##Tr([Q,P])## and Ballentine Problem 6.3

[EE18]

I am trying to solve the following problem from Ballentine:

(a) For finite-dimensional matrices $A$ and $B$, show that $Tr[A, B] = 0.$
(b) Paradox. From this result it would seem to follow, by taking the trace of the commutator $[Q, P] = i\hbar$, that one must have $\hbar = 0$. Use the infinite-dimensional matrices for $Q$ and $P$, found in the previous problem, to calculate the matrices $QP$ and $PQ$, and hence explain in detail why the paradoxical conclusion \hbar = 0 is not valid.

a) is not a problem, it's b). Now I know intuitively that the "resolution" is that $QP$ and $PQ$ are not trace-class. But I wanted to see (as I think is suggested by Ballentine) how to compute that trace and "see the problem".
We compute
$\textrm{Tr(QP - PQ) \equiv \langle n|QP -PQ|n\rangle = \sum_{n'} \langle n|Q|n'\rangle \langle n'|Q|n\rangle - \langle n|P|n'\rangle \langle n'|Q|n\rangle$
$= \sum_{n'} \left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n'} \delta_{n,n'-1}+ \sqrt{n'+1} \delta_{n,n'+1})\right) \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n} \delta_{n',n-1}- \sqrt{n+1} \delta_{n',n+1})\right)$
$- \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n'} \delta_{n,n'-1}- \sqrt{n'+1} \delta_{n,n'+1})\right)\left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n} \delta_{n',n-1}+ \sqrt{n+1} \delta_{n',n+1})\right)$
$= -i\hbar \sum_{n'} (-\sqrt{n'} \delta_{n,n'-1}\sqrt{n+1} \delta_{n',n+1}) + (\sqrt{n'+1} \delta_{n,n'+1}\sqrt{n} \delta_{n',n-1})= -i\hbar \sum_{n'} (-\sqrt{n'} \delta_{n,n'-1}\sqrt{n+1} \delta_{n',n+1}) + n \delta_{n,n'} = = i\hbar \delta_{n,n'}(n - n) = 0$

Obviously some of these manipulations aren't valid, but I want to be clear on where?

Edit: I see that it's not formatting but I'm not sure why? I've used basic Latex and nothing else?

 

[PeterDonis]

I've used basic Latex

There is no single definition of "basic LaTeX". It depends on the specific implementation.

The "LaTeX Guide" link at the bottom left of each post window should help you with what LaTeX syntax is supported here.

 

[malawi_glenn]

$\textrm{Tr}(QP - PQ) \equiv \langle n|QP -PQ|n\rangle = \sum_{n'} \langle n|Q|n'\rangle \langle n'|Q|n\rangle - \langle n|P|n'\rangle \langle n'|Q|n\rangle=$

$\sum_{n'} \left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n'} \delta_{n,n'-1}+ \sqrt{n'+1} \delta_{n,n'+1})\right) \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n} \delta_{n',n-1}- \sqrt{n+1} \delta_{n',n+1})\right)- \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n'} \delta_{n,n'-1}- \sqrt{n'+1} \delta_{n,n'+1})\right)\left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n} \delta_{n',n-1}+ \sqrt{n+1} \delta_{n',n+1})\right)=$

$-i\hbar \sum_{n'} (-\sqrt{n'} \delta_{n,n'-1}\sqrt{n+1} \delta_{n',n+1}) + (\sqrt{n'+1} \delta_{n,n'+1}\sqrt{n} \delta_{n',n-1})= -i\hbar \sum_{n'} (-\sqrt{n'} \delta_{n,n'-1}\sqrt{n+1} \delta_{n',n+1}) + n \delta_{n,n'} =$

$i\hbar \delta_{n,n'}(n - n) = 0$

I fixed it for you. You need ## not $$ to enclose LaTeX. And, you have forgotten some } {

Note that you need paranthesis for the sums, also there are some typos. But perhaps you can find them yourself.

