Commutator and hermitian operator problem

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 :rolleyes: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
  • #1
p2bne
3
0
Hi all, i cannot find where's the trick in this little problem:

Homework Statement


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...
 
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  • #2
Welcome to Physics Forums,
p2bne said:
Hi all, i cannot find where's the trick in this little problem:

Homework Statement


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...
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
Hootenanny said:
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.

Thank you for the answer.
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
Hootenanny said:
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.

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

[tex]\langle a|(AB-BA)|a'\rangle = (a-a')\langle a|B|a'\rangle.[/tex]

Using [itex]AB-BA=cI[/itex], this becomes

[tex]c\langle a|a'\rangle = (a-a')\langle a|B|a'\rangle.\qquad\qquad \rm eq(1)[/tex]

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

For a generic state [itex]|\psi\rangle[/itex], we have a wave function in the A basis,

[tex]\langle a|\psi\rangle = \psi(a).[/tex]

In this basis, the operators are given by

[tex]\langle a|A|\psi\rangle = a\psi(a),[/tex]

[tex]\langle a|B|\psi\rangle = -c{\textstyle{d\over da}}\psi(a).[/tex]

For [itex]|\psi\rangle = |a'\rangle[/itex], we have

[tex]\langle a|a'\rangle = \delta(a-a'),[/tex]

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

So eq(1) becomes

[tex]c\,\delta(a-a') = -c\,(a-a'){\textstyle{d\over da}}\delta(a-a').[/tex]

This equation is mathematically valid in the sense of distribution theory.
 
Last edited:
  • #5
Thanks a lot.
 
  • #6
Avodyne said:
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

[tex]\langle a|(AB-BA)|a'\rangle = (a-a')\langle a|B|a'\rangle.[/tex]

Using [itex]AB-BA=cI[/itex], this becomes

[tex]c\langle a|a'\rangle = (a-a')\langle a|B|a'\rangle.\qquad\qquad \rm eq(1)[/tex]

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

For a generic state [itex]|\psi\rangle[/itex], we have a wave function in the A basis,

[tex]\langle a|\psi\rangle = \psi(a).[/tex]

In this basis, the operators are given by

[tex]\langle a|A|\psi\rangle = a\psi(a),[/tex]

[tex]\langle a|B|\psi\rangle = -c{\textstyle{d\over da}}\psi(a).[/tex]

For [itex]|\psi\rangle = |a'\rangle[/itex], we have

[tex]\langle a|a'\rangle = \delta(a-a'),[/tex]

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

So eq(1) becomes

[tex]c\,\delta(a-a') = -c\,(a-a'){\textstyle{d\over da}}\delta(a-a').[/tex]

This equation is mathematically valid in the sense of distribution theory.
Nice! :approve: 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 :rolleyes:
 
  • #7
What this example also shows is that in *finite* dimensions you cannot have two hermitian operators A and B satisfying [A,B]=cI
 
  • #8
... 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
Hootenanny said:
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 :rolleyes:
That doesn't avoid all of your problems. e.g. you could make the mistake this way:

[tex]
\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[/tex]
[tex]
= \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[/tex]

Another way of stating the whole problem is that the expression
[tex]\langle a | AB - BA | a \rangle[/tex]
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
Any good sources on operators and kets? Tends to be a bit disparate in the QM books I've seen.
 
  • #11
Hurkyl said:
Another way of stating the whole problem is that the expression
[tex]\langle a | AB - BA | a \rangle[/tex]
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.
Can you elaborate on this, please.
 

1. What is a commutator?

A commutator is a mathematical operator used in quantum mechanics to measure the degree of noncommutativity between two other operators. It is defined as the difference between the product of the two operators in one order and the product in the reverse order.

2. How is a commutator represented mathematically?

The commutator of two operators A and B is denoted by [A, B] and is defined as [A, B] = AB - BA.

3. What is the significance of the commutator in quantum mechanics?

The commutator is used to determine the uncertainty in the measurement of two operators. If the commutator is zero, the two operators are said to commute, and their measurements can be made simultaneously with no uncertainty. However, if the commutator is non-zero, the operators do not commute, and their measurements are subject to uncertainty.

4. What is a Hermitian operator?

A Hermitian operator is a special type of operator in quantum mechanics that represents an observable physical quantity. It has the property of being equal to its own Hermitian conjugate, which means that the operator and its adjoint have the same eigenvalues and eigenvectors.

5. What is the relationship between a commutator and a Hermitian operator?

The commutator of two Hermitian operators is always a Hermitian operator. This means that if two operators represent measurable physical quantities, their commutator also represents a measurable physical quantity. Additionally, if the commutator of two operators is zero, it means that the operators commute and can be measured simultaneously with no uncertainty.

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