no, i'm afraid not.

Then we get $\delta \omega = 1 / (1 \, \mu\text{s})$, which makes you think that $\delta \omega = 1 \, \text{MHz}$.

i don't see what's different.

wouldn't you just substitute in what $a^\dagger$ equals?

This is nonsense. $\delta \omega$ is supposed to be an angular frequency, it can't have dimensions of 1/time.

It has to be rad/time.

If we had put the rad in the $Q$, we would be less puzzled by this.

@heather you're misreading the initial relation

I need to get my laundry...

it's not what $a^\dagger$ 'equals'

it's how it acts on states

so if you give it a state with 3 photons it will return $\sqrt{4}|4⟩$

and so on

so, take $a^\dagger |n⟩ = \sqrt{n+1}|n+1⟩$

and add some primes for clarity

well, yeah, but if we don't know $n$ then isn't it kind of the same?

$a^\dagger |n'⟩ = \sqrt{n'+1}|n'+1⟩$

(acting/equality)

and then set $n'=n-1$

oh...

Wow! I got five new hats!!!

but...so $a^\dagger = \sqrt{n-1 +1}|n-1+1\rangle$

::octopus is happy::

or $a^\dagger = \sqrt{n}|n\rangle$

@heather no

you mean $a^\dagger |n-1⟩ = \sqrt{n}|n⟩$

oh.

i guess saying $a^\dagger = $ something is like saying $f = $ something when it's really $f(x) = $ whatever?

it's like taking $f(x) = x^2$ and writing it as $f=x^2$. The problem comes when you try to work out $f(y)$ - it's $y^2$, not $x^2$.

$$a^\dagger = \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ \sqrt{1} & 0 & 0 & 0 \\ 0 & \sqrt{2} & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{array} \right) $$

@heather exactly

Does that help?

@DanielSank needs more \cdots and \vdots

@heather Sort of

$f$ is a thing. It's just not true that $f=5$ (most of the time).

@DanielSank ::blinks::

@DanielSank the helium answer is back at 24 =|

I really need to get my laundry. Please excuse me if I added confusion instead of reducing it.

but that doesn't make sense. @EmilioPisanty just said $a^\dagger$ has to be dependent on what it acts on.

and $n$ is nowhere in that matrix.

@heather that matrix has all the possible $n$'s

or at least, it would, if Dan had put in the dots

okay

but for me, would it just be a 3x3 matrix because n = 0 or 1

it's not a three-by-three matrix, it's an infinity-by-infinity matrix

you're just looking at the first two-by-two block

if you're doing single photons

only 2x2?

@heather yes

per mode

but also, you have multiple modes

but isn't it $\sqrt{n+1}$ for the creation operator...?

or am i just mixing things up

@heather yes

the first row and column correspond to $n=0$

well...$\sqrt{0+1}$ -> $\sqrt{1}$, the second row.

@heather yes

that's perfectly consistent to me

do you see why now?

wait what

second row is $n=1$

you just said n = 0 was the first row.

you're counting up from zero

@EmilioPisanty right...?

in this representation, $$|0⟩ = \begin{pmatrix}1\\0\\0\\0\\ \vdots\end{pmatrix} \text{ and } |1⟩ = \begin{pmatrix}0\\1\\0\\0\\ \vdots\end{pmatrix}$$

does that help?

multiply that matrix times the matrix representation of $|0⟩$

i think i'm just being stupid, because i can't see what i'm doing wrong.

i guess the |0> state times the $a^\dagger$ matrix would give (0, sqrt(1), 0, ..., 0)

@heather yeah

how does that square with the matrix representation of $|1⟩$?

|1> times a dagger would give (0, 0, sqrt(2), 0, ...)

i guess that makes sense, i don't know.

@heather that wasn't the question ;-)

the claim is that $a^\dagger|n⟩ = \sqrt{n+1}|n+1⟩$ specializes to $a^\dagger |0⟩ = \sqrt{1}|1⟩=|1⟩$, and that this is perfectly showcased in the matrix representation

oh

wait

that actually makes sense

this is the creation operator!

so would $a^\dagger|n\rangle$ where n = 1 become |2> then?

