Commute an operator with a constant

[widderjoos]

Homework Statement

Suppose we had an operator A that multiplied a vector by it's norm:
$A \mid \psi \rangle = \langle \psi \mid \psi \rangle \mid \psi \rangle$
I wanted to know what it's commutator with a constant would be.

Homework Equations

$\left[A,B\right] = AB - BA$

The Attempt at a Solution

Suppose b is a real number greater than 1, then
$\left[A,b\right] \mid \psi \rangle =(Ab-bA)\mid \psi \rangle$
$=A (b\mid \psi \rangle ) - b (A \mid \psi \rangle ) = (b^2 - b)\langle \psi \mid \psi \rangle \mid \psi \rangle$

I'm think the second equality is wrong since I read that the commutator of an operator with a constant should be 0 and the operator A looks nonlinear. But what should be done instead? Thanks

 

[latentcorpse]

Homework Statement

Suppose we had an operator A that multiplied a vector by it's norm:
$A \mid \psi \rangle = \langle \psi \mid \psi \rangle \mid \psi \rangle$
I wanted to know what it's commutator with a constant would be.

Homework Equations

$\left[A,B\right] = AB - BA$

The Attempt at a Solution

Suppose b is a real number greater than 1, then
$\left[A,b\right] \mid \psi \rangle =(Ab-bA)\mid \psi \rangle$
$=A (b\mid \psi \rangle ) - b (A \mid \psi \rangle ) = (b^2 - b)\langle \psi \mid \psi \rangle \mid \psi \rangle$

I'm think the second equality is wrong since I read that the commutator of an operator with a constant should be 0 and the operator A looks nonlinear. But what should be done instead? Thanks

Just looking quickly at this, shouldn't you have got

$\left( b^2-1 \right) \langle \psi | \psi \rangle \left( b | \psi \rangle \right)$?

 

[widderjoos]

Just looking quickly at this, shouldn't you have got

$\left( b^2-1 \right) \langle \psi | \psi \rangle \left( b | \psi \rangle \right)$?

oh you're right, but it still doesn't equal 0 though unless I'm missing something again

 

[dextercioby]

How did you get the b^2 ? It shouldn't be there.

 

[mathfeel]

oh you're right, but it still doesn't equal 0 though unless I'm missing something again

Why should it equal to zero? A is non-linear, so there is no guarantee that $A (b|\psi\rangle) = b A |\psi\rangle$.