# Proof of Heisenberg's Uncertainty Principle

1. Mar 8, 2005

### Ed Quanta

Is there any mathematical or physical proof or derivation of Heisenberg's Uncertainty principle out there? Can someone send me a link to one or provide a proof if it isn't too complicated? I know that in quantum mechanics if two operators don't commute then we can't measure both of these simulataneously. Why does this correlation exist?

2. Mar 8, 2005

3. Mar 8, 2005

### marlon

Because Mother Nature has chosen to behave like this

marlon

4. Mar 8, 2005

### marlon

Besides, check out the original manuscript in my last entry of my journal

marlon

5. Mar 8, 2005

### Ed Quanta

I don't really follow this derivation. Why does he have to show that (deltaA')^2=(deltaA')^2?

And how do we know |<A'psi,B'psi>-<B'psi,A'psi>|=2|Im<A'psi,B'psi>|

6. Mar 8, 2005

7. Mar 8, 2005

### dextercioby

The nonmathematized version (the one involving wave-mechanics formalism) is in the beginning of Davydov's book,IIRC.Anyway,every book on QM has the proof for the generalized version:

$$\Delta \mathcal{A}\cdot\Delta \mathcal{B}\geq\frac{1}{2} |\langle[\hat{A},\hat{B}]_{-}\rangle _{|\psi\rangle}|$$

Daniel.

Last edited: Mar 8, 2005
8. Mar 8, 2005

### masudr

Daniel, why is there a minus sign as the subscript of the commutators? I never understood that.

9. Mar 8, 2005

### dextercioby

Because it is the commutator...?
$$[\hat{A},\hat{B}]_{-}=:\hat{A}\hat{B}-\hat{B}\hat{A}$$

and that's how it's elegantly specified the fact that one speaks about commutators of (linear) operators...

Daniel.

10. Mar 8, 2005

### masudr

Oh... OK. I thought the commutator automatically implied a negative sign, and we have the anti-commutator for the version with the plus sign. Ok never mind about that.

I have another question though. If we reverse the order of the commutator, i.e.

$$[\hat{A},\hat{B}] = -[\hat{B}, \hat{A}]$$, we get a minus sign in the uncertainty. But is that of any significance if the product of the variances is negative as opposed to positive?

11. Mar 8, 2005

### dextercioby

If you haven't seen so far,there's a modulus after performing the average of the commutator on the (pure) quantum state $|\psi\rangle$

Daniel.

12. Mar 8, 2005

### masudr

Oh yes, of course. Sorry, it's 04:30AM here in England, and my mind's not thinking straight!

13. Mar 9, 2005

### Ed Quanta

If I can understand this, I understand the derivation of the Uncertainty Principle. But I can't make sense of this one line in the original link provided by Marlon.

14. Mar 9, 2005

### Galileo

Generally:

$$\langle \psi|\phi \rangle = \langle \phi|\psi \rangle^*$$
where the * denotes complex conjugation.

So

$$\langle A\psi|B \psi \rangle=\langle B\psi|A \psi \rangle^*$$

For any complex number $z$ we have $z-z^*=2i\Im(z)$.

15. Mar 9, 2005

### dextercioby

This is very simple.

$$\langle \psi|\hat{A}\hat{B}|\psi\rangle =:u\in \mathbb{C}$$ (1)

Then,using the property:

$$\langle \psi|\hat{A}\hat{B}|\psi\rangle = \langle \psi|\hat{B}\hat{A}|\psi\rangle ^{*}$$ (2)

,we can write the LHS of the equality u wish to prove as:

$$|u-u^{*}|$$ (3)

The RHS of the equality you want to prove is

$$2|Im \ u |$$ (4)

Take the generic algebraic for "u"

$$u=:a+ib$$ (5) $$\Rightarrow u^{*}=a-ib$$ (6)

and then

$$|u-u^{*}|=|2ib|=2|b|=2|Im \ u|$$ (7)

q.e.d.

Daniel.

Last edited: Mar 9, 2005
16. Mar 9, 2005

### Tom Mattson

Staff Emeritus
He's not showing that. He's showing that $( \Delta A )^2=< \psi , A'^2 \psi >$. The fact that he ended up with $( \Delta A )^2=( \Delta A )^2$ simply means that he completed the proof.

17. Mar 9, 2005

### Ed Quanta

Thanks a lot everyone. Much appreciated.

18. Mar 16, 2005

### Antiphon

The very simplest answer is this.

If you want to determine the frequency of a signal by counting
pulses, your frequency determination gets more accurate the longer
you count. In QM, frequency determies energy so the less time you
have to count the frequency, the less certain you will be of the
exact energy.

19. Mar 16, 2005

### dextercioby

Physicists are not really fond of descriptive explanations.They need maths.

Daniel.

20. Mar 16, 2005

### Antiphon

I don't use maths because some of the people on this board are in dire need of descriptive
explanations. I'll leave the maths to the Physicists amoung you.

Edit: ..and let's not forget that there are many who could understand the
basics of much of physics but are confounded by the equations. It's partly
for them that descriptives are healty. The other group who needs it are the
naive budding physicists who have been made to beleive that descriptive
explanations are at odds with good science. There are times when only
the math will do and descriptives aren't helpful, like the nature of electronic
spin. But there are other times (like the uncertainty principle) when
whole philosopies get built on the opacity of the equations. Quantum Uncertainty
is simple to comprehend visually for simple configurations and it is healthy, instructive
and right to do so.

Last edited: Mar 16, 2005