@heather not quite

but it doesn't.

@DanielSank but shouldn't it..?

You forgot the prefactor.

@heather yes. but with an amplitude prefactor of $\sqrt{2}$

$$a^\dagger |n\rangle = \sqrt{n+1}|n+1 \rangle$$

Bye, everybody

but why is there an amplitude prefactor to it?

@EmilioPisanty shouldn't it just be |2>?

@DanielSank have a good day; thank you =)

@heather it could, but then it'd be a different operator

it has a prefactor because it has a prefactor ¯\ _(ツ)_/¯

¯\_(ツ)_/¯ feels like my response to a lot of physics =P

okay.

so is $a|n\rangle$ where $n = 0$ invalid? (sorry, semi-random question)

@heather no, it's valid

it's a good test of how well you understand the formalism ;-)

and so it gives |-1> !?

@heather but what about the prefactor?

hoo boy, let me see.

$i$?

$a|0\rangle = i|-1\rangle$?

@heather $a|n⟩=\sqrt{n}|n-1⟩$

ah, here it is

oh.

on what happens if you take away the prefactors

oh, wait! it's $0$, so...there's nothing?

@heather yes

that's correct

there's a difference between $|0⟩$ (the state vector that describes exactly zero photons in the mode) and $0$ (the zero vector in the state space)

@EmilioPisanty gosh that's confusing

$|0⟩$ is a normalized state, so $⟨0|0⟩=1$

@heather yes, yes it is

there are good reasons for the notation

so wait, if I apply the addition operator to |0>

then I get |1>

and it pays off later by simplifying a bunch of other stuff

but that bit does end up being a bit confusing

@heather you never call it 'addition' operator, really. it's the creation operator.

and...huh, okay, i guess this is kind of making some sort of sense.

let me try to do this: $H=\hbar \omega \, a^\dagger a$

$H = \hbar\omega a^\dagger \sqrt{n}|n-1\rangle$

$H = \hbar\omega \sqrt{n+1-1}|n+1-1\rangle\sqrt{n}|n-1\rangle$?

$H =\hbar\omega\sqrt{n}|n\rangle\sqrt{n}|n-1\rangle$?

$H = \hbar\omega n|n\rangle|n-1\rangle$?

@heather nope

you can't have double kets

::groans::

for one, the left-hand side should always be $H|n⟩$

okay

$H |n⟩= \hbar\omega \sqrt{n+1-1}|n+1-1\rangle\sqrt{n}$

oh

that makes more sense

so then $H|n\rangle = \hbar\omega n|n\rangle$?

@heather correct

can you make out what it means?

okay, so a hamiltonian represents the total energy of the system

and then this total energy is the amplitude of the number of photons

@DanielSank from that paper I just linked to

> When a comet or asteroid strikes a planet, it creates an enormous explosion that throws rock, water, dinosaurs and air into space.

I guess that makes sense?

an impact that big would definitely launch a good bit of rock into space

so

and it is equal to, um

presumably there are dinosaurs floating somewhere in space?

@davidphysics. San Diego is a pretty cool area too. Are you in the computational math program over there? I am assuming because you mentioned SD on your profile, and that area is very well known for computaional sciences.

@heather if $|n>$ is a way of saying you have $n$ particles in one of the available states of your system, and $\hat{H}|n>$ is measuring the total energy of the system being in that state, and since each particle has the same energy, what is the above result saying

uh

i have no idea, tbh.

Why in lightcone quantization would you begin by defining a timelike vector $n^{\mu}$ so that $n_{\mu} x^{\mu} = \lambda \tau$ becomes a new variable

@heather are you familiar with the Schrodinger equation?

@heather $H|n⟩ = n\hbar\omega|n⟩$ tells you two very important things

@bolbteppa not really.

one is that states with well-defined numbers of photons are one and the same as states with well-defined energy

As I said earlier, it is possible for states to not have a well-defined number of photons, in the same way that Schrödinger's cat doesn't have a well-defined aliveness value. This property tells you that well-definedness goes hand-in-hand for energy and photon number.

