# Why not sharp positions?

1. May 18, 2010

### japrufrock

What are the theoretical and practical reasons that no particle can have an absolute sharp position?

Is it because the position operator has no eigenvectors, i.e. position eigenstates are never descriptions of actual physical states (because they are Gaussian vectors with a certain width)?

And what is the physical limitation in practice, is it that we would require an infinite mean energy (because of the uncertainty principle)?

2. May 19, 2010

### ansgar

since the wave function is a continous function

and that the position operator does not commute with the hamiltonian, i.e. it is not a constant but will change as time goes.

3. May 19, 2010

### tom.stoer

For free particles you can prepare (mathematically) a sharply peaked = localized state using a delta function, but it will spread with time due to the Hamiltonian

4. May 19, 2010

### Fredrik

Staff Emeritus
5. May 19, 2010

### ZapperZ

Staff Emeritus
This is a bit puzzling. When I make a position measurement, I have a "sharp" position, which is where I measure it. The uncertainty in that measurement is not part of QM. Rather it is part of the experimental instrument,which continues to be refined.

Furthermore, in the Drude model, the single-particle spectral function of quasiparticles in a metal is a delta function. This is what gives you many of your beloved properties of conductors such as Ohm's Law. So even in principle, and certainly in the practical sense, one can certainly have "sharp positions".

Zz.

6. May 19, 2010

### Fredrik

Staff Emeritus
The best you can hope to accomplish with a "position measurement" is to confine the particle to a region of finite size, determined by the interaction between the particle and the measuring device. That region is never just a single point. So the measurement might give us a wavefunction with compact support, but not a delta function.

I'm not familiar with this model, but it would be very surprising to me if the use of the delta function isn't just an idealization intended to make the calculation easier. I suppose you could argue that everything in physics is an idealization intended to make the calculation easier, but...uh...I'm going to have to think about how I'm going to finish this sentence.

OK, these are two of my thoughts:

1. The argument in the post I linked to seems to imply that only the "really nice" wavefunctions are consistent with translation invariance, which is a part of both Galilei and Poincaré invariance.

2. If you start with one of those "really nice" wavefunctions, I don't see any way for an interaction to change it into a delta function state. This is definitely ruled out for single-particle QM, where time evolution is just multiplication from the left by an operator of the form exp(-iHt), where H=p^2/2m+V and V is a smooth function of position.

Last edited: May 19, 2010
7. May 19, 2010

### Staff: Mentor

You're talking about the wavefunction of the particle after the measurement is made. But the uncertainty principle (mentioned in the OP) just restricts the distribution of (many) measurements based on the particle's original wavefunction before the measurement. Correct?

8. May 19, 2010

### ZapperZ

Staff Emeritus
But the uncertainty in the position is due to my instrument. It has nothing to do with anything "fundamental". QM says nothing about the accuracy of my single measurement of the position. This is not the HUP.

The Drude model is the free electron gas model for metals. It is how you derive things such as Ohm's law.

If you look at the many-body theory of electron-electron interaction, the single-particle spectrum is represented by the imaginary part of the Green's function, which is the propagator. Now, how can we derive the drude model out of such a thing? We set the single-particle spectrum to be a delta function (i.e. infinite quasiparticle lifetime, etc.)! Voila! We get the Drude model and all of the familiar properties of metals (see, for example, Mattuck's Dover text on many-body physics). Such a thing may or may not be an "approximation", but this is, in principle, what is involved in one of the most common application of QM.

Zz.

9. May 19, 2010

### DrChinese

I read ZapperZ as simply answering the above question: the only thing preventing an absolutely sharp position is observational accuracy. There is no other theoretical issue according to standard physics.

10. May 19, 2010

### Fredrik

Staff Emeritus
Right.

Does it matter what the cause is? You suggested that your measurement will put the particle in a delta function state, and I posted a partial explanation of why it won't. Sounds like we agree that we can get arbitrarily close but not all the way.

It doesn't put a lower bound on it, that much is true, but it does seem to say that the accuracy can't be zero.

I don't believe that's correct. Translation invariance seems to require nice states (see #4), and applying a time evolution operator exp(-iHt) with a finite t to a "nice" state can't give us a delta function state. Also, the delta function isn't a member of $L^2(\mathbb R^3)$, so the axioms of the standard version of QM don't consider delta functions to be state vectors.

The question is, when we use the rigged Hilbert space formalism to extend the state vector space to include position "eigenstates", does this also give us a representation of the the relevant group (Galilei/Poincaré) on the "extended" state vector space? If the answer is yes, then I might change my mind...as long as we use a rigged Hilbert space to define what we mean by "QM", instead of a Hilbert space.

11. May 19, 2010

### ZapperZ

Staff Emeritus
Say you have a superposition of u1 and u2. Each is an energy eigenstate. I make an energy measurement, and I measure E1, corresponding to eigenstate u1. What's the uncertainty in my measurement of E1 based on QM?

