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I am a bit confused about the idea of measurement in QM. As far as I understand, if you measure the position of a particle, the wavefunction of that particle changes into a delta function, and thus the particle gets localized.

Now, let's say we have a particle in a box in a state with 3 main peaks ($\psi_3$). If we look at the main peak let's say (so the center of the box) and see the particle, the wave function collapsed at that point. But what happens if we don't see it there? Obviously, the wavefunction changes, as we know for sure that the probability of the particle being at that point is 0 now, but how is the wave function changed? Does it turn into a delta function at a random point, different from the one where we measured, or what?

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  • $\begingroup$ What do you mean by "if we don't see it there?". $\endgroup$ – ZeroTheHero Jun 24 '17 at 19:05
  • $\begingroup$ I mean, the particle is not in the position where we measured it. So, if the particle was there, we have $\psi(x_0)=\delta(x_0)$. In the case I asked about, after the measurement, we get $\psi(x_0)=0$ $\endgroup$ – Silviu Jun 24 '17 at 19:14
  • $\begingroup$ @ZeroTheHero it is described by a probability density before measurement, but after the measurement it can be zero, right? So if you have at a point $\psi(x_0) = \frac{1}{\sqrt{3}}$, so the probability of finding it there is 1/3, if you have 300 ensembles, in 200 (on average) of them, the particle will not be at $x_0$ which means that, after the measurement $\psi(x_0)=0$, right? $\endgroup$ – Silviu Jun 24 '17 at 19:21
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There are two basic points you should understand.

First, in quantum theory you really should pay more attention to the statements made about observations rather than what happens with objects like wavefunction. The nature of the collapse lies in the conditional probabilities. If you have (in the Heisenberg picture) $|\alpha,t\rangle$ as some eigenstate of the observable $A(t)$ with discrete spectrum you also have the projection operator, \begin{equation}\hat{\mathcal{P}}_\alpha(t)=|\alpha,t\rangle\langle\alpha,t|,\quad\hat{\mathcal{P}}_\alpha(t)=\Big(\hat{\mathcal{P}}_\alpha(t)\Big)^\dagger,\quad \hat{\mathcal{P}}_\alpha(t)\hat{\mathcal{P}}_\beta(t)=\delta_{\alpha\beta}\hat{\mathcal{P}}_\alpha(t) \end{equation} The probability for the measurement $A(t)=\alpha$ is given by, \begin{equation} P_\psi\Big[A(t)=\alpha\Big]=\langle\psi|\hat{\mathcal{P}}_\alpha(t)|\psi\rangle=|\langle\alpha,t|\psi\rangle|^2 \end{equation} The collapse is a statement that the conditional probability for the next measurement is a probability computed for the projected state, \begin{equation} P_\psi\Big[B(t_2)=\beta|A(t_1)=\alpha\Big]=P_{\hat{\mathcal{P}}_\alpha(t_1)\psi}\Big[B(t_2)=\beta\Big] \end{equation} which is connected with the formula for the probabilities of two consequental measurements, \begin{equation} P_\psi\Big[A(t_1)=\alpha,B(t_2)=\beta\Big]=\langle\psi|\hat{C}^\dagger\hat{C}|\psi\rangle,\quad \hat{C}=\hat{\mathcal{P}}_\beta(t_2)\hat{\mathcal{P}}_\alpha(t_1) \end{equation}

Second, only normalizable states are physical. Wavefunction $\psi(x)=\delta(x)$ has infinite norm $\int_{-\infty}^{+\infty}dx\,|\psi(x)|^2$ so it can't be realized in nature. Mathematically it's said that the "eigenstates" of the observables with continuous spectrum like coordinate or momentum, $|x,t\rangle$ or $|p,t\rangle$ don't belong to the Hilbert space. Instead they work like distributions - you can construct the wavepackets $\int_{-\infty}^{+\infty}\psi(x)|x,t\rangle$ that will belong to the Hilbert space.

That's connected with probability properties for continuous spectrum even in the classical case. We can't consider the probability of the outcome $X(t)=x$! We can only talk about probability density functions $p(x)$ and probabilities of the outcome being in some interval $P(x_1<x<x_2)=\int_{x_1}^{x_2}dx\,p(x)$.

In quantum theory $|x,t\rangle\langle x,t|$ is not a projector rather it's so called projective valued measure that can form actual projector if you take some interval, \begin{equation} \hat{\mathcal{P}}_{x_1<X<x_2}(t)=\int_{x_1}^{x_2}dx|x,t\rangle\langle x,t| \end{equation} Then when you consider conditional probabilities you should take the condition of $X$ belonging to some interval. Then, \begin{equation} P_\psi\Big[B(t_2)=\beta|x_1<X(t_1)<x_2\Big]=P_{\hat{\mathcal{P}}_{x_1<X<x_2}(t_1)\psi}\Big[B(t_2)=\beta\Big] \end{equation} Then in some sense you can say that for measurement giving $x_1<X(t_1)<x_2$ the wavefunction "collapsed" to the $\hat{\mathcal{P}}_{x_1<X<x_2}(t_1)|\psi\rangle$.

It's often said about this question that real measurements have finite error. Then naively instead of delta-functions the wavefunction actually collapses to some smooth lumps. The issue here is that such lumps will overlap i.e. different outcomes don't exclude each other. Then the standard concept of the projective measurements (that lead to the notion of the "collapse") doesn't work and you should use instead positive operator valued measures (POVM) replacing projectors with general positive self-adjoint operators. Basically that "solution" is that your initial observable doesn't adequately describe the actual measurement process and the resulting probability can actually differ from the naive $|\psi(x)|^2$ and doesn't circumvent the above considerations about conditional probabilities.

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