Secondly, it tells you that the actual energy in those well-defined-energy states is $E_n =n\hbar \omega$

i.e. an $n$-photon state has $n$ times $\hbar\omega$ of energy

where if you fiddle with the $2\pi$s, $\hbar\omega=h\nu$, with $\nu=\omega/2\pi$ the frequency of the mode

so each photon has $\hbar\omega$ energy?

@heather exactly

what is $\nu$?

@heather the frequency

Greek letter nu

what's $\omega$ then?

on high-school texts it's normally written as $f$ but that phases out to $\nu$ as you get to more serious physics literature

@heather $\omega = 2\pi \nu = 2\pi f$ is the angular frequency of the mode

it's got exactly the same information, but it's more useful when doing e.g. calculus and so on

okay.

anyways

$H|n⟩ = n\hbar\omega|n⟩$ is really Planck's formula $E_1=hf$ in disguise

@Blue are A and B polynomial spaces in two variables?

Doesn’t look like T will be surjective...

whoa, that's kinda cool @EmilioPisanty

@heather ;-)

so for SPDC though, I need to have three bosonic operators?

@heather yes

at least

like, as in $H = \hbar\omega a_p a_i^\dagger a_s^\dagger$, or a different hamiltonian altogether?

if you have multiple polarizations, then you'll need more

@heather that formula doesn't make sense

@EmilioPisanty yep, it probably doesn't.

there's a state on the left and an operator on the right

@EmilioPisanty that better?

@heather what's $n$? yes, that's it

though using $\hbar\omega$ in that context isn't great

@EmilioPisanty why not?

you do put in some initial prefactor, but you normally call it whatever else

what do you mean?

@Blue right, the linear indep thing I know (my thesis involves PDE on plynomial spaces)

@heather it looks too similar to the quadratic hamiltonian in something that decidedly isn't

the...quadratic hamiltonian?

@heather $a^\dagger a$

oh.

@Blue uhhh, no explanation on future slides?

it's quadratic, as in, it has two operators

alright, haven't heard that definition of quadratic before.

you normally use something like $H= \varepsilon \, a_p a_i^\dagger a_s^\dagger$, where $\varepsilon$ is some constant

but how do i know what that constant is?

@heather it'll depend on the experiment

oh.

it measures the nonlinear susceptibility of the crystal and its length, among other things

@Blue I don’t, because the spans of those two sets are not the same space!

@heather there exists a nice pair of coordinate operators for the harmonic oscillator, $q=(a+a^\dagger)/2$ ("position") and $p=(a-a^\dagger)/2i$ ("momentum"), normally called "quadratures", for which $H=\hbar \omega \, a^\dagger a = \hbar \omega \, \frac12 (p^2+q^2 -1)$.

does that make it look more quadratic?

As you said, they’re linearly independent. There’s no way to get xy from x and y.

@EmilioPisanty sort of?

=)

anyways, it's a name that's used

don't lose sleep over why

okay

now

in your example

you might have multiple polarizations

so you have multiple modes

per frequency

what do you mean by multiple polarizations?

like, different polarization of signal vs idler?

At Veggie Grill having one of those vegan burgers

so you need to do stuff like $$H= \varepsilon_1 \, a_{p,V} a_{i,H}^\dagger a_{s,H}^\dagger + \varepsilon_2 \, a_{p,H} a_{i,V}^\dagger a_{s,V}^\dagger$$ or something

@heather yes

and the pump as well

@EmilioPisanty aah

that looks nuts.

as to which combinations go into the hamiltonian, that will depend on the crystal

@heather well, you did want to do SPDC with multiple polarizations

it does look bad, but that's on a fundamental level about as bad as it gets

@Blue only if you can multiply the things

if it gets worse it's just a matter of degree and not kind

@EmilioPisanty i just wanted to find a matrix for SPDC =P it wasn't my particular desire to do it with multiple polarizations; that's just how it works.

