Frequency operators

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  • #1
Fredrik
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The fraction of members of a set of numbers [itex]\{k_i\}_{i=1}^N[/itex] that are equal to a specific number k can be written as

[tex]\frac 1 N\sum_{i=1}^N\delta_{kk_i}[/tex]​

Now consider an ensemble of N identical systems. Because of the above, the operator [itex]f_k^{(N)}[/itex] defined by

[tex]f_k^{(N)}|a_{k_1}\rangle\cdots|a_{k_N}\rangle=\left(\frac 1 N\sum_{i=1}^N\delta_{kk_i}\right)|a_{k_1}\rangle\cdots|a_{k_N}\rangle[/tex]​

where the [itex]|a_j\rangle[/itex] are eigenvectors of some observable A, is telling us what fraction of the systems in the ensemble are in state [itex]|a_k\rangle[/itex]. (The eigenvalue is equal to that fraction). It's also telling us what fraction of the results will be [itex]a_k[/itex] if we measure A once on each system in the ensemble (and it's prepared in this particular state). That fraction is equal to the eigenvalue, which is equal to the expectation value of [itex]f_k^{(N)}[/itex] in this state.

Now consider an ensemble such that all the systems are prepared in the state [itex]|s\rangle[/itex], which isn't an eigenvector of A. Does the last interpretation above still hold? I mean, is the expectation value of [itex]f_k^{(N)}[/itex] in an arbitrary state

[itex]|s^{(N)}\rangle=|s\rangle\cdots|s\rangle[/itex]​

still equal to the fraction of measurements of A that will yield the result [itex]a_k[/itex]? Is it possible to justify this without using the axiom that the probability of measuring [itex]a_k[/itex] when the state is [itex]|s\rangle[/itex] is [itex]|\langle a_k|s\rangle|^2[/itex].
 

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  • #2
strangerep
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(How am I supposed to get any work done if you keep posing interesting questions? :-)

I'm not sure I understand your notation properly...

The fraction of members of a set of numbers
[itex]\{k_i\}_{i=1}^N[/itex] that are equal to a specific number k can be written as

[tex]\frac 1 N\sum_{i=1}^N\delta_{kk_i}[/tex]​
It might be better (at least for me) to write this as

[tex] \sum_{i=1}^N \frac{\delta_{kk_i}}{Nk}\; k_i [/tex]​

since then it's more reminiscent of the trace of an operator acting on a state.
That makes it closer to the familiar [itex]Tr(A\rho)[/itex].

[tex]f_k^{(N)}|a_{k_1}\rangle\cdots|a_{k_N}\rangle=\left(\frac 1 N\sum_{i=1}^N\delta_{kk_i}\right)|a_{k_1}\rangle\cdots|a_{k_N}\rangle[/tex]​

where the [itex]|a_j\rangle[/itex] are eigenvectors of some observable A, is telling us what fraction of the systems in the ensemble are in state [itex]|a_k\rangle[/itex]. (The eigenvalue is equal to that fraction).
Is this a tensor product notation for the state? I'm not clear about what you mean.
Maybe it should be an N-fold product of the operators?

It's also telling us what fraction of the results will be [itex]a_k[/itex] if we measure A once on each system in the ensemble (and it's prepared in this particular state). That fraction is equal to the eigenvalue, which is equal to the expectation value of [itex]f_k^{(N)}[/itex] in this state.

Now consider an ensemble such that all the systems are prepared in the state [itex]|s\rangle[/itex], which isn't an eigenvector of A. Does the last interpretation above still hold? I mean, is the expectation value of [itex]f_k^{(N)}[/itex] in an arbitrary state

[itex]|s^{(N)}\rangle=|s\rangle\cdots|s\rangle[/itex]​

still equal to the fraction of measurements of A that will yield the result [itex]a_k[/itex]? Is it possible to justify this without using the axiom that the probability of measuring [itex]a_k[/itex] when the state is [itex]|s\rangle[/itex] is [itex]|\langle a_k|s\rangle|^2[/itex].
Without a more precise version of your formulae, I'm not sure how to answer.
Ballentine does it as follows:

Postulate 1:
To each dynamical variable (physical concept) there corresponds a linear operator
(mathematical object), and the possible values of the dynamical variable are the
eigenvalues of the operator.

Postulate 2:
To each state there corresponds a unique state operator. The average value of a dynamical
variable R, represented by the operator R, in the virtual ensemble of events that
may result from a preparation procedure for the state, represented by the operator [itex]\rho[/itex], is
[tex]
\langle R \rangle ~=~ \frac{Tr(\rho R)}{Tr\; \rho}
[/tex]
the state operator is also referred to as the statistical operator.


He then shows (after a couple more technical points) that from these 2 postulates,
one can extract a notion of "probability distribution". I.e., the formula above for the
average determines the entire probability distribution (modulo a couple more technical
points involving spectral representations). That's not quite what you're asking, but I
think it's close (or at least it might take you a little closer to where you're trying to go).
 
