. – gj255 Jan 9 '14 at 14:45