# Heisenberg Uncertainty Principle - Why Is It Real?

1. May 13, 2010

### mjacobsca

From what I understand, the Heisenberg Uncertainty Principle means that we can never know the position and velocity of a particle at the same time, as the measurement of one necessarily affects the measurement of the other. Great. I understand that theory, and it makes perfect sense.

But what I can't understand is why this principle is used to define degenerate states, the inability of two electrons to occupy the same space at the same time, and other exclusive properties?

As I see it, the principle is a problem with measurement. OUR measurement. Or the limitations of our measuring strategies. But just because we can't measure something doesn't mean it can't exist, can it? Just because we can't measure position and velocity together doesn't mean the particle doesn't have both of these at the same time. Why does the fact that we can't measure two variables simultaneously mean that particles can't occupy the same space, measurable or not? How does this all expand to electron degeneracy?

2. May 13, 2010

### ansgar

no that is Paulis exclusion principle! you must have mixed those two concepts :)

3. May 13, 2010

### alxm

No, it's a fundamental property of quantum mechanics.

4. May 13, 2010

### A Dhingra

'' "As I see it, the principle is a problem with measurement. OUR measurement.""
even i agree with u... that calling this to be the base of quantum mechanics is a good excuse so that science is not questioned.....and true if we can't measure doesn't really mean it is not there.......like we cannot measure anything about god....but people still believe it exists ....
but as far as i know it cannot explain exclusion because it says velocity and position of a particle cannot be determined simultaneously....and exclusion principle of pauli is about these two quantities........

5. May 13, 2010

### dx

Actually, it's a little more subtle than that. "We cannot know the position and momentum of a particle simultaneously" is a sentence that we often use, for convenience, but it should not be interpreted in terms of classical pictures, where you imagine the particle having such attributes that we cannot "know". The problem is not about knowing, but about the meaning that we can attach to that knowledge. The point of the uncertainty principle, and the closeley related notion of complementarity, is that in any situation where one can make meaningful use of the notion of 'momentum conservation' necessarily excludes the use of concepts related to space-time coordination of the phenomenon.

A good example is the double slit experiment. Here, we let a large number of electrons fall on a photographic plate, with a double-slit screen in between. In any situation where the interference pattern appears, there is no possibility of application of momentum conservation laws to the exchange of momentum between the slit-screen and the electron. This is because in such situations (i.e. when the interference pattern appears), one must necessarily give up a control of momentum of the slit-screen by ridigly attaching it to the space-time frame. It is important to keep in mind that any object that we consider to be 'part of the system that is being observed' are subject to the uncertainty relations. If we want to control exchange of momentum with the plate, then it is part of the observed system, and the extent to which that control is possible is related reciprocally to the extent to which at the same time a space description is possible.

In the end, its all about wave-particle duality. The propogation of radiation in free space needs the wave picture, but the interaction of radiation with matter needs the pacticle picture (as was shown beyond doubt in the photo-electric and compton effects). The Heisenberg uncertainty relations in a sense show that this type of complementary description is consistent with the mathematical formalism of quantum mechanics.

Last edited: May 13, 2010
6. May 13, 2010

### mikeph

Two different concepts. the HUP does not say you cannot have two particles in the same space. That is a strange way of saying the Pauli exclusion principle which is something different.

HUP is a result of the non-commutativity of the position and momentum operators; it comes directly from the maths of operating on wavefunctions. All it says is that position and momentum cannot be simultaneously known because an eigenfunction of the position operator must have an undetermined momentum, and vice versa.

The Pauli exclusion principle is something very different, and most often relates to steady states which are eigenfunctions of the energy operator, and therefore have non determined position NOR momentum. So HUP would not apply/be needed.

7. May 13, 2010

### Born2bwire

I would not say that they can't be known. You can perfectly well measure both at the same time. However, there will be a statistical spread in the measurements that occurs regardless of the accuracy and precision of your measurements. It comes down to the fact that, say with position and momentum, you can have multiple momentum measurements associated with a given position measurement (or set of position measurements).

8. May 13, 2010

### mjacobsca

Quoting specifically from the book I am currently reading, "The Five Ages of the Universe", pg 48:

Is this quote wrong? I don't get how a problem with measuring things actually translates to electron degeneracy.