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  • #3
Fredrik
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I'm not sure I understand your notation properly...

It might be better (at least for me) to write this as

[tex] \sum_{i=1}^N \frac{\delta_{kk_i}}{Nk}\; k_i [/tex]​
I apologize. I talked about a "set" of numbers [itex]\{k_i\}_{i=1}^N[/itex], but this terminology and notation is misleading. For example, if we're dealing with sets, we have {3,7,7,7,4}={3,4,7} but that's not what I had in mind. I meant that there are 5 numbers in {k1,k2,k3,k4,k5}={3,7,7,7,4} and that 60% of them are 7. In that example, N=5, and if we choose k=7, we have

[tex]\frac 1 N\sum_{i=1}^N\delta_{kk_i}=\frac 1 5\sum_{i=1}^5\delta_{7k_i}=\frac 1 5(0+1+1+1+0)=\frac 3 5[/tex]

Is this a tensor product notation for the state?
Yes, when I wrote

[tex]|a_{k_1}\rangle\cdots|a_{k_N}\rangle[/tex]

I meant

[tex]|a_{k_1}\rangle\otimes |a_{k_2}\rangle\otimes\cdots\otimes|a_{k_N}\rangle[/tex]

The best way to explain the rest is probably to put it in the appropriate context. There are many claims that the probability rule of QM can be derived from the other axioms. Wikipedia cites this 1968 article by James Hartle, and claims that Hartle's only additional assumption is that if a system is already in an eigenstate of an observable when it's measured, the outcome is certain. (There is at least one more assumption in there: He assumes that the Hilbert space of a "combined" system is the tensor product of the Hilbert spaces of the individual systems).

That's the article I'm trying to understand. Hartle defines the frequency operator as I defined it above, but he uses a different notation. Maybe it was dumb of me to use a different notation, since I could have anticipated that I would have to link to his article later.

Hartle shows that in the limit [itex]N\rightarrow\infty[/itex], the eigenvalue goes to [itex]|\langle a_k|s\rangle|^2[/itex], but for that to be significant, there needs to be some other reason to interpret [itex]f_k^{(N)}[/itex] (which he writes as [itex]f_N{}^k[/itex]) as a frequency operator. I'm trying to understand what that reason is. I think I understand it when the operator is acting on a tensor product of eigenvectors, but not when it's acting on |s>|s>...|s>, which is the case we're interested in.
 
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  • #4
strangerep
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Wikipedia cites this 1968 article by
James Hartle, [...]
I skimmed Hartle's paper many years ago, but didn't know enough then
to appreciate it properly. I've now read it again more carefully.
It's an interesting paper, making a case that "probability" is an
objective property of an infinite ensemble of identically prepared
systems.

Before I (try to) followup on your question, I'll abbreviate some of
Hartle's notation. For a state vector in an N-fold tensor product
space Hartle writes
[tex]
|S\rangle ~:=~ |s_1, 1\rangle |s_2,2\rangle \dots |s_N,N\rangle
[/tex]
where the 2nd parameter inside each ket merely denotes which
component space of the tensor product it's from. So we could probably
write this (with a slight abuse of notation) as
[tex]
|S\rangle ~:=~ \otimes_{\alpha=1}^N |s_\alpha, \alpha\rangle
[/tex]
The scalar product between any two such states can be written
[tex]
\langle S|S'\rangle
~:=~ \otimes_{\alpha=1}^N \langle s_\alpha|s'_\alpha\rangle
[/tex]

With these preliminaries, let's go back to your post #1:

Now consider an ensemble of N identical systems. [...]
the operator [itex]f_k^{(N)}[/itex] defined by

[tex]f_k^{(N)}|a_{k_1}\rangle\cdots|a_{k_N}\rangle
~=~ \left(\frac 1 N \sum_{i=1}^N\delta_{kk_i}\right)
|a_{k_1}\rangle\cdots|a_{k_N}\rangle[/tex]​
Writing it that way is not quite what Hartle does. He defines
the operator in his eq(5) as
[tex]
f_N^{~k} ~=~ \sum_{i_1,\dots,i_N} \Big(
(\otimes_{\alpha=1}^N |i_\alpha, \alpha\rangle)~
~(\frac 1 N \; \sum_{\alpha=1}^N\delta_{k i_\alpha})~
~(\otimes_{\alpha=N}^1 \langle i_\alpha, \alpha| )
\Big)
[/tex]
I've compactified Hartle's notation. Note that the last tensor
product occurs in reverse order to the first.

Also note that all the sums over each [itex]i_1,i_2,\dots,\i_N[/itex]
in the leading summation symbol are intended to be over all the
eigenvalues of A. E.g., considering the sum of [itex]i_1[/itex] alone,
it would be written (I think) as
[tex]
\sum_{i_1 = a_1}^{a_\lambda}
[/tex]
where [itex]\lambda[/itex] is the number of eigenvalues of A.

