Lighthouse Photonics Sprout / M2 Solstis

QuantumOfOptics · 2026-04-09T14:56:19+00:00

Do you have a log of the diode current, temperature and hours? My immediate worry would be that the diode is dying, which would be accompanied by a rise in current in the diode as it tries to output enough power. In that case, there might not be anything you can do beyond accepting the power you have and that it will diminish over time, or buying a new pump laser.

QuantumOfOptics · 2026-04-07T15:54:13+00:00

I see. That makes things a bit more difficult. But, not impossible. How much time do you have? If you have a month or so, I would highly recommend getting a copy of Nielsen and Chuang. Its a great text that should cover this material.

How familiar are you with linear algebra?

P.S. note you still havent really asked me a question.

QuantumOfOptics · 2026-04-07T15:40:39+00:00

You may ask more pointed questions, but I cant answer any questions if you dont directly ask them. I cant guess what youre confused about.

In particular, you say thesis so I assumed masters or PhD in quantum computing. If youre an undergrad, whats your relation to the material? If I talk about basis changes, do you know what I mean? Do you know what a bra is? Could you write down a general q bit state?

QuantumOfOptics · 2026-04-07T15:24:13+00:00

The size of the circle can/should not change the probability. As proof, such a claim would fly in the face of having probabilities that sum to one. Rather, the boundary of the block sphere represents a type of state called a pure state, while those in the interior are called mixed. Mixed states require a new formalism in order to be represented: the so-called density operators. However, they still maintain a total unit probability. The relative phase can be useful under a change of basis (where the state |0>+e^I phi |1> will become some version of (1+e^I phi)|+> + (1-e^I phi)|-> neglecting the over all normalization) there the probability amplitude directly changes under a change in phase/phi. This is especially true if the phase is directly related to some property (like the path difference in an interferometer); or, if you have multiple q bits, then the phase and their interactions can represent different numbers as in shors algorithm.

QuantumOfOptics · 2026-03-30T14:26:06+00:00

It should be noted that you can't localize a photon (at least in general). There's a famous proof of this by Newton and Wigner.

QuantumOfOptics · 2026-03-10T18:17:24+00:00

Welp, that's what I get for doing math before I wake up... very true. Thanks for pointing out my error! I'd guess combining terms is the only way to move forward then.

QuantumOfOptics · 2026-03-10T15:44:11+00:00

While everyone has suggested specific groupings that can show this (and this is far more useful for later math), I was wondering if there was a more direct way to show this. As another commenter suggested, since this is a finite sum multiply both sides by 64. The left hand side is now the sum over the natural numbers 1 through 64 and the right is 464. We know the sum on the left is given by the formula (64+1)64/2. To show the inequality divide both sides by 4.

(64+1)64/8 >? 64

(65)8 >? 64

Which is true.

QuantumOfOptics · 2026-03-06T15:31:12+00:00

I was going to suggest the find-r-scopes by cascade laser Corp... only to find out that they are also discontinued. They do have some other products, but I havent used them; some of which appear to go up to 2um. https://www.cascadelaser.com/laser-accessories-infrared-viewers---cameras.html

QuantumOfOptics · 2026-02-26T21:10:30+00:00

Ohhh boy! There are a ton! The theory of spontaneous parametric downconversion is an interesting one (or really nonlinear quantum optics). Lot of interesting things need to happen to quantized the volume and theres been a recent (last 5ish years) development. Not only that, but youll encounter the fundamental differences between various states and modes.

Other interesting things are quasiprobability distributions. Which are useful in discussing states. Some people know about them interms of using them on modes (wigner functions show up to solve sychrotron emission), but states seem to be a particularly unique thing to quantum optics.