Not for linear combinations

@heather well, if you fix your initial polarization and you use the right crystal, you can just have it restricted to a single term

@Blue a vector space with multiplication is an algebra

and that will give you a matrix with a whole lot of zeroes and then a single one

if you want to be a bit more flexible in what you want to feed to your system, then you do need those multiple terms

books on algebra

hmm, i'll have to think about that.

@Blue what Sam said :P

You know what deals with algebra?

Bourbaki.

I’m not sure that’s what you want though. It seems like a really vague statement that the author thought was trivial but is actually impossible to interpret

These are surprisingly common

@heather ok

i guess it actually depends on whether i can reach any state from the state after applying SPDC more efficiently than using, say, the KLM protocol.

@Blue I admit to not knowing statistics

@heather you want to implement SPDC just to benchmark something that will produce entangled photons?

or do you want a state-preparation procedure that will produce an arbitrary multimode quantum state?

@Blue I wouldn’t expect any of us to know this

if you want the latter, might I suggest that it's maybe a bit of an overambitious project?

Balarka and I know manifolds and some other stuff :P

I'm a manifold

@EmilioPisanty an arbitrary quantum state.

I'm an orbifold

@heather in which state space?

single-photon?

multiple photons.

single photon entangled over two modes?

single photon entangled over $n$ modes?

i believe the latter.

@Blue multivariate regression, multi-variable regression, or something?

@heather by 'multiple' photons, do you mean a state that has a well-defined total photon number bigger than one?

or do you mean a cat state of the form |(one photon in mode A⟩ + |(one photon in mode B)⟩?

also, is this something you want to understand mathematically? or is this something you want to actually implement in a lab?

i'm sorry, i clearly don't know what i'm talking about, but i think i mean a state that has a well defined total photon number bigger than one.

as for the latter question, i'd like to implement it in the lab, but i would implement it on a small scale.

@heather can you give examples of the types of states you're hoping to get?

the bell states could be an example, i suppose.

i think i'm explaining what i mean very poorly.

i'm trying to create a realization of quantum computing which uses SPDC as its nonlinear component, analogous to using the Fabry-Perot cavity.

and then build a small version of a quantum computer using that realization in the lab.

@heather you realize that a successful realization of quantum computing is some years away on bleeding-edge research labs, right?

particularly as regards photon-based quantum computing

@EmilioPisanty i was just about to say this all probably sounds very stupid and naive. so, yes.

if you set your goal as using SPDC to produce an entangled state of a single photon over two modes, and to conclusively demonstrate that entanglement, then I would say that that is a ridiculously ambitious goal but it is still worth pursuing

and that I would be extremely (and very pleasantly) surprised if you achieved it, but also that you would learn all sorts of useful things along the way

hoping for a home-built implementation of anything that resembles a photon-based quantum computer sounds implausible to me

however

if you're shooting for the big goals

then why not do like research labs (and also research disciplines) do

and shoot first for the small goals, and then use the lessons learned there (and the hardware developed for that) to go for something bigger?

Also

note the foreground role of "conclusively demonstrate" in that first comment. Creating entangled states isn't that hard. (that's what decoherence is, really.) The really challenging aspect is coming up with and implementing a detection procedure that will unambiguously prove that what you've created is entanglement between the modes that you'd set out to entangle, and nothing more.

that's often months of work on research-grade labs

(though they are spending a lot of that work into implementations that will then be useful for other fancier things. it doesn't mean that you need their resources or their combined expertise time to do it.)

this is all to say

it's hard to underplay just how impressive a feat it'd be just to demonstrate entanglement at home

@heather no worries

keep up the good stuff, though!

Just bough some super expensive nice smelling, natural body wash and matching lotion. Presumable now i will start smelling like a male god.

@EmilioPisanty Well, the battered, singed, vacuum dried, and irradiated remains of dinosaurs, any way.

But that's almost the same thing, right?

@dmckee hey, I have a career question

ok then. . . ha nvm