# Commutator and hermitian operator problem

1. Dec 5, 2008

### p2bne

Hi all, i cannot find where's the trick in this little problem:

1. The problem statement, all variables and given/known data
We have an hermitian operator A and another operator B, and let's say they don't commute, i.e. [A,B] = cI (I is identity). So, if we take a normalized wavefunction |a> that is eigenfunction of the operator A so that A|a> = a|a>, we should have
<a|[A,B]|a> = c.
But if i write
<a|AB|a> - <a|BA|a>
and, since A is hermitian, i make it act on the bra for the first term and on the ket for the second one i get
a<a|B|a> - a<a|B|a> = 0.
I really don't see where is the problem...

2. Dec 5, 2008

### Hootenanny

Staff Emeritus
Welcome to Physics Forums,
The problem is that you must preserve the order, i.e. you must operate on the first term using B first and then using A afterwards.

3. Dec 5, 2008

### p2bne

Anyway, I'm sorry, but I don't get why. I mean, i'm just operating with A*=A on the bra <a|.
As far as I remember, for example, all the funny harmonic oscillator eigenstates derivation is about playing on the fact that there's no difference acting on the right or on the left.
Isn't it true that i can write <a|CD|b> as (C*|a>)*(D|b>) (where the star is the hermitian conjugate)??

4. Dec 5, 2008

### Avodyne

That's not it. The operator A can act to the left, and be replaced with its eigenvalue a. The resolution is that <a|B|a> is infinite, so you are multiplying zero by infinity.

To see how it works in more detail, start with two different A eigenstates with eigenvalues a and a'. Then we have

$$\langle a|(AB-BA)|a'\rangle = (a-a')\langle a|B|a'\rangle.$$

Using $AB-BA=cI$, this becomes

$$c\langle a|a'\rangle = (a-a')\langle a|B|a'\rangle.\qquad\qquad \rm eq(1)$$

The right-hand is apparently zero if we set a=a', but in fact we must be more careful.

For a generic state $|\psi\rangle$, we have a wave function in the A basis,

$$\langle a|\psi\rangle = \psi(a).$$

In this basis, the operators are given by

$$\langle a|A|\psi\rangle = a\psi(a),$$

$$\langle a|B|\psi\rangle = -c{\textstyle{d\over da}}\psi(a).$$

For $|\psi\rangle = |a'\rangle$, we have

$$\langle a|a'\rangle = \delta(a-a'),$$

$$\langle a|B|a'\rangle = -c\,{\textstyle{d\over da}}\delta(a-a').$$

So eq(1) becomes

$$c\,\delta(a-a') = -c\,(a-a'){\textstyle{d\over da}}\delta(a-a').$$

This equation is mathematically valid in the sense of distribution theory.

Last edited: Dec 5, 2008
5. Dec 5, 2008

### p2bne

Thanks a lot.

6. Dec 6, 2008

### Hootenanny

Staff Emeritus
Nice! Thanks for the explanation, I've never seen distributional theory used in this way before.

I guess I should have just kept my mouth shut. I'm not a fan of operators acting both to the left and right, it tends to get confusing, so I just stick to letting them operate to the right

7. Dec 6, 2008

### borgwal

What this example also shows is that in *finite* dimensions you cannot have two hermitian operators A and B satisfying [A,B]=cI

8. Dec 6, 2008

### turin

... and thus the space used for QM must be ... infinite. Tada! Another way of saying it is that the seeming problem assumes that you can commute the summations (b/w the outer indices and the inner indices); however, this is not allowed (see conditional vs. absolute convergence).

9. Dec 6, 2008

### Hurkyl

Staff Emeritus
That doesn't avoid all of your problems. e.g. you could make the mistake this way:

$$\langle a | AB - BA | a \rangle = \langle a | AB | a \rangle - \langle a | BA | a \rangle = \left( \langle a | BA | a \rangle)^* - a \langle a | B | a \rangle$$
$$= \left( a \langle a | B | a \rangle)^* - a \langle a | B | a \rangle = a \langle a | B | a \rangle - a \langle a | B | a \rangle = 0$$

Another way of stating the whole problem is that the expression
$$\langle a | AB - BA | a \rangle$$
makes the mistake of binding the same indeterminate, a, to both a generalized bra and a generalized ket. That's a big nono -- each generalized object needs to have its own independent variable.

10. Dec 6, 2008

### Confundo

Any good sources on operators and kets? Tends to be a bit disparate in the QM books I've seen.

11. Dec 7, 2008

### turin

Can you elaborate on this, please.