# I Eigenstate & probability

1. Nov 9, 2016

### bluecap

Why is the probability of finding a particle in an eigenstate of position zero and not one?

When we say we have located a particle at a particular position - why is it always in a superposition of position eigenstates about that position. But still the probability should not be zero.

I need the math. Thank you.

2. Nov 9, 2016

### Staff: Mentor

The underlying reason is that there is no such thing as a "position eigenstate". More precisely, a state with a precisely defined position is not normalizable; it is not contained in the set of square integrable functions, which is the applicable Hilbert space.

Mathematically, a position eigenstate is described by a Dirac delta function; a position eigenstate with eigenvalue $x$ is $\delta(x)$. But the Dirac delta "function" is not actually a function; heuristically, this is because it has zero value at all points but one, and an infinite value at that one point, whereas a true function can only have finite values at all points. To see why $\delta(x)$ has an infinite value at one point, consider the general rule that probabilities must add up to one; if we apply this to a position eigenstate, we find that we must have (heuristically)

$$\int_{- \infty}^{\infty} \delta(x) dx = 1$$

If you dig into how integrals are defined, you will find that, if $\delta(x)$ only has a nonzero value at one value of $x$, it cannot have any finite nonzero value at that $x$ and still satisfy the above.

More here:

https://en.wikipedia.org/wiki/Dirac_delta_function

Strictly speaking, we can never say that. Physically, the reason is the uncertainty principle. Mathematically, that principle says, heuristically,

$$\Delta x \Delta p \ge h$$

where $h$ is Planck's constant, i.e., a nonzero real number. But if we could locate a particle at a single definite position, then $\Delta x$ would be zero, and the above uncertainty relation could not possibly be satisfied. So we can't actually locate a particle at a single definite position. The best we can do is to measure its position to be within some small but finite range.

3. Nov 9, 2016

### bluecap

Thanks. Simon wrote it "When we say we have located a particle at a particular position - it is always in a superposition of position eigenstates about that position. That is how we describe it in maths right?" Well.. let's say we rewords it into "When we say we have located a particle within some small but finite range - it is always in a superposition of position eigenstates about that some small but finite range". Is this correct statement?

Why is it always in a superposition of position eigenstates about that small but finite range.. its related to your statement above "But the Dirac delta "function" is not actually a function; heuristically, this is because it has zero value at all points but one, and an infinite value at that one point, whereas a true function can only have finite values at all points"? Or not since we can't actually locate a particle at a single definite position and this "always in a superposition of position eigenstates??" no longer hold?

4. Nov 9, 2016

### Staff: Mentor

Since any state which is not an eigenstate is a superposition of eigenstates, then yes, this is true.

Because of the uncertainty principle.

5. Nov 9, 2016

### bluecap

Ok. In classical macroscopic objects, this is also true about positions of quantum particles being in some finite range and hence in superposition of eigenstates? So macroscopic objects are coarse graining of these microscopic degrees of freedom and we can say macroscopic objects are coarse graining of superposition of position eigenstates?

6. Nov 9, 2016

### Staff: Mentor

This is related to the uncertainty principle in the sense that in any mathematical model that is consistent with the uncertainty principle, the "position eigenstates" must have properties like those of the Dirac delta function.

7. Nov 9, 2016

### Staff: Mentor

Since positions of quantum particles are always in superpositions of position eigenstates (because the eigenstates themselves are not normalizable), then yes, this is true.

What do you mean by "coarse graining"? Can you give a mathematical expression of this?

8. Nov 9, 2016

### dextercioby

The coordinate observable has a purely continuous spectrum, hence the probabilistic interpretation we owe to M. Born needs adjustment.
For a point a in the continuous spectrum of A, the absolute probability of measuring it is ill-defined, as the theory of quantum mechanics only tells you the probability of measuring a in the open interval (b,c). If the interval (b,c) is the whole spectrum, then the probability is 1, of course. This could be rephrased to saying that the coordinate observable is "not sharp" (the b is strictly smaller than c, can never be equal). To obtain a true probability value, you need not only a spectral value, but also a spectral interval containing that value. In case of a discrete (countable) spectrum, the interval is not needed anymore, the spectral value is enough to obtain absolute theoretical probabilities.

To say "we have located the particle at a particular position" (call it "a" the value and A the coordinate observable having "a" in its spectrum) means that, if you measured A a 1.000.000 times on 1.000.000 identically prepared systems, you've got a statistical spread of measurement values around "a" with >99% probability of measuring "a" in a very small (yet non zero) interval around it. The larger the interval becomes (thus increasing even more the >99% probability towards 100%), then the less true it is that "we have located the particle at a particular position".

