# Position eigenstates

1. Aug 6, 2014

### Staff: Mentor

When a particle's position is measured, does the wavefunction collapse to the eigenstate of the measurement (a delta function), or do you account for the accuracy and precision of the measurement device by making the state be a mixture of eigenstates corresponding to the device's imperfect performance (e.g. a gaussian spike)?

2. Aug 6, 2014

### atyy

Although loosely we say that after a position measurement, the state collapses into a position eigenstate, this is strictly speaking not true since the position eigenstate is unphysical (not normalizable). One way to treat this is to take the resolution of the detector into account.

An example of such a treatment is given in http://arxiv.org/abs/1211.4169 (section 4.1).

Last edited: Aug 6, 2014
3. Aug 6, 2014

### Matterwave

To me, the only sensible option is the latter. A delta function itself can not be physically achievable since its position uncertainty is 0, meaning its momentum uncertainty is infinite. An infinite momentum uncertainty is just as unphysical to me as an infinite position uncertainty. And no matter how much introductory physics books like to maintain that the "uncertainty principle is still maintained because you get a zero times infinity", I do not buy such arguments since they have no backing behind them to take this limit.

However, as I have not read any peer reviewed articles on this particular subject, I can't vouch for my answer.

4. Aug 6, 2014

### WannabeNewton

The measuring device is assumed to be ideal just as in classical mechanics.

5. Aug 6, 2014

### dextercioby

There are valid formulations without collapse and never in the theory are measuring devices supposed to be imperfect.

6. Aug 6, 2014

### jostpuur

I don't believe that a perfect answer to the question would exist because the usual philosophical problems (though sometimes of practical kind...) are related to this. However, I believe that a somekind of answer, which probably hints in a right direction, does exist, and it is that in a quantum mechanical measurement the state collapses through an orthogonal projection to a subspace spanned by some collection of the eigenvectors. In other words, the guess about Gaussian weight looks little suspicious, but if it was replaced by a characteristic function (indicator function), then I would say yes.

For example, suppose that some particle detecting camera has pixels, and that the pixels can be modeled as some sets $\Omega_1,\Omega_2,\Omega_3,\ldots\subset{R}^3$. They could be little cubes or balls, for example. Then the camera works so that it will tell if a particle was found in some pixel. If a particle was found in a pixel $\Omega_k$, in the measurement process the wave function was projected like this:

$$\psi \mapsto \chi_{\Omega_k}\psi$$

where

$$\psi:\mathbb{R}^3\to\mathbb{C}$$

is some wave function, and

$$\chi_{\Omega_k}:\mathbb{R}^3\to\{0,1\},\quad \chi_{\Omega_k}(x) = \left\{\begin{array}{c} 1,\quad & x\in \Omega_k \\ 0,\quad & x\notin\Omega_k \\ \end{array}\right.$$

is the characteristic function corresponding to the spatial region of the pixel.

The usual explanation about the quantum mechanical collapse is that the state is projected like this:

$$|\psi\rangle \mapsto |a\rangle \langle a |\psi\rangle$$

where $|a\rangle$ is some normalized eigenstate. The normalization means $\langle a'|a\rangle = \delta_{aa'}$ with a Kronecker delta. I think it is very reasonable to generalize this by stating that perhaps the collapse happens like this:

$$|\psi\rangle\mapsto \sum_{a\in\Omega} |a\rangle \langle a|\psi\rangle$$

where the set $\Omega$ is related to the practical resolution of the measurement device. Also I might ask that if some measurement device cannot separate some eigenvalues, which are too close to each other, what else would the collapse be? Then with observables of continuous spectra this is naturally interpreted as

$$|\psi\rangle\mapsto \int\limits_{\Omega}|x\rangle\langle x|\psi\rangle dx$$

where the normalization is achieved with a Dirac delta: $\langle x'|x\rangle = \delta(x-x')$.

Last edited: Aug 6, 2014
7. Aug 6, 2014

### tom.stoer

the well-know proof for the uncertainty relation does not work for position eigenstates b/c it uses expressions like |ψ|2, <ψ|ψ> and similar expressions which are ill-defined for ψ(x) ~ δ(x); I do not know whether this proof can be generalized, e.g. using rigged Hilbert spaces

8. Aug 6, 2014

### atyy

Do you mean something like the model in http://arxiv.org/abs/quant-ph/0307057 (section VII.B, Eq 310-338)?

9. Aug 6, 2014

### Staff: Mentor

atyy, I think that I understood your point about position states not being normalizable, I had forgotten about that when I asked. Does the same thing apply for any non-quantized eigenvalues. E.g. if you are measuring energy for an unbound state (so that it is not quantized), presumably the same issue would arise that the detector would give a number with limited precision. Or does that problem only arise for position since the eigenstate is not a good function?

10. Aug 6, 2014

### USeptim

If the particle has a delta-function position, the momenta would be spread over the infinite, when you integrate the square of this you will get an infinite kinetic energy which is un-physical.

