The h Bar

Slereah

3:00 PM

I guess the basis in QFT is just number states + spin states

which is fairly discrete

Semiclassical

yeah, right up until you pass to the continuum limit

bolbteppa

If it's in a finite volume it's discrete

Slereah

Hm

but that's just if you pick the number states as indexed by the momentum

which is probably not a good idea

Semiclassical

i mean, momentum basis is bloody useful

bolbteppa

At least non-relativistically, if it was on all of free space and you had a discrete spectrum, the energy level would have to be negative

Slereah

3:01 PM

Each part of the Fock state is just an $n$-particle Hilbert space, which is discrete

So it will be like uuuuuuh

I'm not sure

IIRC the KG Hilbert space is discrete so surely EM should be too?

bolbteppa

I think it extends to the relativistic case where we have already established anti-particles

Semiclassical

on a side note, i recently graded a few problems involving how to write stuff like the particle current density operator in the Fock space momentum basis

most people did it right

but

seeing people in a grad level course write down expressions that don't have sound units

2

Slereah

It should be in between $\mathbb{N}$ and $\mathbb{N}^\mathbb{N}$, but that doesn't really narrow it down

Semiclassical

is painfully depressing

bolbteppa

Again KG is discrete because of the finite volume assumption before taking the continuum limit right, now there's probably some comments in that PCT book I'm not sure about

Slereah

3:03 PM

I'm guessing the answer for KG is probably in Reed & Simon

although really, I think it may just be in non-relativistic books

Semiclassical

i mean, in fairness, i blame the error they made on Schwabl's notation for plane waves and how he states the orthonormality condition

but, you still need to be able to see that your final result will not have sound units if it contains the volume as a factor

Slereah

philsci-archive.pitt.edu/18363/1/…

3

Ah, Earman

you usually find him when there's something weird

bolbteppa

"When Mathematical Foundations was published in 1932 it was certainly true that separable spaces sufficed for all of the extant applications of ordinary non-relativistic QM." :(

That's a good find, it brings up Streater etc

Slereah

Earman is usually good

He writes a lot about weird physics aspects

Semiclassical

he talks about the thermodynamic limit starting on bottom of page 10

Slereah

3:15 PM

"If, as is conventionally assumed, the one-particle Hilbert space H is separable, then F(H), Fs(H), and Fa(H) are separable."

I am not feeling physics enough to read this paper rn

It will have to wait

Semiclassical

Earman discusses LQG starting on page 13

bolbteppa

It even quotes the Rovelli paper discussed above

Slereah

"If one wants to present these inequivalent representations of the free field all at once using an external direct sum to stitch them together then the direct sum Hilbert space is non-separable. But again the superselection story renders the non-separability innocuous."

so the answer is "It's complicated"

bolbteppa

Brilliant, "A basis for $\mathcal{H}_{diff}$ is provided by s-knot states which are labeled by continuous moduli parameters, resulting in a non-countable basis."

> "Various problems with doing quantum physics in non-separable Hilbert spaces have been discussed. These problems show that non-separable spaces disappoint expectations formed from operating in the separable arena and force deviations from the usual ways of doing business in this arena, but none of them presents a crippling roadblock to quantum physics. In any case, it seems shortsighted to lay the blame for perceived problems at the feet of non-separable Hilbert spaces"

I knew it!

As weird as all this is, it's staring one in the face with maybe the first problem one likely encounters, a free particle

Semiclassical

@glS my reading of those answers would right now be: if you know a state is separable, then finding the states which minimize the entanglement of formation is tractable if not easy

Semiclassical

3:44 PM

i think i've found a paper which gives the minimal decomposition for two-qubit Werner states explicitly. see the bottom of page 19 here: arxiv.org/pdf/quant-ph/9604024.pdf

the |e_i>'s are defined two pages earlier in eqn (21). just Bell states up to phases

glS

@Semiclassical I actually just remembered I asked a question about exactly that problem

3

Q: What is a separable decomposition for the Werner state?

