Graduate Commutation and Non-Linear Operators

[MisterX]

Suppose $A$ is a linear operator $V\to V$ and $\mathbf{x} \in V$. We define a non-linear operator $\langle A \rangle$ as $$\langle A \rangle\mathbf{x} := <\mathbf{x}, A\mathbf{x}>\mathbf{x}$$

Can we say $\langle A \rangle A = A\langle A \rangle$? What about $\langle A \rangle B = B\langle A \rangle$ ?

More generally if we have $Q\mathbf{x} = q(\mathbf{x}) \mathbf{x}$ with scalar function $q$, when does $AQ=QA$ ? I assert we can choose $A, Q$ such that this is false.

 

[andrewkirk]

Can we say $\langle A \rangle A = A\langle A \rangle$? What about $\langle A \rangle B = B\langle A \rangle$ ?

The way to test this is to expand $(\langle A\rangle B)\mathbf x = \langle A \rangle (Bx)$ and $(B\langle A\rangle)\mathbf x = B( \langle A \rangle x)$ using your definitions above to see whether they give exactly the same result.

 

[MisterX]

I guess I am not sure about the use of the associative property here. But ignoring this it seems
$\langle A \rangle (B\mathbf {x}) = <B\mathbf {x}, AB\mathbf {x}>B\mathbf {x}$ and $B( \langle A \rangle \mathbf {x})= <\mathbf {x}, A \mathbf {x}>B\mathbf {x}\neq \langle A \rangle (B\mathbf {x})$.

This is troubling to me because this step seems to be used in the proof of Heisenberg's uncertainty principle. Sources claim $$[A-\langle A \rangle, \, B- \langle B \rangle ] = [A,\,B] $$
I guess now am not sure if this is true and why.

 

[andrewkirk]

That's because in that Heisenberg formula $\langle A\rangle$ is a scalar, not a nonlinear operator. The only similarity with this problem is the use of angle bracket symbols, but you have used them to mean something completely different from what Heisenberg means by them.

Where did you get the notion that $\langle A\rangle$ is a nonlinear operator?

 

[MisterX]

That's because in that Heisenberg formula $\langle A\rangle$ is a scalar, not a nonlinear operator. The only similarity with this problem is the use of angle bracket symbols, but you have used them to mean something completely different from what Heisenberg means by them.

Where did you get the notion that $\langle A\rangle$ is a nonlinear operator?

Are you asserting that $\langle A\rangle$ is not 2nd order in $\mathbf{x}$? Doesn't the average of an operator depend on the state on which it acts?

 

[andrewkirk]

Doesn't the average of an operator depend on the state on which it acts?

That's right. To be precise, $\langle A \rangle$ denotes the scalar value $\langle \mathbf x\ |\ A\ |\ \mathbf x\rangle$ where $\mathbf x$ is the current state.

But $\langle A \rangle$ denotes the scalar result, not an operator. It is only an operator in the sense that scalar multiplication is a linear operation. If that is the meaning in your OP then the answers are, treating $\langle A \rangle$ as the linear operator that pre-multiplies any vector $\mathbf y$ by the scalar $\langle A \rangle$ (which is $\langle \mathbf x\ |\ A \ |\ \mathbf x\rangle$), that both of the following are true:

• $\langle A\rangle A= A\langle A\rangle$
• $\langle A\rangle B= B\langle A\rangle$

since scalar multiplication commutes with any linear operator.

But we need to note that $\langle A\rangle A \mathbf y$ is
$\langle \mathbf x\ |\ A \ |\ \mathbf x\rangle A\mathbf y$
not
$\langle \mathbf y\ |\ A \ |\ \mathbf y\rangle A\mathbf y$

It has to be admitted, the standard notation is a little confusing because $\langle A \rangle$ depends on $\mathbf x$ even though that does not appear anywhere in the notation.