11. Aug 6, 2014

### atyy

In my post I was only thinking that the post-measurement state cannot be a position eigenstate because it is not normalizable and hence not in the Hilbert space, and not that position is a continuous variable.

In the first link I gave http://arxiv.org/abs/1211.4169 Distler and Paban say that when an observable has a continuous spectrum, the resolution of the detector must be specified as part of what it means to measure the variable. However, WannabeNewton and dextercioby seem to disagree with this, as I think http://arxiv.org/abs/quant-ph/0307057v1 Ozawa and http://arxiv.org/abs/0706.3526 Busch (citing Ozawa) also do. So I'd like to know whether WannabeNewton and dextercioby would agree with Ozawa and Busch, and disagree with Distler and Paban (or whether they agree with neither, and prefer some other formulation for continuous variables).

Last edited: Aug 6, 2014
12. Aug 6, 2014

### Staff: Mentor

For most problems it doesn't matter, as the forward evolution in time of the delta function leads immediately to more reasonable states (such as that gaussian spike). You don't have to ascribe physical significance to "the wave function at the exact moment of collapse" (phrasing it this way makes it easier to understand the question as an interpretational issue) to get sensible results for any time after that.... and that delta function does make for a simple initial condition.

13. Aug 7, 2014

### vanhees71

Despite the fact that I don't think that one needs any collapse postulate and that the minimal statistical interpretation of quantum theory is the only solid interpretation we have about this most successful theory invented by man so far, it's also mathematically clear that there are no position eigenvectors in the literal sense but only generalized ones in the sense of distribution theory. Since the spectrum of the position operator is continuous only, there is no true eigenvalue and no true eigenstate. This is made explicit in the position representation, where the generalized eigenvector is represented by the Dirac $\delta$ distribution:
$u_{\vec{x}'}(\vec{x})=\delta^{(3)}(\vec{x}-\vec{x}').$
This is not a square integrable function and thus cannot represent a physically realizable state for a particle.

This is also reflected in the already mentioned Heisenberg-Robertson uncertainty relation $\Delta x_j \Delta p_k \geq \hbar/2 \delta_{jk}$: There is no state, where the particle has sharp position or sharp momentum.

14. Aug 7, 2014

### strangerep

First, one must let go of the standard "collapse" nonsense. Although it makes approximate sense (well, sort of) for filter-type measurements, it does not hold water in general. Think of a photon striking a detector screen: the photon is absorbed and does not even exist after the "measurement".

A more realistic approach is to model the measurement via an interaction between system and apparatus, such that the final state of the apparatus is correlated with the initial state of the system. See Ballentine ch9 for a discussion of this.

Regarding an observable with continuous spectrum, one can a construct suitable model using POVM (positive operator-valued measure). The resolution of unity provided by a complete set of (usually non-orthogonal) coherent states is an example of this sort of thing.

15. Aug 7, 2014

### Staff: Mentor

16. Aug 7, 2014

### tom.stoer

I think it makes no sense to "discuss away" the position eigenstates.

You do have this problem in QM simply b/c in practice you use position- or momentum eigenstates to do calculations. Yes, you may argue that a momentum measurement does not collapose your state to a plane wave; this is certainly correct. But fapp you use plane waves ;-)

17. Aug 7, 2014

### atyy

Yes. If we want a concrete proposal to discuss, how about http://arxiv.org/abs/quant-ph/0307057? Ozawa proposes a measurement model for a POVM for sharp position in section VII.B, Eq 310-338, with the post-measurement state of the system given in Eq 325. As far as I can tell, the post-measurement state is not a delta function.

Busch http://arxiv.org/abs/0706.3526 agrees that Ozawa's measurement model does sharp position measurement, but Busch does not explicitly comment on the post-measurement state given by Ozawa.

There is an interesting comment in the discussion of Ozawa's paper. "(I) Can every measuring apparatus be described by an indirect measurement model? ... Although some measuring apparatuses, especially in the attempts for quantum nondemolition measurements [26], allow indirect measurement model descriptions, it is still difficult to convince any schools of measurement theory of the affirmative answer to question (I) above."

Last edited: Aug 7, 2014
18. Aug 7, 2014

### vanhees71

I heavily disagree. There are no eigenstates for spectral values of a self-adjoint operator in the continuous part of the spectrum. That's a mathematical fact and has nothing to do with the physical interpretation of quantum theory. The statement that pure states are represented by rays defined by true normalizable Hilbert-space vectors must be taken seriously.

Of course, also "momentum eigenstates", i.e., "plane waves" in the position representation
$$u_{\vec{p}}(\vec{x})=\frac{1}{(\sqrt{2 \pi})^{3}} \exp(\mathrm{i} \vec{p} \cdot \vec{x})$$
are obviously no wave functions representing a pure state in quantum mechanics, because they are not normalizable.

They are "generalized" states, i.e., they belong to the dual of the nuclear space, which is an appropriate dense subspace, where the self-adjoint operators are defined. The dual of this smaller subspace of Hilbert space is always much larger and contains generalized functions (distributions).

