# Trace of operator with continuous spectrum

1. Nov 6, 2013

### Delta Kilo

Greetings,

I must be missing something obvious but how is Tr{} defined exactly in case of contunuous spectrum operators? Everywhere I look I see it defined as a sum of [possibly infinite sequence of] eigenvalues. Is the following correct:
Given $Q = \int f(q) \left| q\right\rangle \left\langle q\right| dq$, where $\left\langle q' | q'' \right\rangle = \delta (q'-q'')$, then $Tr \{\rho Q \} = \int f(q) \left\langle q | \rho | q \right\rangle dq$ ?

Thanks, DK

2. Nov 6, 2013

3. Nov 6, 2013

### Delta Kilo

The definition here (and elsewhere) is only for discrete sequences of basis vectors, not for continuum. Is it just a matter of replacing sums with integrals and that's it? What about convergence criteria and suchlike?

4. Nov 6, 2013

### kith

I would say yes. If the integral doesn't converge, the operator is simply not in the trace class.

The density operator is in the trace class by definition. Since tr{AB}=tr{BA} => tr{[A,B]}=0 but [x,p]=i*hbar, at least xp and px can't be in the trace class.

5. Nov 6, 2013

### DrDu

A necessary condition for a hermitian operator to be trace class is that it has only a discrete spectrum.

6. Nov 6, 2013

### Delta Kilo

What about $\rho = \int_0^1 |x\rangle \langle x | dx$, where $\langle x' |x'' \rangle = \delta(x'-x'')$? Is this a valid state operator? What about its spectrum?

7. Nov 6, 2013

### dextercioby

The trace of an operator which is not necassarily self-adjoint is defined only wrt a countable orthonormal set in a Hilbert space. There's always a self-adjoint operator for which this set is its (true) eigenvectors set.

8. Nov 6, 2013

### rubi

The definition also works in case of an uncountable basis. In this case, one can show that at most countably many terms of the sum can be non-zero (otherwise, the sum diverges).

9. Nov 6, 2013

### Delta Kilo

Thanks for your replies but I am even more confused now.
Can someone please show me (or point me to) an example of state operator ρ for a spin-0 particle in free space which corresponds to a gaussian wavepacket?

10. Nov 6, 2013

### Staff: Mentor

Dextercioby gave the correct answer IMHO.

To the original poster this is a VERY mathematically non trivial issue (when you hear those words the translation is its HARD).

The correct way to handle continuous spectrums, while it can be done the way Von Neumann did in his classic Mathematical Foundations using Stieltjes integrals and so called resolutions of the identity, is not the way its usually done in physics, and you need what are called Rigged Hilbert Spaces:
http://arxiv.org/pdf/quant-ph/0502053v1.pdf

But my advice is, if you are just starting out in QM, is not to get caught up in this stuff. I did, and it lead me on a sojourn into some very esoteric areas of pure math. I emerged with a good understanding of this stuff, but its not really germane to the physics which is what you should be concentrating on at the start.

Thanks
Bill

Last edited: Nov 7, 2013
11. Nov 6, 2013

### kith

A wavepacket corresponds to a pure state so the state operator is simply ρ=|ψ><ψ|.

12. Nov 6, 2013

### Delta Kilo

Bill,

It's actually my third attempt. First time was long long time ago in a galaxy far far away and I skipped most of the lectures as I had more important stuff to occupy myself with :) Second time I skimmed through some introductory texts and then Griffith but I took a great deal of it for granted and it didn't sink in.
This time, following suggestions on this forum and yours in particular, I'm going thought the Ballentine text pretending I forgot everything I read on the subject before. I do not mind gory details, in fact I'm looking for them.

Ballentine starts with postulating the existence of a state operator ρ with Tr{ρ} = 1. Nothing at all is said about the vector space on which the operator is hmm... operating, apart from it is being some sort of [rigged] Hilbert space. OK. Then at some point positional operator Q is introduced as Q|ψ> = x|ψ>. This makes ψ a continuous function of x: ψ=ψ(x). I take it as a hint that we are now operating in a space of functions of x with inner product defined as $\langle \phi | \psi \rangle = \int \phi^* (x) \psi(x) dx$. Hence the question how do I calculate trace for an operator on this space.

