quantum mechanics - Meaning of the anti-commutator term in the uncertainty principle

What is the meaning, mathematical or physical, of the anti-commutator term?

$$\langle ( \Delta A )^{2} \rangle \langle ( \Delta B )^{2} \rangle \geq \dfrac{1}{4} \vert \langle [ A,B ] \rangle \vert^{2} + \dfrac{1}{4} \vert \langle \{ \Delta A, \Delta B \} \rangle \vert^{2},$$ where $\Delta A, \Delta B, A$ and $B$ are operators.

The inequality is still true, and the anti-commutator term "strengthens" the inequality, but why does it appear?

Answer

Dear Rodrigo, it's an interesting stronger version of the uncertainty principle for general operators $A,B$ that I've never seen before but I just verified it holds. Just to be sure, the anticommutator is simply

$$\{A,B\}\equiv AB+BA.$$ I like when the braces are only used for pairs of Grassmannian objects but people use it as a bookkeeping device to simplify $AB+BA$ in all situations. Nothing difficult about the notation. Note that the commutator and anticommutator appear totally symmetrically in the inequality, a fact we will derive.

To see why the stronger inequality holds, open Wikipedia here

http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations

where only the simpler version of the inequality (without the squared anticommutator) is proved by combining two inequalities. The first one,

$$||A\psi||^2 ||B\psi||^2 \geq |\langle A\psi|B\psi\rangle|^2$$ remains unchanged. However, the second inequality from the Wikipedia article may be strengthened to a full-fledged equality $$|\langle A\psi|B\psi\rangle|^2 = \left| \frac{1}{2i} \langle \psi | AB-BA | \psi \rangle \right|^2 + \left| \frac{1}{2} \langle \psi | AB+BA | \psi \rangle \right|^2$$ This identity simply says that the squared absolute value of a complex number is the sum of the squared real part and the squared imaginary part (which was omitted on Wikipedia). Combining the previous two inequalities, one gets your "stronger" uncertainty principle.

(Of course, the equation derived above is uselessly weak unless the expectation values of $A,B$ vanish themselves. It can be strengthened into yours by repeating the same prodedure for $\Delta A = A-\langle A\rangle$ and similarly $\Delta B = B-\langle B\rangle$ instead of $A,B$.)

I wrote what the anticommutator means mathematically and why the inequality is true. Now, what does the anticommutator term mean physically? I don't know what this question mean. It's a term in an equation that I can read and explain for you again. The precise answers in physics are given by mathematics. So I guess that the answer you want to hear is that it means nothing physically, it's just pure mathematics. This fact doesn't mean that it can't be useful.

Well, in normal cases, the stronger version is not "terribly" useful because the anticommutator term is only nonzero if there is a "correlation" in the distributions of $A,B$ - i.e. if the distribution is "tilted" in the $A,B$ plane rather than similar to a vertical-horizontal ellipse which is usually the case in simple wave packets etc. Maybe this is what you wanted to hear as the physical explanation of the anticommutator term - because $AB+BA$ is just twice the Hermitean part of $AB$, it measures the correlation of $A,B$ in the distribution given by the wave function - although the precise meaning of these words has to be determined by the formula.