Some inconsistency on operator expectation value

[karlzr]

Consider two Hermitian operator A, B; Define
[A,B]=iC,
then operator C is also Hermitian.
we calculate the expectation value with respect to |a>, one eigenstate of A with the eigenvalue a.
From the left side, we have:
<a|[A,B]|a>=<a|(AB-BA)|a>=(a-a)<a|B|a>=0,
while on the right side, <a|iC|a> does not necessarily vanish.
That is :<a|[A,B]|a> ≠ <a|iC|a>, which is absurd !
So what is wrong in my calculation?

 

[Chopin]

What makes you say that \langle a | iC | a \rangle need not vanish? I think you've just shown that if you have an operator C which is defined using that equation, then its expectation value must vanish for any eigenstate of the other operators.

You can check this by looking at angular momentum, whose values are defined in exactly this way: [L_x,L_y] = iL_z. An eigenstate of L_x will always have an expectation value of 0 for L_z--it must, because if it didn't, then it would break rotational symmetry. So you've just confirmed mathematically a fact that must already be true geometrically.

 

[karlzr]

What makes you say that \langle a | iC | a \rangle need not vanish? I think you've just shown that if you have an operator C which is defined using that equation, then its expectation value must vanish for any eigenstate of the other operators.

You can check this by looking at angular momentum, whose values are defined in exactly this way: [L_x,L_y] = iL_z. An eigenstate of L_x will always have an expectation value of 0 for L_z--it must, because if it didn't, then it would break rotational symmetry. So you've just confirmed mathematically a fact that must already be true geometrically.

What about [x,p]=i\hbar here C is the Planck constant.
The expectation value should be a delta function.

 

[Jorriss]

What about [x,p]=i\hbar here C is the Planck constant.
The expectation value should be a delta function.

<a|[A,B]|a>=<a|(AB-BA)|a>=(a-a)<a|B|a>=0,

Use the example of [p,q] = ih.

There's a lot of interesting mathematical subtlety here I do not understand... so I'll contribute to this in hopes of getting more input from those who know more.

Two things come to mind. [p,q]=ih can not be realized for a finite dimensional vector space and either p or q must be unbounded. <a|[A,B]|a>=<a|(AB-BA)|a>=(a-a)<a|B|a>=0 then becomes illdefined, I believe, as < a | B | a > may not be finite as B is unbounded so we are getting something of the form 0 * infinity.

What may be happening is you are attempting to take the trace of an infinite dimensional matrix and that does not make sense.

I think this is the right track, but clearly there is much I do not know...

 

[Ilmrak]

What about [x,p]=i\hbar here C is the Planck constant.
The expectation value should be a delta function.

Hello!

I'll do to you some questions I think could be useful.

On which space are X,P hermitian?
Do this space include their eigenstates?
What's the value of \langle x|i\hslash|x\rangle?

Ilm

 

[tom.stoer]

you should have a look at

http://arxiv.org/abs/quant-ph/9907069v2
Mathematical surprises and Dirac's formalism in quantum mechanics
F. Gieres
(Submitted on 22 Jul 1999 (v1), last revised 21 Dec 2001 (this version, v2))
Abstract: By a series of simple examples, we illustrate how the lack of mathematical concern can readily lead to surprising mathematical contradictions in wave mechanics. The basic mathematical notions allowing for a precise formulation of the theory are then summarized and it is shown how they lead to an elucidation and deeper understanding of the aforementioned problems. After stressing the equivalence between wave mechanics and the other formulations of quantum mechanics, i.e. matrix mechanics and Dirac's abstract Hilbert space formulation, we devote the second part of our paper to the latter approach: we discuss the problems and shortcomings of this formalism as well as those of the bra and ket notation introduced by Dirac in this context. In conclusion, we indicate how all of these problems can be solved or at least avoided.

 

[karlzr]

Use the example of [p,q] = ih.

Two things come to mind. [p,q]=ih can not be realized for a finite dimensional vector space and either p or q must be unbounded. <a|[A,B]|a>=<a|(AB-BA)|a>=(a-a)<a|B|a>=0 then becomes illdefined, I believe, as < a | B | a > may not be finite as B is unbounded so we are getting something of the form 0 * infinity.

What may be happening is you are attempting to take the trace of an infinite dimensional matrix and that does not make sense.

I think this is the right track, but clearly there is much I do not know...

Hello!

I'll do to you some questions I think could be useful.

On which space are X,P hermitian?
Do this space include their eigenstates?
What's the value of \langle x|i\hslash|x\rangle?

Ilm

you should have a look at

http://arxiv.org/abs/quant-ph/9907069v2
Mathematical surprises and Dirac's formalism in quantum mechanics
F. Gieres

Perhaps you are right. Just as Jorriss mentioned, the infinity might be the key to this problem. In the article tom.stoer listed, it is pointed out that P is hermitian only to square integrable wavefunctions. Thus, P is not Hermitian when it comes to the eigenfunction of X, which is delta function.
Actually, I really don't know how to normalize this delta function?

 

[arojo]

Hi,

I do not know the general answer, but if we limit ourselves to the [p,q] = ih case, as it was remarked earlier in the discussion you must know what does it mean \langle x|i\hslash|x\rangle?

Actually if \langle x| x\rangle is the probability for the particle of being at x, then the function must be normalized to 1 for x \in\Omega. We can do this multiplying the wave function by another function like e^{-\alpha x^2} where 1 &gt;&gt; \alpha &gt; 0. This let us to integrate the function keeping it normalized to 1 and in the limit of \alpha → 0 after a contour integration we get the expected result.

How to easily generalize this argument to more complex cases is a good question. I suppose in any case the problem would be present every time that the set in question is non-compact or the spectrum is non localized.

Cheers