Can you be more specific? For a simple wavepacket $|\psi \rangle = e^{-x^2/2a}$, and $\rho | \phi \rangle = (|\psi \rangle \langle \psi|) | \phi \rangle = e^{-x^2/2a} \int \phi(x') e^{-x'^2/2a} dx'$, right? How do I apply textbook definition of Tr{ρ} to it, where do I get the required countable orthonormal basis $|e_k\rangle$ for the space I'm in?

13. Nov 6, 2013

### kith

As I wrote in my first post, I would simply use the integral for the trace. tr{ρ} = tr{|ψ><ψ|} = ∫dx<x|ψ><ψ|x> = ∫dx|ψ(x)|² = 1 if ψ(x) is normalized.

Obviously, the integral doesn't converge if ψ(x) is not a square-integrable function. So the correct Hilbert space is L². The states |x> above are not basis states of this space, they do not even belong to it. So there's no contradiction to what DrDu and rubi said.

Last edited: Nov 6, 2013
14. Nov 6, 2013

### rubi

You just pick some random orthonormal basis for your Hilbert space. In the case of $\mathcal H = L^2(\mathbb R)$, you can choose the harmonic oscillator eigenstates $\left|n\right>$ for example. Then you just compute $\mathrm{Tr}\rho = \sum_n \left<n\right|\rho\left|n\right> = \sum_n \left<n|\psi\right>\left<\psi|n\right> = \sum_n \left|\left<n|\psi\right>\right|^2$. The trace is independent of the basis you choose.

15. Nov 7, 2013

### Staff: Mentor

Well since the test space of the Gelfland triple has a basis of elements for the Hilbert space one can define it in the usual way ie Ʃ <bi|A|bi> which is well defined since A can be applied to the |bi>.

There is no problem there.

What you may be worried about is expressions like <x|X|x'> where the state x is represented by the Dirac delta function and you end up with the dreaded square of the Delta function ie ∫ δ (x'-x'') X δ (x - x'') dx''. Making things like this rigorous is far from trivial and you will have to consult advanced treatments on it. The following Phd thesis is the standard treatment:
http://physics.lamar.edu/rafa/webdis.pdf [Broken]

The reason you haven't got a straight forward answer, other than what I said above, which is likely to be unsatisfactory to you, is because such doesn't exist - it's a very difficult issue.

Thanks
Bill

Last edited by a moderator: May 6, 2017
16. Nov 7, 2013

### Delta Kilo

OK thanks, I think I got it (sort of). I'm trying to make sense of this quote:
So, in the expression <Q> = Tr{ρQ}, ρ=ƩAk|Ek><Ek| is over L2(R) with discrete basis |Ek> but Q=∫x|x><x|dx is over Ω× with continuous basis |x>, do I get this right?

To compute the trace I can write Tr{ρQ}=ƩAk<Ek|Q|Ek>, then express each |Ek> as an integral in terms of |x> and then massage it until I get Tr{ρQ}=∫x<x|ρ|x>dx. Is this what happens when Ballentine computes the trace in the unnumbered expression just before (2.29)?

17. Nov 7, 2013

### DrDu

I think rigorous texts are avoiding to use the trace. However, this is quite a world on it's own.
The standard method to obtain representations of operators for mixed states is the GNS construction:
http://en.wikipedia.org/wiki/GNS_construction
For thermal states, which are maybe of most interest, the Kubo Martin Schwinger construction is standard:
http://en.wikipedia.org/wiki/KMS_state

There are quite some text available explaining this C* algebraic approach to QM.

18. Nov 7, 2013

### Staff: Mentor

Indeed.

I suspect the trace form of the Born rule still holds even for continuous observables, but its trickier. It cant involve the form I posted previously because if x=x' you get the square of the delta function which is a massive can of worms.

Conceptually I think it's best to think of such cases as a discrete dust and take limits. Stuff inside the limit cause all sorts of issues like the square of the delta function but outside is OK.

Thanks
Bill

19. Nov 7, 2013

### DrDu

Nevertheless this KMS condition forms the basis e.g. for the Matsubara formalism in QFT which is quite standard.

