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?


2 Answers 2


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


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.

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    $\begingroup$ This "strong equality" is used as a stepping stone in proving the more common expression for HP (in position-momentum terms for example). That's how I was taught it at least. $\endgroup$
    – Noldorin
    Commented Jan 28, 2011 at 16:41
  • $\begingroup$ What do you mean by HP? $\endgroup$ Commented Jan 28, 2011 at 18:15
  • $\begingroup$ I guess that Noldorin means the Heisenberg principle but I don't understand why the comment should be right... $\endgroup$ Commented Jan 28, 2011 at 21:11
  • $\begingroup$ @Luboš Motl Could you give some example of this correlation? $\endgroup$ Commented Jan 29, 2011 at 18:40
  • $\begingroup$ Dear @Rodrigo Thomas, for example, the classical probability distribution for $x,p$ may be $C'\exp(-K(p-cx)^2-Lx^2)$. Note that it is centered at $x=0$ and $p=0$. However, for a fixed value of $x$, the most likely value of $p$ is $p=cx$. So $p$ and $x$ like to increase with each other. Imagine a typical graph with many dots showing statistical evidence that $p$ is correlated with $x$. If that's the distribution, there will be a nonzero expectation value of $px$ - whose best (Hermitian) definition in QM is $\{p,x\}/2$. $\endgroup$ Commented Feb 2, 2011 at 14:49

Dear Rodrigo, I am not aware of any direct physical meaning of the anti-commutator term, but it is useful when you want to pin down the states that saturate the inequality in the Heisenberg principle. Obviously, two conditions must be met if an equality in the usual uncertainty principle should occur: the anti-commutator term must vanish and the Cauchy-Schwarz inequality (see the comment by Luboš Motl) must be saturated. The latter happens if and only if the vectors $\Delta A|\psi\rangle$ and $\Delta B|\psi\rangle$ are collinear, let's say $\Delta A|\psi\rangle=\lambda\Delta B|\psi\rangle$. This is equivalent to $(A-\lambda B)|\psi\rangle=(\langle A\rangle-\lambda\langle B\rangle)|\psi\rangle$, that is, $\psi$ is an eigenvector of $A-\lambda B$. But then the expectation value of the anti-commutator becomes $\langle\{\Delta A,\Delta B\}\rangle=(\lambda+\lambda^*)\langle(\Delta B)^2\rangle$, which vanishes only if $\lambda$ is purely imaginary unless, of course, $|\psi\rangle$ is an eigenvector of $B$ in which case the whole inequality is trivial. So in the end, the state $|\psi\rangle$ saturates the inequality in the (usual formulation of the) uncertainty principle if and only if it is an eigenstate of $A-\lambda B$ for some purely imaginary $\lambda$. This happens for example for the coherent states of the harmonic oscillator.


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