I was reading:


Where an inequality is presented:

$$ \sigma_A \sigma_B = | \frac{1}{2} \langle \lbrace \hat{A}, \hat{B} \rbrace \rangle - \langle{\hat{A}}\rangle \langle \hat{B} \rangle | ^2 + | \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle ^2 $$

I found the notation hard to understand:

$$ \langle \lbrace \hat{A}, \hat{B} \rbrace \rangle$$

Is bra-ket expression, but it contains a single expression inside$\lbrace \hat{A}, \hat{B} \rbrace $ (so it can't be a dot product). How do I interpret this? Similar problem arises here:

$$ \langle{\hat{A}}\rangle \langle \hat{B} \rangle$$

where the $A, B$ are single-entities contained in a closed bra-ket expression.

  • $\begingroup$ mathworld.wolfram.com/Anticommutator.html $\endgroup$ Mar 28, 2018 at 16:10
  • 3
    $\begingroup$ $\{\hat A,\hat B\}:= \hat A\hat B+\hat B\hat A$ is the anticommutator. $\endgroup$ Mar 28, 2018 at 16:10
  • $\begingroup$ Ok so, $ \lbrace \hat{A}, \hat{B} \rbrace = \hat{A} \hat{B} + \hat{B} \hat{A} $. The bra-ket syntax (from my understanding) is of the form $ <vector>\rangle | \langle <vector> $ or $ \langle <vector> | <vector>\rangle $. The expression: $ \hat{A} \hat{B} + \hat{B} \hat{A} $ I think is a vector (or a scalar which I'll accept as 1 dimensional vector), so to me $ \langle \hat{A} \hat{B} + \hat{B} \hat{A} \rangle$ is missing an argument $\endgroup$ Mar 28, 2018 at 16:14
  • $\begingroup$ :), @ZeroTheHero Does my response in the comments makes sense? I don't understand why $\langle \lbrace \hat{A}, \hat{B} \rbrace \rangle $ is at all well defined since, I'm under the impression whenever I see a $\langle, \rangle$ pair, there should be an inner or outer product occurring (with at least 2 arguments), but the expression written earlier has only 1 argument, the anticommutator $\endgroup$ Mar 28, 2018 at 16:27
  • 1
    $\begingroup$ $\langle \{\hat A,\hat B\}\rangle$ is the average value of the product $\hat A\hat B+\hat B\hat A$. $\hat A\hat B+\hat B\hat A$ is an operator constructed from the symmetric product of $\hat A$ and $\hat B$. $\endgroup$ Mar 28, 2018 at 17:24

3 Answers 3


The angled brackets here are not bra-kets. For any operator $\hat{O}$, the expression $\langle \hat{O}\rangle$ denotes the expectation value of the operator with respect to a state - here implicitly assumed to be arbitrary, but the same state for all such expressions occurring.


So the expression has three different pieces. First, you have the operators which are represented here by hats, e.g. $\hat A$ and $\hat B$. Second, the curly brackets or braces represent the anti-commutator operator $$\{\hat A, \hat B\}\equiv\hat A\hat B + \hat B\hat A.$$ The final piece is the "enclosed bra-ket" notation "$\langle\cdot\rangle$" you are asking about which simply represent the average or expectation value for whatever is enclosed in the braket. So $\langle \hat A\rangle$ is the expectation value for the operator $\hat A$.

Now the expectation value generally depends on what your state is, and in quantum mechanics (especially in the Schrödinger picture), this information is contained in state vectors such as $|\psi\rangle$ and it's Hermitian dual $\langle\psi|$. So the expectation value of some operator $\hat A$, given that the state of your system is $|\psi\rangle$, is given by $$\langle \hat A\rangle = \langle\psi|\hat A|\psi\rangle.$$ Note that in terms of notation, the left hand side looks like an abbreviated from of the right side. This is no accident, and is historically why the Dirac notation was developed to look the way it does.


In addition to the other good answers directly explaining the meaning, it would be constructive to see how it is derived:

The inequality is the generalized uncertainty principle, which is the most complete form of uncertainty principle.

Given Hermitian operators $A,B$, we define vectora $\left|q\right> \& \left|p\right>$ with an arbitrary wavefunction $\left|\phi \right>$ as:

$$\left|q\right>=(A-\left< A\right>)\left|\phi\right> \ ;\ \left|p\right>=(B-\left< B\right>)\left|\phi\right>\tag{1} $$

By Schwarz inequality:

$$\left<q|q\right>\left<p|p\right> \ge \left<q|p\right>\left<p|q\right>\tag{2} $$

$$(\left<A\right>^2-\left<A^2\right>)(\left<B\right>^2-\left<B^2\right>)\ge \Re^2[\left<p|q\right>]+ \Im^2[\left<p|q\right>]\tag{3} $$

After some work:

$$(\Delta A)^2(\Delta B)^2\ge [\frac{1}{2}(\left<p|q\right>+\left<q|p\right>)]^2+[\frac{1}{2i}(\left<p|q\right>-\left<q|p\right>)]^2\tag{4} $$

$$(\Delta A)^2(\Delta B)^2\ge (\left<\phi\right|\frac{1}{2}[AB+BA]\left|\phi\right>+\left<A\right>\left<B\right>)^2+(\left<\phi\right|\frac{1}{2i}[AB-BA]\left|\phi\right>)^2\tag{5} $$

Simplifying it, we have:

$$(\Delta A)^2(\Delta B)^2\ge (\left<\phi\right|\frac{1}{2}\{A^*,B^*\}\left|\phi\right>)^2+(\left<\phi\right|\frac{1}{2i}[A,B]\left|\phi\right>)^2\tag{6} $$

Where $\{A,B\}$ is anticommutator: $AB+BA$ ; $A^*$ and $B^*$ are defined as $A-\left<A\right>$ and $B-\left<B\right>$ respectively. In general, the anticommutator part is too complicated to be useful.

As an inequality, the generalized uncertainty principle always suggests:

$$(\Delta A)^2(\Delta B)^2\ge (\left<\phi\right|\frac{1}{2}\{A^*,B^*\}\left|\phi\right>)^2+(\left<\phi\right|\frac{1}{2i}[A,B]\left|\phi\right>)^2 \ge (\left<\phi\right|\frac{1}{2i}[A,B]\left|\phi\right>)^2 $$

This is the usual uncertainty principle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.