20. Nov 7, 2013

### kith

I don't know the details of this rigged Hilbert space business but isn't the main result that all pathological functions like exp(ikx) or δ(x) behave nicely if we simply replace H by the extended space Ω× in the formalism? We just have to keep in mind, that state functions need to lie in the Hilbert space which is a subspace of Ω×. If the |x> are eigenstates of the self-adjoint operator X, they should form a basis of Ω×. I don't see a problem with using these states in the trace directly, so I guess this is what Ballentine did in the equation you are asking about. But maybe there are some mathematical subtleties I'm missing.

21. Nov 7, 2013

### rubi

Everything happens in $L^2(\mathbb R)$ here. $Q = \int x \left|x\right>\left<x\right|\mathrm d x$ is just physics notation for the spectral theorem for unbounded self-adjoint operators. $\left|x\right>\left<x\right|\mathrm d x$ is a projection-valued measure. There is no continuous basis involved here (every orthonormal basis of $L^2$ is countable!). The formula basically tells you that $Q$ is unitarily equivalent to a multiplication operator on some Hilbert space.

Getting the integral expression is actually not so hard and you don't need any rigged Hilbert spaces for it at all. For example if $\rho=\left|\psi\right>\left<\psi\right|$, then $\mathrm{Tr}\rho = \sum_n \left<n|\psi\right>\left<\psi|n\right> = \sum_n \left<\psi|n\right>\left<n|\psi\right> = \left<\psi\right| \left(\sum_n\left|n\right>\left<n\right|\right) \left|\psi\right> = \left<\psi|\psi\right>$, which is just defined using an intergal. (This is not yet rigorous, but can be made rigorous easily, using some continuity arguments.)

You have to remember that we are actually using $L^2(\mathbb R)$ as our concrete Hilbert space, where the inner product is defined using an integral. There is no need to express most operators using the spectral theorem, because we know how they act on elements of $L^2$. For example $Q:\mathcal S\rightarrow\mathcal S$, $(Q\psi)(x) = x\psi(x)$. This is way clearer than writing $Q=\int x\left|x\right>\left<x\right|\mathrm d x$. The $\theta(q-Q)$ operator, Ballentine is using, acts on states simply by $(\theta(q-Q)\psi)(x) = \theta(q-x)\psi(x)$. You can derive most of these integral expressions by writing out the standard inner product of $L^2$ and possibly applying standard theorems like Fubini. Rigged Hilbert spaces are pretty much overkill here.

22. Nov 7, 2013

### Delta Kilo

Thanks for the answer, but I just don't see how is it possible. Operator Q takes a perfectly valid member of $L^2$ and flings it over the fence into some other space. We need to either restrict the domain of Q or expand the range, in either case we are no longer operating on $L^2$. As I understand it, both choices have merits and result in two different schools of thoughts. The latter leads to RHS and Ω×, where $|x\rangle$ are valid kets forming continuous orthonormal basis. Please correct me if I'm wrong.

"I swear, it's not what it looks like!"

23. Nov 8, 2013

### rubi

Of course, $Q$, being an unbounded operator, has a restricted domain. (There aren't two schools of thought. Even if you use rigged Hilbert spaces, the domain and range of $Q$ doesn't change.) One usually uses the Schwartz space $\mathcal S$. It is however a subspace of $L^2(\mathbb R)$ and $Q:\mathcal S\rightarrow S$, so everything happens in $L^2(\mathbb R)$. $\left|x\right>$ is a valid mathematical object if one introduces rigged Hilbert spaces, but it isn't in $L^2(\mathbb R)$ anymore, so it can't form a basis for it. It doesn't even form a basis for $\Omega^\times$ ($=\mathcal S^*$), especially not an orthonormal one (since there is no notion of orthogonality in $\Omega^\times$).

It is really just the spectral theorem expressed in a different notation. A mathematician would write $Q=\int x \mathrm dE$, with $E$ some projection-valued measure. We can just introduce the notation $\mathrm dE = \left|x\right>\left<x\right|\mathrm d x$. This doesn't require $\left|x\right>$ to be an individual object. Of course, you can make it into one by introducing RHS, but this doesn't enable you to prove things that you couldn't prove without them. It just enables you to prove these things using Dirac notation. The price you have to pay is that you have to learn the formalism, while the standard mathmatician way of doing it requires just some basic functional analysis.