 

[WernerQH]

Now I know intuitively that the "resolution" is that $QP$ and $PQ$ are not trace-class. But I wanted to see (as I think is suggested by Ballentine) how to compute that trace and "see the problem".

Have you done Problem 6.2 ? Have you seen Eqn. 6.35 ?

It may be easier to first look at the operators $a^\dagger a$ and $a a^\dagger$. The commutator should be $1$, but the trace is tricky.

 

[EE18]

Have you done Problem 6.2 ? Have you seen Eqn. 6.35 ?

It may be easier to first look at the operators $a^\dagger a$ and $a a^\dagger$. The commutator should be $1$, but the trace is tricky.

Yes, I have done Problem 6.2 which is where the matrix elements which I quote in my OP are obtained from. I guess I'm just not sure where in the trace computation things are meant to go wrong, other than the infinite sum in the first place?

 

[WernerQH]

Yes, $\operatorname{Tr} 1$ doesn't seem to be well defined.

 

[strangerep]

@EE18 : is the above enough for you? I.e., that $(\infty - \infty)$ is ill-defined -- it could be anything.

 

[EE18]

@EE18 : is the above enough for you? I.e., that $(\infty - \infty)$ is ill-defined -- it could be anything.

I'm not sure it is -- but what do you think? Is that what you think Ballentine had in mind?

 

[strangerep]

I'm not sure it is -- but what do you think? Is that what you think Ballentine had in mind?

I can't read his mind, so I have no idea what he intended.

This is perhaps a good example in which to investigate why such an explanation seems unsatisfying.

Let's denote by $Tr_N$ a trace restricted to an N-dim Hilbert space. One might try to imagine taking a sequence of (finite) N-dimensional Hilbert spaces, and showing $Tr_N(QP) - Tr_N(PQ) = 0$ for all $N$, and then claiming that $Tr_N(QP) - Tr_N(PQ) = Tr_N(QP - PQ)$ still holds for $N\to\infty$. But this is equivalent to (falsely) claiming that $\infty-\infty$ is well-defined.

So then we must examine carefully what $\,\lim_{N\to\infty} Tr_N(QP)$ actually means. Let $a_i := Tr_i(QP)$. Usually, when we ask for the infinite sum $\Sigma_{i=0}^\infty a_i$ to be meaningful, we want the sequence $$(a_0, a_1, a_2, a_3, \dots)$$ to be a Cauchy sequence. But this is not so in our case. The sequence of partial sums increases with $N$.

Another property of any (good, i.e, sensible) limit is that it must be independent of how you approach it. E.g., we expect commutativity of addition to apply, so that $1 + 2 - 3 + 4 - 5 + 6 \dots$ "should" be the same as $1 + 2 + 4 - 3 + 6 + 8 + 10 - 5 \dots$. But if you stop at any particular finite number of terms, the two partial sums are not equal, in general, even though the full sequences obviously consist of the same elements. So what does "limit of partial sums" mean in this case??

Long story short: if one allows ill-defined expressions into one's math, one should not be surprised when nonsense inevitably emerges.

[But perhaps the math experts here can give better explanations and/or more compelling examples of the nonsense that can arise.]

 

[PeterDonis]

One might try to imagine taking a sequence of (finite) N-dimensional Hilbert spaces

How would one define $Q$ and $P$ for such Hilbert spaces? The whole point of asking the question about $Q$ and $P$ is that they are only defined on the infinite dimensional Hilbert space of functions of real numbers (or n-tuples of real numbers).

Since we have $[Q, P] = i \hbar I$, we would expect the commutator of the infinite dimensional matrices $Q$ and $P$ to be a diagonal matrix with $i \hbar$ as each diagonal element. The trace of this matrix is obviously not zero since each term has the same sign so there are no cancellations possible. Whether you want to express this as the trace not being well-defined (because of the infinite number of terms in the sum) or not is immaterial; the point is that it is definitely not zero. The key detail in the problem is simply to compute both matrices and show explicitly how their commutator works out to be the diagonal matrix described above.