The physicist's sloppy use of these concepts of the "rigged Hilbert space" works so well, because the Hilbert space is extremely nice to us ;-)). My math prof. used to say that the separable Hilbert space is so well-mannered that it is almost like a unitary space of finite dimension. The emphasis lays on "almost". A very nice paper on the fact, that sometimes you can get nonsense with the naive physicist's handling of these issues can be found here

http://arxiv.org/abs/quant-ph/9907069

An example, where any serious student should get worried once in his/her quantum mechanics lecture when it comes to the evaluation of cross sections from the S matrix. There, the physicists happily square the "momentum-conserving" $\delta$ distribution and then discussing this mathematical nonsensical result away with a lot of handwaving (sometimes called "Fermi's 2nd trick"; nobody could tell me so far, what's Fermi's 1st trick then ;-)). Here, it's easily cured by using wave packets, i.e., true states, in the initial state, and this elucidates a lot the physical meaning of what a cross section is and how real-world scattering experiments are to be understood quantum mechanically. You find a very good explanation in the good old textbook by Messiah and (for the relativistic case) in Peskin/Schroeder, Intro to Quantum Field Theory.

Other sloppyness is not cured yet: There is no mathematical proof for the mathematical existence realistic relativistic interacting field theories like QED or the Standard Model although these are among the most successful theories concerning the agreement between theory and observations ;-)).

19. Aug 7, 2014

### tom.stoer

I agree in principle, but I disagree in practice.

We use plane waves, that's a fact!

You are invited to re-write 99% of the QM literature.

20. Aug 7, 2014

### jostpuur

The standard collapse explanations are often too distant from reality. I'll put forward a real question of practical relevance here: We assume that a time evolution of a wave function is known under the assumption, that there is nothing on its way (nothing that would cause a wave function collapse). The wave function is

$$\psi:[0,\infty[\times\mathbb{R}^3\to\mathbb{C},\quad (t,x)\mapsto \psi(t,x)$$

We assume that this is a non-relativistic case. We can also assume that at time $t=0$ we have $\langle x\rangle = 0$ and $\langle p\rangle = (0,0,p_3)$ with $p_3>0$ so we know roughly where the packet is going. Then we modify the situation by placing a two dimensional plate $\mathbb{R}^2\times \{R\}$ on the way with $R>0$, and we assume that this plate functions as a measurement device. It absorbs the hitting particle, and informs us about the hitting point. What is the probability density

$$\rho:\mathbb{R}^2\to [0,\infty[$$

that describes the particle's hitting point? What principles imply what this probability density ends up being?

Does Ballentine answer my question about the probability density on a two dimensional detector plate?

21. Aug 7, 2014

### Staff: Mentor

The point of what Strangerep and Vanhees said is that the formalism QM does not have collapse. That is an interpretational thing.

That's trivial.

Its a two dimensional wavefunction, so you take the absolute value squared it to get the probability density.

Here is a correct QM analysis of the situation you are getting at - in one dimension - but easily extended to two
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

But before going any further you need to understand what a wave-function is and the exact postulates that go into QM.

For that you need to read the first 3 chapters of Ballentine.

Since you are a math type Gleason's theorem may also help in understanding the underlying principles:
http://kof.physto.se/cond_mat_page/theses/helena-master.pdf [Broken]

Basically if you require non contextuality Gleason's theorem shows Born's rule is the only measure that can be defined on a Hilbert space.

For a simpler more modern version see post 137:

Thanks
Bill

Last edited by a moderator: May 6, 2017
22. Aug 7, 2014

### strangerep

The best reference for that (afaik) is Mandel & Wolf.

There was a long thread about detector clicks and single photons in which more precise reference within M+W were given. See in particular my posts #7 and #73 -- but the whole thread is probably worth (re-)reading.

BTW, (Bhobba), I don't think this is trivial. M+W spend a lot of time on this sort of thing. Especially, the extra detailed stuff in ch14, iirc. I think I'll augment my signature line to include a mention of M+W.

Last edited: Aug 7, 2014
23. Aug 8, 2014

### jostpuur

The wave function has the form $\psi:[0,\infty[\times\mathbb{R}^3\to\mathbb{C}$, and the probability density $\rho:\mathbb{R}^2\to [0,\infty[$. It makes no sense to "set" $\rho=|\psi|^2$.

24. Aug 8, 2014

### tom.stoer

Both the wave function and the probability density depend on time; so it's simply

$$\rho(x,t) = |\psi(x,t)|^2;\;\forall x \in \mathbb{R}^3\,\wedge\,\forall t \in \mathbb{R}^3$$

If the wave function is a time-dependent wave packet, then the probability density is time-dependent as well. Associated with this time-dependent probability density you get a time-dependent probability current as well.

This is standard textbook QM (which is the translation of bhobba's "trivial)

25. Aug 8, 2014

### jostpuur

Recall this:

That is what the observations look like. If you want a theory to describe that, you need a theory that gives some probability density $\rho:\mathbb{R}^2\to [0,\infty[$. Your probability density $\rho:\mathbb{R}^4\to [0,\infty[$ is something else, and doesn't look very useful.

Last edited: Aug 8, 2014