Consider the two-qubit Werner state, defined as $$\rho_z = z |\Psi_-\rangle\!\langle \Psi_-| + \frac{1-z}{4}I, \quad |\Psi_-\rangle\equiv\frac{1}{\sqrt2}(|00\rangle-|11\rangle),$$ for $z\ge0$. Using the PPT criterion, one can see that this state is separable iff $0\le z\le 1/3$. I couldn't, howe...

Semiclassical

ooooooo

i'm a little surprised by that first answer tho

$\rho_z=(I+zZ\otimes Z-zX\otimes X-zY\otimes Y)/4.$

i'd have expected $-zZ\otimes Z$

easy to check tho

Slereah

@bolbteppa is any of us truly free

bolbteppa

Imagine how free free string field theory is

Semiclassical

oh, but the bell state that question is $(|00\rangle-|11\rangle)/\sqrt{2}$

glS

3:49 PM

@Semiclassical well, you want upper-left and bottom-right components of the matrix to be $z/4$, so the plus sign on $ZZ$ makes sense I'd say

Semiclassical

yeah

i was expecting rotation invariance, but that's only when you mix with the singlet state

glS

not that I have any idea where they came up with the ansatz in the first place though

Semiclassical

actually, i think i can shed light on that

b/c i saw this coming out from the averaging on another approach

over the sphere, the terms linear in $\alpha$ average to zero whereas $\alpha_j \alpha_k$ averages to $\frac13 \delta_{jk}$

Slereah

I should look up at Earman's bibliography rly

he makes a lot of good articles

Semiclassical

so after averaging the state becomes $\frac14(1-\frac13 \sum_j \sigma_j \otimes \sigma_j)$

Slereah

3:53 PM

sites.pitt.edu/~jearman

Semiclassical

and since this averaging gives us a werner state, we get that ansatz

@glS small question: is there a standard notation for bell states? Wikipedia has $|\Psi_-\rangle=(|01\rangle-|10\rangle)/\sqrt{2}$ for instance

whereas yours has $|00\rangle-|11\rangle$ instead

bolbteppa

philsci-archive.pitt.edu/2673/1/earmanfraserfinalrevd.pdf

That (Haag) is probably good too

Slereah

Haag's theorem is one of those theorem that philosophers of science love

Haag's theorem and Malament's theorem

Semiclassical

and QFT practitioners love to roll their eyes at

glS

@Semiclassical I usually try to abide with wikipedia's one. I probably just forgot which one was which in this case. Not that it ever really matters

Semiclassical

3:58 PM

well, it matters a little insofar as Werner states need the right symmetry

but in terms being separable? yeah, irrelevant

but anyways, i think the above argument gives $$\frac13(I/4)+\frac23 |\Phi_-\rangle\langle \Phi_-|=\frac14\left(I-\frac13 Z\otimes Z-\frac13 Y\otimes Y-\frac13 Z\otimes Z\right)$$

glS

@Semiclassical fair. But I don't know what properties of the Werner would change if you instead considered states that are locally unitarily equivalent to them

Semiclassical

not a lot, i reckon

bolbteppa

This guy is literally quoting specific sentences I know, incredible

Semiclassical

you'd basically just have $(1\otimes V)W$ states instead of $W$ states

so now invariant under $U\otimes (UV^\dagger)$ not $U\otimes U$

but, i mean

that's a pretty boring change all things considered

oh. derp. in my earlier equation, should've been $X\otimes X$ for the second term

woops

so now group it as $\frac{1}{12}(I-X\otimes X)+\frac{1}{12}(I-Y\otimes Y)+\frac{1}{12}(I-Z\otimes Z)$

at which point each of them is diagonal in the X/Y/Z basis

so yay, separable

glS

@Semiclassical right, I can see that. Then it's just a matter of playing with XX, YY, ZZ to make them into known projections. It's cute, but I don't really like this method. It's very specific to this particular case

Semiclassical

4:06 PM

ya

no argument there

glS

anyway, I half-remember that separable decompositions can be found numerically without too much trouble via semidefinite programming. Not that I've every actually done that

but also, deciding separability is computationally hard. So I'm not sure how these two propositions match. Probably the complexity of the algorithm finding a separable decomposition can be arbitrarily high, even if it's usually not

