# General Uncertainty Relation between 2 operators

1. Dec 31, 2006

### Reshma

The general uncertainty relation between two observables A and B.
$$(\Delta A)^2(\Detla B)^2 \geq -{1\over 4}<[A, B]>^2$$

I have to prove the above relation using the definition of expection values etc.
The reference I use (Liboff) have this relation given as an exercise. But Gasiorowicz's book has given some useful hints on this relation. But I couldn't go all the way to prove this result.

Here is my attempt at it.
<Takes a deep breath>
$$(\Delta A)^2 = <(A - <A>)^2>$$
$$(\Delta B)^2 = <(B - <B>)^2>$$

Let U = A - <A> & V = B - <B> and consider $\phi = U\psi + i\lamba V \psi$
The uncertainties in A & B would be correlated only if the two operators do not commute.
Let A & B be Herimtian so U & V will also be Hermitian.
$$I(\lambda) = \int dx \phi^* \phi \geq 0$$
$$I(\lambda) = \int dx(U\psi + i\lamba V \psi)^*(U\psi + i\lamba V \psi) = \int dx \psi^* [U^2 + \lambda^2 V^2 +i\lambda[U,V]\psi$$
Using the defintion of $\Delta A & \Delta B$
$$I(\lambda) = \left((\Delta A)^2 + \lambda^2 (\Delta B)^2 + i\lambda<[A, B]>\right) \geq 0$$

I have to get rid of $\lambda$ to get my result. Anyone has good idea here??

BTW, Happy New Year to all the hardworking homework helpers in PF. Keep up the great work!

Last edited: Dec 31, 2006
2. Dec 31, 2006

### Gokul43201

Staff Emeritus
Looks like theres a few $\lambda$s gone missing. I see them in the source, though - you've omitted a "d" in "lambda". In the last line, you're missing a couple of expectation value brackets. Also, it's easier to do this whole thing in the Dirac notation (if you're familiar with that) - it avoids the need for unnecessary integrals.

First of all, I'd make one simplification, and write $\phi = U\psi - \lambda V \psi$, absorbing any factors of i into $\lambda$. It reduces the possibility of making sign errors down the way.

Next, choose $\lambda=UV*(V^2)^{-1}$. Plugging that in should land you with the correct form of the Cauchy-Schwarz inequality. And if you ask how you were supposed to anticipate this substitution, remember the goal - you want something that has a product of expectation values of the uncertainty (or variance) operators.

From there it's tricks with commutators and anti-commutators, noticing whether these are Hermitian or anti-Hermitian, and using the fact that you can write any operator as a sum of Hermitian and anti-Hermitian operators.

Last edited: Dec 31, 2006
3. Jan 2, 2007

### Reshma

I tried the above integral using Dirac's notation. I retained the same function though.

$$I(\lambda) = \int dx \phi^*\phi \geq 0$$
$$I(\lambda) = \int dx \phi^*\phi$$
$$= <\phi|\phi>$$
$$= <(U+ i\lamba V)\psi|(U+ i\lamba V)\psi>$$
$$= <U\psi|U\psi> + <U\psi|i\lambda V\psi> + <i\lambda V\psi|U\psi> + < i\lambda V\psi|i\lambda V\psi$$
$$= <U\psi|U\psi> + i\lambda<U\psi|V\psi> - i\lambda<V\psi|U\psi> + (i\lambda)(-i\lambda)<V\psi|V\psi>$$
$$= <U\psi|U\psi> + \lambda^2 <V\psi|V\psi> + i\lambda<[U, V]>$$
$$= (\Delta A)^2 + \lambda^2 (\Delta B)^2 + i\lambda<[A, B]>$$

I think it's correct, I double checked it. The hint given in my text is: Minimise $I(\lambda)$ and substitute the value of $\lambda$ in the last equation.
So it means taking ${{dI}\over {d\lambda}} = 0$ right?

Last edited: Jan 2, 2007
4. Jan 2, 2007

### Gokul43201

Staff Emeritus

If $$| \phi \rangle \equiv U | \psi \rangle +i \lambda V | \psi \rangle$$

then $$\langle \phi | =\langle \psi |U -i \lambda^* \langle \psi | V$$

since U, V are Hermitian. This then gives me :

$$\langle \phi | \phi \rangle = \langle U^2 \rangle + | \lambda |^2 \langle V^2 \rangle -i \lambda ^* \langle VU \rangle + i \lambda \langle UV \rangle$$

Do you understand where the requirement that $\langle \phi|\phi \rangle \geq 0$ comes from?

Now what I've got here will look like what you've got*, only if you assume $\lambda$ is real. This requires that any $\lambda$ that you plug back in also better be real.

I'm not sure how it's going to go (I suspect it will lead to the weak form of the uncertainty relation) but go ahead and take the derivative.

* Your first two terms are still missing expectation value brackets. The way you've written it, these first two terms are operators while the third term is a scalar.

Last edited: Jan 2, 2007
5. Jan 5, 2007

### Reshma

If $\phi$ is Hermitian, the object $<\phi>$ must be real and hence $<\phi|\phi> \geq 0$

$$I(\lambda)= (\Delta A)^2 + \lambda^2 (\Delta B)^2 + i\lambda\langle[A, B]\rangle$$
Minimum will occur when:
$$2\lambda(\Delta B)^2 + i \langle [A, B] \rangle = 0$$
$$\lambda = \frac{-i\langle[A, B]\rangle}{2(\Delta B)^2}$$

Substituting this value of lambda in my last equation:
$$(\Delta A)^2 - \frac{\langle[A, B]\rangle^2}{4(\Delta B)^2} + \frac{\langle[A, B]\rangle^2}{2(\Delta B)^2} \geq 0$$

$$(\Delta A)^2(\Detla B)^2 \geq -{1\over 4}\langle[A, B]\rangle^2$$

I seem to get my required result. But I am still unsure of the nature of $\lambda$. Apart from being real, $\lambda$ has also been treated as a variable. Can you suggest an alternate technique for this proof?

P.S. I am still at the beginner's stage as far as Dirac's notations are concerned. I think I will stick to the integral form till I get more comfortable will Dirac's conventions.

Last edited: Jan 5, 2007
6. Jan 5, 2007

### Gokul43201

Staff Emeritus
You're still making the same mistake here. You need to use $\langle \psi | \Delta A \cdot \Delta A | \psi \rangle = \langle ( \Delta A)^2 \rangle$ by definition of the expectation value, $\langle H \rangle \equiv \langle \psi | H | \psi \rangle$

$$I(\lambda)= \langle (\Delta A)^2 \rangle + \lambda ^2 \langle (\Delta B)^2 \rangle + i\lambda\langle[A, B]\rangle$$

Note the important difference: $(\Delta A)^2$ is an operator, while $\langle (\Delta A)^2 \rangle$ is a scalar (a number).

Incorporate changes made above (division by an operator is not defined, but division by a scalar is).

Well, looks like $\lambda$ is actually imginary, not real.

Doesn't it look like the RHS has the wrong sign? You have a negative number in the RHS! Clearly the product of 2 positive reals must be greater than a negative real, so you've ended up with a trivial truth. I suspect the error comes from the assumption that $\lambda$ be real, which later didn't pan out.

https://www.physicsforums.com/showpost.php?p=692973&postcount=52

Use any of the following for the proof of the Cauchy-Schwarz inequality, that I skipped the steps of, in the above.

http://mathworld.wolfram.com/SchwarzsInequality.html
http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality
http://ccrma.stanford.edu/~jos/mdft/Cauchy_Schwarz_Inequality.html

Last edited: Jan 5, 2007
7. Jan 6, 2007