9. May 14, 2010

### Born2bwire

I would say that it is although I do not know what they mean by "electron degeneracy." The Pauli Exclusion principle deals with the fact that electrons, among other fermions, cannot occupy the same state and thus cannot be degenerate. Of course there are other degeneracies, like energy degeneracies which occur in atoms. The Pauli Exclusion principle is not derived from Heisenberg's Uncertainty Principle. As far as I recall, it isn't derived.

And again, the uncertainty principle is not a measurement problem. It precludes any kind of error or interaction with the system by measurement. It is a consequence of the mathematics of quantum mechanics. If I am observing different colored balls roll up and down a hill, at a given measurement I might find that there is a green ball 1 m up the hillside. Another measurement of an identical system may give me a blue ball 1 m up the hillside. In this system, we can observe different colored balls at the same position because the colors indicate properties of the balls that do not preclude them from being at the same position on the hill. When we observe a particle in a given energy eigenstate, there are a range of valid position and momentum states that the particle can be in. However, the eigenspace of the position is not the same as the eigenspace of the momentum. This means that a given position state can have multiple valid momentum states and vice-versa. So when you measure a system and find that the system has a certain position state, there are multiple momentum states that you can measure that are associated with the position state. This gives rise to a spread in the simultaneously measured momentum measurements. The uncertainty principle gives a relationship between the spreads in the measurements of these non-commuting observables. If they were commuting, then they would share the same eigenspace which means that if I find observable A in state X, then observable B would only be seen in state X.

10. May 14, 2010

### alxm

It's derived from the (defining) antisymmetry requirement from fermions, which in turn can be shown to be a consequence of having half-integer spin.

11. May 14, 2010

### bapowell

With respect to the HUP, I agree with dx that the issue transcends one of measurement. I've always found a simple explanation by Griffiths to be especially satisfying:

Consider a pulse traveling down a rope. The pulse is a localized disturbance, hence, it has a reasonably well-defined position. But, what is its wavelength (momentum)? A localized particle thus has an uncertain momentum.

Now consider a standing wave. Now there is a well defined wavelength (momentum). But...now where is it? A particle with well-defined momentum has uncertain position.

12. May 15, 2010

### Born2bwire

See, I thought that half-integer spin particles requiring anti-symmetry was the Exclusion Principle, or is the Exclusion Principle just an application of this to electrons specifically?

13. May 15, 2010

### alxm

Well the exclusion principle is usually expressed as "No two electrons in an atom can have the same quantum numbers"
or more generally "No two fermions can occupy the same state at the same time".

Antisymmetry is a more fundamental property, which is relatively easy to show (if you're not too strict):
If you exchange the coordinates (including spin) of two identical particles, all observables must remain the same, so at most, the wave function has changed by a phase factor (let's just call it q):
$$\psi(x_2,x_1) = q\psi(x_1,x_2)$$ but two exchanges have to give back the original wave function so: $$q\psi(x_2,x_1) = q^2\psi(x_1,x_2) = \psi(x_1,x_2)$$, so $$q^2 = 1, q = \pm 1$$
The only possibilities are that the wave function changes sign on exchange (fermions) or not (bosons). With relativistic theory, one can show that this comes from having half-integer spin and bosons integer spins.

So this is a boundary condition that must be enforced on the solutions to the Schrödinger equation. Now, if you consider (for the sake of simplicity) a system of two non-interacting particles,
the S.E. is separable and so the total wave function can be expressed as a product of two single-particle wave functions (Hartree product):
$$\Psi = \phi_1(x_1)\phi_2(x_2)$$ But this does not satisfy anti-symmetry!

So you have to form an anti-symmetrized product (Slater determinant):
$$\Psi = \phi_1(x_1)\phi_2(x_2) - \phi_1(x_2)\phi_2(x_1)$$
Per the principle of superposition, this is also a valid solution to the S.E. if the former equation was. You also have here that $$\phi_1 \neq \phi_2$$.

So this gives you the more general form of the Exclusion principle.