 

[EE18]

How would one define $Q$ and $P$ for such Hilbert spaces? The whole point of asking the question about $Q$ and $P$ is that they are only defined on the infinite dimensional Hilbert space of functions of real numbers (or n-tuples of real numbers).

Since we have $[Q, P] = i \hbar I$, we would expect the commutator of the infinite dimensional matrices $Q$ and $P$ to be a diagonal matrix with $i \hbar$ as each diagonal element. The trace of this matrix is obviously not zero since each term has the same sign so there are no cancellations possible. Whether you want to express this as the trace not being well-defined (because of the infinite number of terms in the sum) or not is immaterial; the point is that it is definitely not zero. The key detail in the problem is simply to compute both matrices and show explicitly how their commutator works out to be the diagonal matrix described above.

Would it be possible then to point where I failed in my algebra seeing as I get that their commutator produces the (infinite) zero matrix?

 

[strangerep]

How would one define $Q$ and $P$ for such Hilbert spaces? The whole point of asking the question about $Q$ and $P$ is that they are only defined on the infinite dimensional Hilbert space of functions of real numbers (or n-tuples of real numbers).

Ah, good! That's the crucial item I was looking for to make the explanation more concrete.

Summary: when confronted by the statement that the trace is ill-defined, a (naive, but persistent) student might ask why they can't just use a sequence of finite-dim Hilbert spaces. The answer is that, although the trace might be finite, there exist no such Q and P as finite-dim matrices which satisfy the canonical commutation relations $[Q,P] \propto 1$.
Hence that sequence of finite-dim Hilbert spaces is irrelevant to the CCRs.

@EE18: This was the point of the exercise, imho. Your algebra gives $\infty-\infty$, which is not necessarily zero. But if you try to prove that it is zero, using some kind of limit argument on a sequence of finite-dim spaces, it's irrelevant since at each step in the sequence you cannot find finite-dim matrices Q and P which satisfy the commutation relations.

 

[vanhees71]

Since $[\hat{q},\hat{p}] = \hbar \mathrm{i} \hat{1}$ and the Hilbert space, where this is realized with self-adjoint operators, is the separable infinitely-dimensional Hilbert space, the trace of this commutator is simply not defined.

 

[PeterDonis]

Would it be possible then to point where I failed in my algebra

Your first step doesn't look right to me:

$$\textrm{Tr} (QP - PQ) \equiv \langle n|QP -PQ|n\rangle$$

The trace is the sum of the diagonal elements, which in the basis of energy/number eigenstates would be $\Sigma_n \bra{n} QP -PQ \ket{n}$. The final answer should be $\Sigma_n i \hbar$.

 

[strangerep]

Your first step doesn't look right to me: $$\textrm{Tr} (QP - PQ) \equiv \langle n|QP -PQ|n\rangle$$
The trace is the sum of the diagonal elements,

Yes, he omitted a $\sum_n\,$, but even after inserting it the apparent paradox persists.

Ballentine, p15, says

A useful characteristic of an operator A is its trace, defined as $$ Tr \,A ~=~ \sum_j \langle u_j | A | u_j \rangle ~,$$ where $\{ |u_j\rangle \}$ may be any orthonormal basis.

But, of course, he goes on to say:

The trace of a matrix is just the sum of its diagonal elements. For an operator in an infinite-dimensional space, the trace exists only if the infinite sum is convergent.

...which is the core of the answer to the current exercise. The step $Tr(QP) - Tr(PQ) = Tr(QP-PQ)$ is mathematically invalid because the 2 traces each do not exist separately.

 

[PeterDonis]

The step $Tr(QP) - Tr(PQ) = Tr(QP-PQ)$ is mathematically invalid because the 2 traces each do not exist separately.

Neither does the trace of the commutator, correct? It's an infinite sum that does not converge.

 

[strangerep]

Neither does the trace of the commutator, correct? It's an infinite sum that does not converge.

Indeed, but the tricky bit (from a student's point of view) is to achieve understanding of why the rule that $Tr(AB) = Tr(BA)$ is not useful here.