Last edited: Nov 9, 2016
9. Nov 9, 2016

### bluecap

By the way. Is this valid also for all observable like spin or energy.. that if the particle state vector is described by a superposition of eigenstates for *any* observable, then the particle does not have a determined value for that observable? No exception?

When you look at a wall, you won't see the individual atoms but the whole macroscopic wall with less degrees of freedom.. is this not called "coarse graining"? What is your definition of coarse graining then?

10. Nov 9, 2016

### Staff: Mentor

Yes.

All of the degrees of freedom are still there, but you are only measuring a very small number of them; and the degrees of freedom you are measuring are not degrees of freedom of any individual atom. For example, if I say the wall has a certain "position", I really mean something like: the center of mass of the wall is at some point in space, and the dimensions of the wall, centered on that center of mass, have some particular set of values--and all values having some small but nonzero range associated with them instead of being single definite numbers, since I can only measure things with a finite accuracy.

Another way of looking at this is: if I make a measurement on a single quantum object (say an electron), I can often pin down its state exactly, or at least to within a very small range. For example, if I measure an electron's spin along a certain axis, I get a definite result--spin up or spin down--corresponding to a definite spin state for that electron.

But if I make a measurement on a wall, I get something like what I described above, which does not pin down the full quantum state of the wall, including all of its atoms, very well at all; there will be a huge number of possible quantum states for the wall that are consistent with the measurement result. And I label this whole huge collection of possible quantum states of the wall with a single "classical" label, something like: the wall is "here" (at a particular position with a particular size). In this sense, yes, by assigning a classical "state" to the wall I am coarse-graining over the space of all possible quantum states, partitioning that space into subspaces, each containing a huge number of individual quantum states, and each subspace labeled with a "classical" state label.

11. Nov 9, 2016

### bluecap

Why can't you pinpoint energy or spin or other observable exactly.. what HUP would be violated?

12. Nov 9, 2016

### Staff: Mentor

That isn't what I said, and it isn't the question you asked. You asked, if the particle is in a superposition of eigenstates for any observable, is it true that the particle does not have a determined value for that observable. That is true, but note carefully that it's a conditional statement: if a particle is in a superposition.

The question it looks like you actually wanted the answer to is this: I said that a particle can't ever be in a single position eigenstate (i.e., not in a superposition of position eigenstates) because that would violate the uncertainty principle. Is this also true of the eigenstates of every observable? The answer to that question is no: there are plenty of observables for which particles can be in eigenstates without violating the uncertainty principle. Spin is one of them.

Energy is more problematic, because the energy observable--the Hamiltonian operator--also governs the time evolution of the system. So strictly speaking, if a system is in an exact eigenstate of the Hamiltonian, it can never change and nothing can ever happen to it. But that means we can never measure it, because measurement would be a happening--something would have to change in order to record the measurement result, and we would have to include that something in our analysis, and that would mean the system as a whole, including both measured quantum object and measuring apparatus, would not be in an eigenstate of the full Hamiltonian, including the interaction involved in the measurement.

Discussions of, for example, the energy levels of electrons in atoms being "stationary states" gloss over this; what is actually meant is that an electron which is in a definite energy level is in a eigenstate of a Hamiltonian $H_0$ which only describes the atom and the electron--not any other interactions involved. But in any real atom, there will be other terms in the full Hamiltionian $H$ besides $H_0$, terms that describe, for example, interactions that allow the electron to emit or absorb photons and jump to a lower or higher energy level. The electron is never in an eigenstate of $H$, because if it were, as above, it could never change. Similar remarks apply to any quantum system.

13. Nov 9, 2016

### bluecap

Thank you. For multi particle systems (such as multiple entangled particles or larger macroscopic object). Which is the case that you can apply one eigenstate to all of them at same time by considering the center of mass (etc.) in some finite range (energy, position or other observable)? I found this statement in Zurek article https://arxiv.org/pdf/0707.2832.pdf
"Therefore, instantaneous eigenstates of the reduced density matrix of the open systems will in general not coincide with the preferred pointer states singled out by einselection".

Note reduced density matrix of the open system is not just one particle but multi particles via decoherence so how can it mentioned "instantaneous eigenstates of the reduced density matrix" (I understand the concepts of density matrix).. does it mean center mass of them or each particle?

14. Nov 10, 2016

### atyy

No. A particle can be at a definite position upon measurement.