So the whole thing is just a mega spectral decomposition on steroids.

To understand why it's reasonable to interpret [itex]f_N^{~k}[/itex]
as a frequency operator, one needs to write it out for some low
value of N (e.g., N=2 or N=3) and see what it looks like. Hartle
just says "The number in brackets (i.e., [itex]N^{-1} \sum
\delta_{k i_\alpha}[/itex]) is just the fraction of states
[itex]|i_1\rangle \dots |i_N\rangle[/itex] which are in the state k."
and obviously expects the reader to get it immediately.

This is easy enough to understand if this operator acts on a tensor
product of identical copies of a state [itex]|i_\alpha\rangle[/itex]
which is an eigenstate of A, but it's less obvious why such an
interpretation is reasonable for (copies of) a more general state.
My take on this is that the interpretation of most operators
(i.e., their correspondence with certain observable dynamical
variables) is usually done via some kind of appeal to the classical
limit. So maybe it's best to think of the [itex]f_N^{~k}[/itex]
operator in more classical terms, acting on definite states. Then
the "frequency" interpretation is easy enough, so we carry this
interpretation over to quantum case.

So I think the answer to your original question, i.e.,

Is it possible to justify this without using the axiom that the
probability of measuring [itex]a_k[/itex] when the state is
[itex]|s\rangle[/itex] is [itex]|\langle a_k|s\rangle|^2[/itex]
is indeed "yes", if you accept Hartle's implicit assumption of
the meaning of "frequency" in classical probability theory.


(I don't know whether of the above actually helps, but at least it
keeps the thread alive. :-)

P.S: I wonder when latex will resume working properly in previews.
Right now, I can only "submit" and then "edit" to see properly what
I've written. :-(
 
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  • #5
Fredrik
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It's an interesting paper, making a case that "probability" is an
objective property of an infinite ensemble of identically prepared
systems.
Maybe we should think of this paper not (just) as a proof of the probability rule in one specific axiomatic formulation of QM, but as the definition of what "probability" means in QM, along with a proof that the definition makes sense.

I guess that's what you're saying here:
My take on this is that the interpretation of most operators
(i.e., their correspondence with certain observable dynamical
variables) is usually done via some kind of appeal to the classical
limit. So maybe it's best to think of the [itex]f_N^{~k}[/itex]
operator in more classical terms, acting on definite states. Then
the "frequency" interpretation is easy enough, so we carry this
interpretation over to quantum case.

So I think the answer to your original question, i.e.,
...
is indeed "yes", if you accept Hartle's implicit assumption of
the meaning of "frequency" in classical probability theory.
I don't know why I didn't think of that. I guess I'm not used to thinking of probability as something that needs a definition. :smile:

I guess this was much easier than I thought. I was trying see how his definition of the frequency operator agrees with my preconceived notion of probability, when I should have taken his definition of [itex]f_N^{~k}[/itex] as a definition of "probability". If that's the case, then I think my question has been answered. (I don't feel a need to understand every detail of his proof. I have too many other things I'd like to understand right now). Thank you again.

Writing it that way is not quite what Hartle does. He defines
the operator in his eq(5) as
[tex]
f_N^{~k} ~=~ \sum_{i_1,\dots,i_N} \Big(
(\otimes_{\alpha=1}^N |i_\alpha, \alpha\rangle)~
~(\frac 1 N \; \sum_{\alpha=1}^N\delta_{k i_\alpha})~
~(\otimes_{\alpha=N}^1 \langle i_\alpha, \alpha| )
\Big)
[/tex]
...
To understand why it's reasonable to interpret [itex]f_N^{~k}[/itex]
as a frequency operator, one needs to write it out for some low
value of N (e.g., N=2 or N=3) and see what it looks like.
I think the definition in #1 is equivalent to his. I just wrote down what you get when you let his [itex]f_N{}^k[/itex] act on a tensor product of eigenstates of A. The result can be used as an alternative definition. The reason I preferred the alternative definition, rather than the explicit formula for [itex]f_N{}^k[/itex], is that the alternative, along with my explanation of the sum [itex]N^{-1}\sum_\alpha\delta_{ki_\alpha}[/itex] makes it clear why [itex]f_N{}^k[/itex] can be interpreted as a frequency operator when it's acting on a tensor product of eigenstates of A.
 
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  • #6
strangerep
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I guess this was much easier than I thought.
I will say that I didn't find Hartle's paper to be "easy" at all. Had to work quite
hard on the appendix to understand the inf-product Hilbert space construction. :-)

I was trying see how his definition of the frequency operator agrees with my
preconceived notion of probability, when I should have taken his definition of
[itex]f_N^{~k}[/itex] as a definition of "probability".
I'm still not entirely confident with my previous "take" on this. But what is
probability in a classical scenario? At some point in any rigorous treatment,
one must construct inf-products of phase spaces, or whatever, to model the
concept of "infinite ensemble". Plenty of other authors approach it in
different ways, however.
 

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