 

[EE18]

Indeed, but the tricky bit (from a student's point of view) is to achieve understanding of why the rule that $Tr(AB) = Tr(BA)$ is not useful here.

Does $Tr(QP)$ exist in the first place?

 

[strangerep]

Does $Tr(QP)$ exist in the first place?

Well, you've already done most of the algebra to answer that. Edit your original calculation so that it's just $Tr(QP)$ (but insert a $\sum_n$ at the front to make a genuine trace). What do you get when you simplify the resulting expression?

 

[EE18]

Well, you've already done most of the algebra to answer that. Edit your original calculation so that it's just $Tr(QP)$ (but insert a $\sum_n$ at the front to make a genuine trace). What do you get when you simplify the resulting expression?

I've got the following but I'm still struggling to point at the exact step where things go wrong as contrasted with the finite dimensional case. In finite dimensions the proof uses linearity and the cyclic property of the trace -- I guess showing $Tr(PQ) \neq Tr(QP)$ is enough to show that particular proof strategy won't work, but not that the statement cant hold in general?

Edit: I also think I do something illegal in the last line when I collapse the delta functions which are multiplied together and off by an index of one in each case. If you're able to elucidate that I would greatly appreciate it.

We compute
$$Tr({QP - PQ}) = \sum_n\sum_{n'} \left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n'} \delta_{n,n'-1}+ \sqrt{n'+1} \delta_{n,n'+1})\right) \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n} \delta_{n',n-1}- \sqrt{n+1} \delta_{n',n+1})\right) $$
$$- \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n'} \delta_{n,n'-1}- \sqrt{n'+1} \delta_{n,n'+1})\right)\left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n} \delta_{n',n-1}+ \sqrt{n+1} \delta_{n',n+1})\right) $$
$$= \sum_n-i\hbar \sum_{n'} (-\sqrt{n'} \delta_{n,n'-1}\sqrt{n+1} \delta_{n',n+1}) + (\sqrt{n'+1} \delta_{n,n'+1}\sqrt{n} \delta_{n',n-1})
$$
$$= \sum_ni\hbar \sum_{n'} (n+1) \delta_{n',n+1} + n\delta_{n,n'+1}
= \sum_ni\hbar$$
which is a sum, as expected, over the diagonal elements of the operator $QP - PQ$ and certainly does not equal zero.

 

[strangerep]

I've got the following but I'm still struggling to point at the exact step where things go wrong as contrasted with the finite dimensional case. In finite dimensions the proof uses linearity and the cyclic property of the trace -- I guess showing $Tr(PQ) \neq Tr(QP)$ is enough to show that particular proof strategy won't work, but not that the statement cant hold in general?

The problem is earlier than that. If $Tr(PQ)$ does not exist then you can't validly go any further. So,... compute $Tr(PQ)$ by itself and see what you get.

Edit: I also think I do something illegal in the last line when I collapse the delta functions which are multiplied together and off by an index of one in each case. If you're able to elucidate that I would greatly appreciate it. We compute
$$Tr({QP - PQ}) = \sum_n\sum_{n'} \left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n'} \delta_{n,n'-1}+ \sqrt{n'+1} \delta_{n,n'+1})\right) \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n} \delta_{n',n-1}- \sqrt{n+1} \delta_{n',n+1})\right) $$
$$- \left(-i\sqrt{\frac{\hbar M\omega}{2}}(\sqrt{n'} \delta_{n,n'-1}- \sqrt{n'+1} \delta_{n,n'+1})\right)\left(\sqrt{\frac{\hbar}{2M\omega}}(\sqrt{n} \delta_{n',n-1}+ \sqrt{n+1} \delta_{n',n+1})\right) $$

You seem to have lost a minus sign on the 1st term.

$$= \sum_n-i\hbar \sum_{n'} (-\sqrt{n'} \delta_{n,n'-1}\sqrt{n+1} \delta_{n',n+1}) + (\sqrt{n'+1} \delta_{n,n'+1}\sqrt{n} \delta_{n',n-1})
$$

