Some inconsistency on operator expectation value

In summary, we discussed the Hermitian operators A and B, defined by [A,B]=iC, and concluded that if C is also Hermitian then the expectation value of [A,B] with respect to an eigenstate of A must be 0. We used the example of [p,q] = ih to illustrate this, noting that this case is not well-defined due to the unboundedness of p and q. We also considered the possibility of taking the trace of an infinite dimensional matrix, which does not make sense. Finally, we discussed the normalization of a delta function eigenstate and its relation to the probability of finding a particle at a specific position.
  • #1
karlzr
131
2
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?
 
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  • #2
What makes you say that [itex]\langle a | iC | a \rangle[/itex] need not vanish? I think you've just shown that if you have an operator [itex]C[/itex] 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: [itex][L_x,L_y] = iL_z[/itex]. An eigenstate of [itex]L_x[/itex] will always have an expectation value of 0 for [itex]L_z[/itex]--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.
 
  • #3
Chopin said:
What makes you say that [itex]\langle a | iC | a \rangle[/itex] need not vanish? I think you've just shown that if you have an operator [itex]C[/itex] 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: [itex][L_x,L_y] = iL_z[/itex]. An eigenstate of [itex]L_x[/itex] will always have an expectation value of 0 for [itex]L_z[/itex]--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 [tex][x,p]=i\hbar[/tex] here C is the Planck constant.
The expectation value should be a delta function.
 
  • #4
karlzr said:
What about [tex][x,p]=i\hbar[/tex] here C is the Planck constant.
The expectation value should be a delta function.

karlzr said:
<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...
 
  • #5
karlzr said:
What about [tex][x,p]=i\hbar[/tex] 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 [itex]X,P[/itex] hermitian?
Do this space include their eigenstates?
What's the value of [itex]\langle x|i\hslash|x\rangle[/itex]?

Ilm
 
  • #6
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.
 
  • #7
Jorriss said:
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...

Ilmrak said:
Hello!

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

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

Ilm

tom.stoer said:
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?
 
  • #8
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 [itex]\langle x|i\hslash|x\rangle[/itex]?

Actually if [itex]\langle x| x\rangle[/itex] is the probability for the particle of being at x, then the function must be normalized to 1 for [itex]x \in\Omega [/itex]. We can do this multiplying the wave function by another function like [itex]e^{-\alpha x^2}[/itex] where [itex] 1 >> \alpha > 0 [/itex]. This let us to integrate the function keeping it normalized to 1 and in the limit of [itex] \alpha → 0 [/itex] 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
 

1. What is an operator expectation value?

An operator expectation value is the average value of a physical quantity, such as position, momentum, or energy, when a system is in a particular state. It is calculated by taking the inner product of the state vector with the operator corresponding to the physical quantity.

2. Why is there sometimes inconsistency in operator expectation values?

Inconsistency in operator expectation values can occur due to several reasons, such as experimental errors, imperfect measurements, or incorrect assumptions about the state of the system. It is important to carefully analyze the experimental setup and consider all possible sources of error to minimize inconsistencies.

3. How can operator expectation values be used in quantum mechanics?

Operator expectation values are a fundamental concept in quantum mechanics and are used to calculate the probabilities of obtaining certain measurement outcomes. They also play a crucial role in understanding the evolution of quantum systems over time.

4. Can operator expectation values be negative?

Yes, operator expectation values can be negative. This is because they are a mathematical average and do not necessarily reflect the physical reality of the system. Negative values can arise when the system has a complex wavefunction, which is a common occurrence in quantum mechanics.

5. How can operator expectation values be experimentally determined?

Operator expectation values can be experimentally determined by performing repeated measurements on a large number of identically prepared systems and calculating the average value of the physical quantity of interest. This average value will give an estimate of the operator expectation value for that particular state.

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