Semiclassical

4:29 PM

@glS one thing i do notice: when $z=1/3$, the sepearable decomposition for Werner in that answer involves 6 density matrices

but in thie case we're dealing with two qubits, so dimensions d=d'=2

so the other answer would claim that at most (dd')^2+1=5 densit matrices are needed.

so while the 6-matrix decomposition is elegant, it's evidently not the minimum possible

ymb1

On the Aviation site: In a banked turn, why does the center of gravity, rather than the center of lift, follow the path of a circular arc?: would love to see a physicist's take if anyone is interesting in posting an answer there

glS

@Semiclassical it's not that that number is needed. That number is sufficient for any state. Many states will need less. Finding the optimal separable decomposition is a different problem entirely

Semiclassical

"the other answer" should have been "the answer to the other question"

sure

but it does mean there's some way to condense the separable decomposition further

whether or not this is easy to find :P

my guess would be doing something clever with $X\otimes X+Y\otimes Y$

glS

@Semiclassical oh, yes, sure, that's true

that bound is just Caratheodory btw, given a decomposition it should be easy to find a more efficient one

Semiclassical

like, using the ladder ops $\sigma_\pm = X\pm i Y$

glS

4:37 PM

Carathéodory's theorem (convex hull)

Carathéodory's theorem is a theorem in convex geometry. It states that if a point x of Rd lies in the convex hull of a set P, then x can be written as the convex combination of at most d + 1 points in P. Namely, there is a subset P′ of P consisting of d + 1 or fewer points such that x lies in the convex hull of P′. Equivalently, x lies in an r-simplex with vertices in P, where r ≤ d {\displaystyle r\leq d} . The smallest r that makes the last statement valid for each x in the convex hull of P is defined as the Carathéodory's number of P...

Semiclassical

yeah

i'm familiar with it in the polytope case

makes sense that it'd work in general convex sets

glS

as always, Watrous proves this stuff:

Semiclassical

nice

$X\otimes X+Y\otimes Y = \frac12 \sigma_+\otimes \sigma_-+\frac12 \sigma_-\otimes \sigma_+$

bleh.

glS

so in practice, you reduce the decomposition in exactly the same way you'd do it to get a more efficient convex decomposition of a vector in a convex hull

@Semiclassical gotta go. Btw, if you want another fun exercise: show that all pure entangled states violate a Bell inequality (and more precisely, they probably all violate the CHSH). Context is physics.stackexchange.com/questions/671268/…

it's probably not too hard, but I haven't had enough time to think about it yet

oh, and here's another question I asked some time ago about separable decompositions:

3

Q: What separable $\rho$ only admit separable pure decompositions with more than $\mathrm{rank}(\rho)$ terms?

As shown e.g. in Watrous' book (Proposition 6.6, page 314), a separable state $\rho$ can always be written as a convex combination of at most $\mathrm{rank}(\rho)^2$ pure, separable states. More precisely, using the notation in the book, any separable state $\xi\in\mathcal X\otimes\mathcal Y$ can...

quantum-state entanglement

Semiclassical

4:53 PM

wave

antimony

11:08 PM

Ahh very interesting, thanks @ACuriousMind & @Slereah. Does this mean in single photon experiments, they are really just very very narrow wavepackets, rather than a true SINGLE photon? OR that the experiment is design such that the momentum/position resolution is sufficient to confine the linewidth within an acceptable margin of error? if so what happens to the position uncertainty in this case? --- i suppose the significance of that really depends on the experiment objectives?

ACuriousMind

@antimony ehhh..."position" is a thorny issue for photons even without the uncertainty/wavepacket issue

there's no good relativistic position operator, so photons don't really have "position"

antimony

i see, is this much more uncertain than say wavelength?

ACuriousMind

they have probability functions to be detected crossing a surface, that behave practically much like position but it's not like position for non-relativistic particles

antimony

ahh right

so is the single photon experiment label just marketing and not actually truly single photon?