Here's what I think the continuation should be. (Note the swap of summation parameters $n,n'$, and an implicit swap of summation order near the end.)
$$\begin{aligned}
=~& i\hbar\, \sum_n \, \sum_{n'} \Big(\sqrt{n' (n+1)} \delta_{n,n'-1} \,\delta_{n',n+1} ~-~ \sqrt{n(n'+1)} \delta_{n,n'+1} \, \delta_{n',n-1})\Big) \\
=~& \hbar\, \sum_n \, \sum_{n'} \sqrt{n' (n+1)} \delta_{n,n'-1} \,\delta_{n',n+1} ~-~ i\hbar\, \sum_n \, \sum_{n'}\sqrt{n(n'+1)} \delta_{n,n'+1} \, \delta_{n',n-1} \\
=~& \hbar\, \sum_n \, \sum_{n'} \sqrt{n' (n+1)} \delta_{n,n'-1} \,\delta_{n',n+1} ~-~ i\hbar\, \sum_{n'} \, \sum_n \sqrt{n'(n+1)} \delta_{n',n+1} \, \delta_{n,n'-1} \\
=~& 0 ~.
\end{aligned}
$$But, as I said, the fatal problem arises earlier when one tries to treat $Tr(PQ)$ by itself as a well-defined number.

 

[PeterDonis]

If $Tr(PQ)$ does not exist then you can't validly go any further.

Is there a theorem that says that, in order for $\text{Tr}([Q, P])$ to exist, $\text{Tr}(QP)$ and $\text{Tr}(PQ)$ must exist? After all, $[Q, P]$ is an operator in its own right.

 

[strangerep]

Is there a theorem that says that, in order for $\text{Tr}([Q, P])$ to exist, $\text{Tr}(QP)$ and $\text{Tr}(PQ)$ must exist?

Not that I know of, but don't let that stop you from looking further.

After all, $[Q, P]$ is an operator in its own right.

Indeed -- a multiple of the identity operator. But here we get into the intricacies of "trace-class" vs "non-trace-class" operators. Alas, I don't remember enough of my functional analysis studies to say much more than that. I sure wish former mentor micromass was still here.

 

[EE18]

The problem is earlier than that. If $Tr(PQ)$ does not exist then you can't validly go any further. So,... compute $Tr(PQ)$ by itself and see what you get.You seem to have lost a minus sign on the 1st term.

Here's what I think the continuation should be. (Note the swap of summation parameters $n,n'$, and an implicit swap of summation order near the end.)
$$\begin{aligned}
=~& i\hbar\, \sum_n \, \sum_{n'} \Big(\sqrt{n' (n+1)} \delta_{n,n'-1} \,\delta_{n',n+1} ~-~ \sqrt{n(n'+1)} \delta_{n,n'+1} \, \delta_{n',n-1})\Big) \\
=~& \hbar\, \sum_n \, \sum_{n'} \sqrt{n' (n+1)} \delta_{n,n'-1} \,\delta_{n',n+1} ~-~ i\hbar\, \sum_n \, \sum_{n'}\sqrt{n(n'+1)} \delta_{n,n'+1} \, \delta_{n',n-1} \\
=~& \hbar\, \sum_n \, \sum_{n'} \sqrt{n' (n+1)} \delta_{n,n'-1} \,\delta_{n',n+1} ~-~ i\hbar\, \sum_{n'} \, \sum_n \sqrt{n'(n+1)} \delta_{n',n+1} \, \delta_{n,n'-1} \\
=~& 0 ~.
\end{aligned}
$$But, as I said, the fatal problem arises earlier when one tries to treat $Tr(PQ)$ by itself as a well-defined number.

But this seems to disagree with @PeterDonis who, like me, gets a sum over $i\hbar$ for all $n$? Also, I agree that $Tr(QP)$ not existing per se is not an objection to $Tr([Q,P])$ existing and equalling 0?

 

[strangerep]

But this seems to disagree with @PeterDonis who, like me, gets a sum over $i\hbar$ for all $n$?