More specifically, since you can consider spin to be an independent variable, you can seperate the single-particle wave functions into a spin-part and spatial part:
$$\phi(x) = \chi(\vec{r})\alpha(s)$$ or $$\phi(x) = \chi(\vec{r})\beta(s)$$ (depending on the spin):
$$\Psi = \chi(\vec{r_1})\alpha(s_1)\chi(\vec{r_2})\beta(s_2) - \chi(\vec{r_2})\alpha(s_2)\chi(\vec{r_1})\beta(s_1)$$

So no two electrons with the same quantum numbers (which define the spatial orbital), and two electrons per spatial orbital, and we have the more specific form.
(and an important result as far as atomic properties are concerned)

14. Jun 12, 2010

### friendkey

Aren't we still dealing with measurement here as....
1. There cannot be any pulse without a wavelength.
2. The standing wave occupies a definite instantaneous location which changes....in most cases, too rapidly.

In either of examples, it is limitation of our eyes(or instrument), in terms of field of view and refresh rate.
Just because we can't see (by which I mean measure) something, doesn't mean it is not there.

I agree with the part that one quantity cannot be measured without changing the other, but it should not be regarded as a fundamental property.

15. Jun 12, 2010

### bapowell

So then what's the wavelength of a pulse? A pulse is not wave-like...how does one ascribe a wavelength to it?

I disagree with this. The energy of a standing wave is distributed across the rope -- it is not localized at any time.

16. Jun 12, 2010

### friendkey

Aren't you beginning with "not a measurement problem" and ending with "spreads in the measurements of..." statement?

17. Jun 12, 2010

### friendkey

On the contrary, transverse pulse on a rope is either a "half wave" or a "wave". There is a significant amount of displacement attached to each point on rope which changes with time.

Some images might help with a little recall of elementary physics.
http://www.tutorvista.com/content/physics/physics-iii/waves/transverse-and-longitudinal-waves.php

Agreed with your disagreement. Total energy of a standing wave can never be localized, but that really doesn't mean I cannot calculate the energy attached to Simple Harmonic Motion of any point on rope at any given instant.

18. Jun 12, 2010

### bapowell

What's the wavelength of a delta function moving down a rope? Thanks for the elementary physics refresher...I really needed that.

Of course you can calculate the energy density at any point at any instant. But the point of this example is that position of a particle cannot be local so long as the function which is describing it has a well-defined wavelength (wavefunction) (ie the energy of the simple harmonic oscillator is never fully localized). That's all. It's a fundamental duality that has nothing to do with measurement.

19. Jun 13, 2010

### friendkey

Do we really need to limit our understanding of HUP in "Delta Potential Well" and Schrodinger Equation? To explain this with our example, we should learn to generate one delta function on a rope first. (Please read next lines before getting upset!)

Please do not get upset. I humbly apologize if you found my words offending. If I will not question, argue or quote references, it will become a religion instead of science. It is a fairly onesided debate where HUP and hence you are right, whereas I am trying to clear my doubts. You are maestro, I am discipulo.
Makes sense with that duality. If rope is straight (Particle nature), there is no uncertainty. "In Comparison", when the rope is carrying "any" wave(Wave nature), there is uncertainty as it could be anywhere within the amplitude and energy distributes itself all over it.

But that stays very fulcrumed around words "in comparison", isn't it? Even if I cannot observe the energy being localized, I know that it is in that one rope and not in space around it, and the rope changes its position.

Is HUP about consequence of introducing dual nature and nothing more?

Pardon me for being a stubborn learner! Thank you for your responses.

20. Jun 13, 2010

### bapowell

Who said anything about delta function potential wells? I'm talking about a pulsed disturbance traveling down a rope as an analogy for the wave function of a particle with a perfectly known position. Such a disturbance does not have a wavelength -- it's a localized form. Therefore, the analogy says that the wave function of a particle with well defined position has no well defined wavelength -- ie has no well defined momentum.

I think you misunderstand the analogy. A straight rope is not meant to represent a particle. In fact, we can do away with the rope all together. A localized particle has a delta function-like wavefunction in the position representation. The Fourier transform brings us to the momentum representation -- it is a constant function in momentum space -- ie has no well defined momentum. The converse is also true -- a particle with well-defined momentum is a delta function in momentum space, and is a constant in position space. The rope is simply a vehicle for making this more tangible and pedagogical -- I read it in an undergrad quantum mechanics text book.

The 'dual nature' is a fundamental property of Fourier transforms, and so, it's built into the formalism even before we start talking about physics! I guess the devil is in the interpretation. The trouble is in understanding the physics behind the fact that quantum wave functions have this property -- some point towards measurement, some suggest it's fundamental to the nature of things.