ACuriousMind

well, the measurement process is that you have some screen that detects a single incident photon by absorbing it and triggering a current at the position of absorption

antimony

11:12 PM

right

ACuriousMind

but of course there's some inherent uncertainty in that position (you can't narrow down the position below the resolution of the "pixels" of whatever detector you're using)

antimony

and that position resolution is naturally finite, as is the detectable wavelength resolution

right

ACuriousMind

and detecting that single photon doesn't tell you whether or not the state before measurement was a single photon or some superposition - measurement changes the state, after all

antimony

ooooh

i see!

so we don't even know if it was a wavepacket or a single photon prior to the detection event?

only that the detection event is consistent with the energy(?) of a single photon rather than a cohort?

ACuriousMind

well...we know that the coherent light that lasers emit is not an eigenstate of the photon number operator

antimony

11:15 PM

> is not an eigenstate of the photon number operator

sorry thats a bit over my head

what does it mean?

ACuriousMind

it means that the pulse that comes out of a laser is not something you could describe as "this laser just emitted 10 photons"

antimony

ahh i see

ACuriousMind

it's something that, when measured, could be 8 photons or 12 photons or whatever - it's a "wavepacket" but not only in the sense of frequency or momentum but also in the sense that it does not have a definite number of photons

antimony

fascinating!

so single photon emission really means, a probability above <threshold> of single photon count?

ACuriousMind

when you turn down the intensity, your detector will start recording single blips, but that doesn't mean that the laser is emitting the photons one by one - it just means it's emitting something where it becomes very probably that we can detect photons one-by-one, but as I said, the state before measurement is different from the single localized photon we detect

@antimony yes

antimony

11:18 PM

wow interesting :)

ACuriousMind

this is also true for non-lasers - the time at which a single excited atom emits a photon is not deterministic

antimony

right

ACuriousMind

so when you put such an excited atom into let's say a sphere of photon detectors, what happens is that the atom evolves into a superposition of the state "excited atom" and the state "de-excited atom + emitted photon" and as time increases the latter part becomes more and more likely and at some non-deterministic point one of the detectors around the atom will detect a photon and collapse that superposition

antimony

incredible

ACuriousMind

at no point in time was there a pure "single photon" state propagating, just a superposition

antimony

11:21 PM

and that superposition was in place all the way during the time the photon was supposedly travelling from the atom to the detector?

ACuriousMind

yes - one part of the state consisted of the photon travelling, but the other (constantly shrinking) part was still just the excited atom sitting there with no photon around

antimony

i see

ACuriousMind

a laser is essentially a scaled-up version of this, and that's why it's not emitting a definite number of photons, but at low intensity you'll still get the detector screen very likely just detecting one single photon after the other

antimony

how do we know its very likely a single photon? is it somewhat discernable from the known responsivity and quantum efficiency of the detector? ie. we know the photocurrent cannot be too many photons?

i guess its unlikely within the detector wavelength bandwidth that a photon group could impart the same power as a single photon at a different wavelength

ACuriousMind

@antimony the laser has a certain intensity and a fixed wavelength and if you attenuate that intensity to the point where there's like one photon of that wavelength per second on average necessary to reach that intensity it becomes very unlikely you'll detect two photons at once

antimony

11:31 PM

ahh gotcha

thats is incredible

ACuriousMind

(caveat: this is very much my theorist account of what happens and the practical implementations may be much more complicated - I don't know much about actual quantum optics)

antimony

understood

oh bit of a silly question, but if re. that case we discussed earlier, where there's a single atom emitting a photon surrounded by a sphere of detectors. and the state of a single atom with no photon is constantly shrinking, does that shrinking theoretically never reach zero? but decays to some point we call "zero"

or at the point of detection it is actually definitely zero?

or not exactly? since some incredibly rare high energy event might excite the current in our detector or some non-zero probability of measurement error basically

ACuriousMind

the lifetime is usually related to the width of a Breit-Wigner distribution which never reaches zero, but of course you're at some point reaching the "this won't happen even if the universe existed fora billion times its lifespan" point

antimony

ahh i see, interesting thanks

ACuriousMind

emitting a photon is just like radioctive decay (gamma decay is actually exactly that) - there's a half-life, but never a point where you can be certain everything has decayed

gamma radiation just usually comes from excitation of the nucleus, not of the atom as a whole

antimony

11:40 PM

ahh i see right

ACuriousMind

I am genuinely delighted how you're just "this is amazing" and not "but how can that be?????" :)