15. Nov 10, 2016

### Staff: Mentor

What measuring device gives us a definite position, without uncertainty?

16. Nov 10, 2016

### atyy

That's a different practical consideration. The use of the HUP by Peter Donis is a wrong application of a principle. The HUP does not prevent a particle from having a particular position.

It is true that the wave function of a particle cannot be a position eigenstate, because position eigenstates are not admissible wave functions. It is also true that the probability of a particle being at any particular position is zero, because position is a continuous variable. However, when the position of a particle is measured, the particle does have a particular position.

The HUP talks about measurements on two different ensembles of particles, and the variance of position in one ensemble, with the variance of momentum in a second ensemble.

Last edited: Nov 10, 2016
17. Nov 10, 2016

### Staff: Mentor

No, it isn't. A "position eigenstate" is not normalizable and hence is not actually a valid state at all.

What state is the particle in after the measurement? A heuristic treatment might say it is in a "position eigenstate" (for an instant, anyway, since such a state quickly evolves into a superposition of position eigenstates), but such a treatment would not hold up under scrutiny since, as above, a position eigenstate is not actually a valid state, and the particle has to be in a valid state after measurement.

18. Nov 10, 2016

### bluecap

I think you missed the message above Peter or it's so hard for this thread.

My question is simply if the system is composed of many components.. do you get the eigenstate of the system by the superposition of all the components or particles? For example for eigenstate of position in some finite range. Do you get superposition of all positions of all particles then get the eigenstate of the superposition?

19. Nov 11, 2016

### vanhees71

You are contradicting youself! Correct is that there is no position eigenstate, and the conclusion is that the position of a particle can never be accurately determined at point. That's indeed reflected in the uncertainty principle, $\Delta x \Delta p_x \geq \hbar/2$, telling you that the standard deviation of the position probability distribution is never 0.

Measuring the position is also always of finite resolution, no matter whether you use oldfashioned devices like a cloud chamber (the droplets condensing due to ionization due to a particle running through) or a modern pixel detector (where the pixels may be small and densely packed the better technology gets, but it's always of finite resolution). True is that you can measure the position of a particle in principle with arbitrary small uncertainty, but it's always a finite non-zero accuracy. The uncertainty relation tells you that you cannot prepare a particle with definite position in principle, but of course you can measure its position with much higher precision than the implied $\Delta x$, and this you must do to resolve the uncertainty $\Delta x$ with high significance.

20. Nov 11, 2016

### Staff: Mentor

This is muddled and I'm not sure I can disentangle it. "Superposition of all positions of all particles" is meaningless; that's not what a superposition is. "Eigenstate of the superposition" is also meaningless.

Let me try to describe how QM models a system with just two particles and see if that helps. We assume both particles have no spin so the Hilbert space only contains information about position. The single-particle Hilbert space has a basis of wave functions $\psi(x)$, i.e., functions of a single position $x$. The two-particle Hilbert space has a basis of wave functions $\Psi(x_1, x_2)$, i.e., functions of two positions $x_1$ and $x_2$. These wave functions give the probability amplitude for one particle to be at position $x_1$ and the other particle to be at position $x_2$.

Now, before we can even talk about eigenstates of this two-particle system, we have to know what observable we are looking for eigenstates of. Suppose we are talking about the "joint position" observable: i.e., we measure, simultaneously, the positions of both particles. Then the eigenstates of this observable will be states like $\delta(x_1) \delta(x_2)$; I'm being somewhat sloppy with notation here, but hopefully the meaning is clear that this is a state with one particle exactly at $x_1$ and the other exactly at $x_2$. And, just as with the single particle case, this state is actually not a valid state at all; it's not normalizable.

But there are also other observables we could talk about: for example, a "center of mass position" observable. Without trying to specify here exactly how we would measure such an observable, we can see that the center of mass position, assuming that both particles have the same mass, is just $(x_1 + x_2)/2$, i.e., the average of the two particles' positions. For a given wave function $\Psi(x_1, x_2)$, we can calculate the probability $P(X)$ of observing a center of mass position $X$. But note that a given center of mass position $X$ does not require any particular positions $x_1$ or $x_2$ for the two particles themselves; all it tells us is their average. So eigenstates of the center of mass position observable will not be eigenstates of position for either individual particle; if we write a center of mass eigenstate down in the position basis we have been using, it will be a superposition of an infinite number of possible wave functions $\Psi(x_1, x_2)$ that satisfy the constraint that $(x_1 + x_2)/2 = X$.