# Uncertainty principle

1. Sep 8, 2009

### ╔(σ_σ)╝

My Solid state physics class just end and my professor referred to the Uncertainty principle about how you can not know the momentum and position of an electron at a given instant but, you can only know one.

My question is why ? Why is this phenomena so ? I was going to ask my prof but I had to go to another class.

I was hoping someone on here can tell me why and show me a proof, if possible. I have gotten into a bad habit of not taking facts until I know why or have seen the proof.

2. Sep 8, 2009

### alxm

It's so because momentum and position operators do not commute.

3. Sep 9, 2009

### olgranpappy

next class, don't go to another class--ask your professor.

4. Sep 9, 2009

### Fredrik

Staff Emeritus
It's not just that you can't know both at the same time. It doesn't have both at the same time. Any state that has a well-defined position doesn't have a well-defined momentum, and vice versa. In classical mechanics, the answer to the question "where is the particle now?" is always an ordered triple (x,y,z) of real numbers. In QM, the best answer is...the wavefunction, i.e. a function from $\mathbb R^3$ to $\mathbb C$ (when the time coordinate is held fixed), and if the wavefunction is sharply peaked (i.e. if the position is well-defined), then its Fourier transform is smeared out over a large region, which makes the momentum ill-defined.

I recommend that you read the first few pages of https://www.amazon.com/Introduction...=sr_1_1?ie=UTF8&s=books&qid=1252499987&sr=8-1 (use the "look inside" feature), if you haven't already, to make sure that you understand wave functions and their interpretation.

There is an argument, which I believe is originally due to Heisenberg, that basically says that the only way to measure the position of a physical system is to have it interact with other particles, which then "nudges" the system, making our knowledge of the momentum less precise. Lots of books include a slightly more detailed version of this argument. Here's one I saw recently. Read the paragraph on page 40 that starts with "As argued by Heisenberg" (and don't try to read the rest of the book until you've had at least one more QM class and 3-4 more math classes, if you want to stay sane).

This argument gets a lot of negative comments in these forums, because people who have studied QM know that the version of the uncertainty principle that actually appears in QM has nothing to do with "nudges" or "knowledge" about the position and momentum. It's much more profound than that. So some of those negative comments are valid, because the argument gives you the wrong idea about what QM actually says. However, I think a lot of people (including me, until recently) have failed to understand the significance of this argument. It plays the same role in the development of QM as Einstein's postulates in the development of SR. They are both somewhat sloppy, but insightful statements that can help you guess what mathematical model to use in the new theory you're trying to find. Einstein's postulates eventually gave us Minkowski space, and Heisenberg's argument eventually gave us Hilbert space.

Let me clarify an important detail. Heisenberg's argument leads to an approximate inequality called the uncertainty principle. The theory that was eventually found using Heisenberg's insights is quantum mechanics. QM includes an exact inequality which is also called the uncertainty principle. That one is what people in this forum have in mind when they talk about the uncertainty principle. That's why Heisenberg's argument gets more negative comments than it deserves. People just see an argument that doesn't use any of the concepts that are defined by QM and a result that has a very different interpretation than what they call the uncertainty principle, so to them it looks like complete nonsense.

This is a statement and proof of the modern version of the uncertainty principle (i.e. the one that actually appears in the theory of QM):

If $A=A^\dagger$ and $B=B^\dagger$, then

$$\Delta A\Delta B\geq\frac{1}{2}|\langle[A,B]\rangle|$$

where the uncertainty $\Delta X$ of an observable $X$ is defined as

$$\Delta X=\sqrt{\langle(X-\langle X\rangle)^2\rangle}$$

Proof:

$$\frac{1}{2}|\langle[A,B]\rangle| =\frac{1}{2}|(\psi,[A,B]\psi)| =\frac{1}{2}|(B\psi,A\psi)-(A\psi,B\psi)| =|\mbox{Im}(B\psi,A\psi)|$$

$$\leq |(B\psi,A\psi)| \leq \|B\psi\|\|A\psi\| =\sqrt{(B\psi,B\psi)} \sqrt{(A\psi,A\psi)}=\sqrt{ \langle B^2\rangle}\sqrt{\langle A^2\rangle}$$

The second inequality is the Cauchy-Schwarz inequality Now replace A and B with $A-\langle A\rangle$ and $B-\langle B\rangle$ respectively. This has no effect on the left-hand side since the expectation values commute with everything, so we get

$$\frac{1}{2}|\langle[A,B]\rangle| \leq \sqrt{\langle (B-\langle B\rangle)^2\rangle}\sqrt{\langle (A- \langle A\rangle)^2\rangle}=\Delta B\Delta A$$

That's how I wrote it down in my personal notes, but I should clarify a few things. I'm writing scalar products as (x,y) here. The expectation value $\langle X\rangle$ is defined as $(\psi,X\psi)$, or in bra-ket notation $(|\psi\rangle,X|\psi\rangle)=\langle\psi|X|\psi \rangle$, so it would be more appropriate to use the notation $\langle X\rangle_\psi$ for the expectation value, and $\Delta_\psi X$ for the uncertainty. We should really be talking about the uncertainty of X in the state $\psi$.

By the way, you have an awesome name.

Last edited by a moderator: May 4, 2017
5. Sep 9, 2009

### ueit

You can know both position and momentum of an electron at a given time. You cannot predict them, but you can find out how they were in the past. Your professor is wrong.

6. Sep 9, 2009

### kote

Seemed like this professor was giving a simple statement of the HUP. Are you denying the HUP?

Fredrik has it right.

7. Jan 30, 2011

### angie_liamzon

can you please help me regarding the mathemamical proof of heisenberg's uncertainty principle using fourier transform? i don't know where to start. thanks.

8. Jan 30, 2011

You know, I'm aware of what you describe, and what I always couldn't understand is that how all this works. We know that Heisenberg's argument(how the observer is entagnled with the observable system) is "correct", from a phycisal standpoint, and it gives a calculation of the momentum-position "uncertainty". Yet we proceed to say that, in modern QM, this uncertainty is not due to effects of measurement(and affecting the system), but an inherent property of quantum systems. Right? Yet, again, Copenhagen's interpretation states simply, that we cannot say anything about a system if it's not observed; that is, all we can calculate are the observables states anyway. It seems to me that this version of QM uncertainty is a general mathematical formulation of the initial Heisenberg argument, as applied to all quantum systems, without having to deal explicitly with the "nature" of each system and the devices used to measure it? Am I right about this?

9. Jan 30, 2011

### Staff: Mentor

Maybe this will help:

Fourier Transforms and Uncertainty Relations