Zz.

12. May 20, 2010

### dx

Quantum mechanics does not rule out a sharply defined position. It only says that such a sharp definition implies that the momentum p must have an infinite uncertainty.

In string theory, the uncertainty relation is modified to

$$\Delta x \geq \frac{\hbar}{\Delta p} + \alpha' \frac{\Delta p}{\hbar}$$

This implies an minimum uncertainty in position of the order of $$~ \sqrt{\alpha '}$$

13. May 20, 2010

### ZapperZ

Staff Emeritus
Certainly. This is similar to having plane wave states with the momentum having delta functions, while the position values can be anything. There's nothing, in principle, that prevents this.

We may be puffing and huffing over this thread for naught. The OP hasn't responded or participated since posting this.

Zz.

14. May 20, 2010

### japrufrock

Not for naught! I think I see it more clearly now. Form a theoretical point of view, there is no limitation in having perfectly sharp position states. But in practice that implies an infinite energy.

What about the creation of a particle anti-particle pair when you try to do that 'sharpening'? Is that a theoretical limitation in relativistic QM?

About delta funtions, for what I can remember they are just idealizations of physical states in the limit of Delta x=0, they are not real functions representing actual states, isn't it?

thanks all

15. May 20, 2010

### ZapperZ

Staff Emeritus
It does? Where do you get this from?

Are you confusing the SPREAD in a value versus the value itself?

Zz.

16. May 20, 2010

### Fredrik

Staff Emeritus
The uncertainty of an observable in one of its eigenstates is 0, as you undoubtedly already know. And QM includes an axiom that says that a measurement of an observable leaves the system in an eigenstate of that observable, so the answer to your question is 0.

But position isn't an observable, at least not according to the books that go into detail about these things. (For example "An introduction to the mathematical structure of quantum mechanics: A short course for mathematicians", by F. Strocchi, and "Mathematical theory of quantum fields", by H. Araki. I have only read a few pages in each). Each measuring device that performs a "position measurement" does so with a finite accuracy, and is therefore represented mathematically not by "the position operator", but by some smoothed out version of it. (I don't know the exact details. I'm going to have to investigate this further).

17. May 20, 2010

### Fredrik

Staff Emeritus
Are my arguments so obviously flawed that they don't even need to be addressed? There are issues with the definition of "state", translation invariance, how to get a "nice" wavefunction to change into a delta function, etc.

At the very least, we'd have to consider a formulation of QM based on the mathematics of rigged Hilbert spaces instead of the mathematics of Hilbert spaces before we can even use terms like "sharp position", and that may not solve all the issues.

Last edited: May 20, 2010
18. May 20, 2010

### ZapperZ

Staff Emeritus
Er.. if momentum is an observable, then why isn't position?

I can easily transform into momentum space, and THEN the status of p and x are now reversed.

Zz.

19. May 20, 2010

### Fredrik

Staff Emeritus
Everything I said about position holds for momentum as well. In particular, they're both unbounded operators, and neither of them is a mathematical representation of an observable. Measuring devices designed to measure position and momentum are represented by bounded operators constructed from the position and momentum operators, not by the position and momentum operators themselves.

20. May 20, 2010

### ZapperZ

Staff Emeritus
Then there's something flawed with your idea. I can easily make a set of plane-wave states that are eigenstates of the momentum operator. Same deal as with the energy eigenstate example that I had earlier. Not only that, I can also make it so that the momentum operator commutes with the Hamiltonian! So if I have a non-degenerate state, I've shown that the eigenvalues of both p and H can be determined simultaneously with equal accuracy.

Zz.

21. May 20, 2010

### Fredrik

Staff Emeritus
Huh? Plane waves aren't state vectors, and if you mean something like

$$\int_{-\infty}^\infty g(p)e^{-ipx}dp$$

with a g that's nice enough for this to be a state vector, then it doesn't have a well-defined momentum.

22. May 20, 2010

### Count Iblis

I agree with Fredrik here. What you can do is put the whole system in a finite volume and impose some appropriate boundary conditions. Then everything is well defined.

23. May 20, 2010

### Count Iblis

Another obvious issue with sharp position measurements is that a single particle formulation of QM will have to break down the moment you attempt to locate a particle to within a region the size of the Compton length.

24. May 20, 2010

### ZapperZ

Staff Emeritus
No, I mean as in the Wannier function.

Zz.

25. May 20, 2010

### dx

I interpreted the question in the thread title to mean "is there a limit to how sharp a position a state can have?", and the answer to that is clearly 'no', since whether you want to include delta functions in the state space or not, the whole 1-paramenter family of functions fε defined by

fε(x) = 1/ε if x is in [0, ε]
fε(x) = 0 otherwise

are well defined, and ε can be as small as you want.

Last edited: May 20, 2010