I should let Peter speak for himself, but I think he was replacing the commutator $[P,Q]$ by $i\hbar$ before computing the trace. That's not the same thing as you were doing. Is $(\infty^2 - \infty)$ equal to 0 or not? The question is meaningless.

 

[PeterDonis]

I think he was replacing the commutator $[P,Q]$ by $i\hbar$ before computing the trace.

Not really, I haven't made any computations in my posts. I was simply observing that, since we already know that $[P, Q] = i \hbar I$, any correct computation of $\text{Tr}([P, Q])$ as the trace of an infinite matrix must give the result of an infinite sum of terms each equal to $i \hbar$, i.e., must equal $\text{Tr}(i \hbar I)$.

 

[strangerep]

Not really, I haven't made any computations in my posts. I was simply observing that, since we already know that $[P, Q] = i \hbar I$, any correct computation of $\text{Tr}([P, Q])$ as the trace of an infinite matrix must give the result of an infinite sum of terms each equal to $i \hbar$, i.e., must equal $\text{Tr}(i \hbar I)$.

But then you're working with equations like "$\infty - \infty \stackrel{?}{=} \infty$" which is meaningless. As soon as one puts $\infty$ into any normal-looking equation, one can derive nonsense. E.g., since $\,\infty + 1 = \infty$, then by subtracting $\infty$ from both sides we get $1 = 0$. One must immediately stop algebraic manipulations as soon as one encounters a divergent quantity, and either retrace one's steps, or find some rigorous way to handle the divergence.

And it gets worse: afaik, there is no such thing as a "correct computation" of the trace of an unbounded operator. IIRC, Conway(or Kreyszig?) gives example(s) of an operator whose trace (wrt to one particular orthonormal basis) is 0, yet wrt to another orthonormal basis its trace diverges to $\infty$.

Googling for "trace-class operator" immediately gives useful expositions of this, e.g., the notes of Christa Hawthorne (based on Conway).

 

[PeterDonis]

then you're working with equations like "$\infty - \infty \stackrel{?}{=} \infty$"

I don't see why. If you have $Q$ and $P$ as infinite matrices, you should be able to formally compute $QP$, $PQ$, and their difference as infinite matrices, term by term. In general, yes, you can encounter ambiguities in how you group the terms, but to say this can happen is not the same as saying it must happen in every single such computation. It seems to me perfectly possible, at least logically speaking, for a computation of the infinite matrix $QP - PQ$ to yield $i \hbar$ times the infinite identity matrix.

 

[strangerep]

I don't see why. If you have $Q$ and $P$ as infinite matrices, you should be able to formally compute $QP$, $PQ$, and their difference as infinite matrices, term by term.

Formally, sure.

In general, yes, you can encounter ambiguities in how you group the terms, but to say this can happen is not the same as saying it must happen in every single such computation. It seems to me perfectly possible, at least logically speaking, for a computation of the infinite matrix $QP - PQ$ to yield $i \hbar$ times the infinite identity matrix.

Yes, but in the above you haven't performed the sums which are required to get athe trace of both sides. (Did you misspeak in the above?)

 

[PeterDonis]

in the above you haven't performed any sum which is required to get a trace.

That's right--I just described computing, formally, the matrix $QP - PQ$ and finding that it is $i \hbar$ times the infinite identity matrix. Once you know that, taking the trace (and seeing that it gives an infinite sum of terms each of which is $i \hbar$) is simple.

 

[strangerep]

[...] taking the trace (and seeing that it gives an infinite sum of terms each of which is $i \hbar$) is simple.

No, it's not. An infinite sum like $(\sum_{n=0}^\infty 1)$ is divergent.

 

[PeterDonis]

An infinite sum like $(\sum_{n=0}^\infty 1)$ is divergent.

Yes, I know that. I'm just saying that that infinite sum is what formally taking the trace of the infinite identity matrix gives you. I think something like that is what Ballentine intends to illustrate with the problem under discussion.