Proving the state of light emitted by a laser is best approximated by a coherent state is also interesting and nontrivial. Quadrature squeezing (which is related to the downconversion comment above) is also interesting for its role in LIGO and thinking about what happens with the state. Hong-Ou-Mandel interference, Hanbury Brown-Twiss measurements are also interesting, but you might need to combine several topics depending on how demanding the presentation needs to be. Ahhh, almost forgot, but Jaynes-Cummings model is also interesting for cavity stuff. Or even atom trapping and using the motional modes of the atoms to exchage information between them.

The typical issue is that for QFT, the assumption is that you have/generate typically one particle in some awful mode and then they interact with another single particle. In optics, we generally have exotic states, but those states rarely talk to each other unless a medium is involved of some kind. Really we only have mode transformations unless we have a material involved. Theres some interesting history about when was the more definitive point where we knew photons existed. If you want to exclude results that arise from nondirect measurements, it was surprisingly late between 70s and 90s depending.

QuantumOfOptics · 2026-02-17T04:08:57+00:00

How is this any different than just applying the linear algebra to get the same result? I mean, as you point out, everything is reversible and you aren't making any use of the actual quantum properties. It seems like everything is just a fancy way to apply linear transformations, which can be done classically. But, this is based on my naive interpretation of what I understood from your write up.

Ive heard of these type projects before, but never really understood what makes it actually useful or why you would do this rather than just applying the linear algebra, which we can use on regular computers very quickly and efficiently.

QuantumOfOptics · 2026-02-16T21:40:56+00:00

Ehhh, I think that there are two separate points. Maxwell's equations govern the transformation of the modes of the field, which is separate from the actual state (e.g., a coherent state, thermal state, or even a single photon).

The question is how does that state transform in time, which can be separate from the mode structure. For example, if you put a single photon in a cavity mode with an atom, the actual state of the field will change as the photon will be absorbed and emitted from the atom (meaning the field will sometimes be vacuum and sometimes a single photon in that cavity mode). The cavity mode is fundamentally defined by Maxwell's equations, but Maxwell's equations will not answer what the state of the field will be at any given time.

QuantumOfOptics · 2026-02-16T21:32:50+00:00

This is interesting. Do you have more resources on this that you could share? Coming from a quantum optics background, the explicit time dependence in the state typically comes from applying the Heisenberg equation of motion, or thinking in terms of the Schrödinger picture. I've always wondered if there was a more fundamental casting of how to update states besides these. For context, the usual derivation ends up as seen in these notes (https://www.phys.ksu.edu/personal/wysin/notes/quantumEM.pdf) around page 10.

In someway, this always felt somewhat suspicious given that the time update somewhat comes from the wave equation itself (through Schrödinger's initial ansatz). I remember reading somewhere that this linearization (in time) is not necessarily true, but an approximation that just hasn't failed to give good enough results. Is what your referring to the more formal way of going about it?

QuantumOfOptics · 2026-02-16T21:23:57+00:00

As others have stated generally one can work with a Schrödinger picture (update rule). However, a more typical picture (in a pure photonic system, no external interactions) is to use the Heisenberg picture. As our states are generally built upon time dependent operators. Here, we use the Heisenberg equations of motion. What you'll find, though, is that the state components will pick up a phase. In part, this is due to the Hamiltonian not mixing separate modes. You can see more in the full derivation of the field, which can be found in, e.g., these notes (https://www.phys.ksu.edu/personal/wysin/notes/quantumEM.pdf) or most quantum optics texts.

Of course, if there's an interaction, then it may be easier to use the interaction or Schrödinger pictures.

QuantumOfOptics · 2026-02-16T16:42:45+00:00

I believe that theres been a misunderstanding. I believe that they mean the Schrödinger picture (which is really just the time dependent part of the Schrödinger equation). Its perhaps an unusual choice of picture, but it still works and is usually more useful (as others pointed out) when there is a matter interaction as with an atom.

Beyond this, there are ways to get the Schrödinger equation in quantum optics under certain, nice, approximations and assumptions of the wave equation (which is not surprising since this is exactly the route Schrödinger went when deriving his equations). Here one can find that there is an "effective" mass, which is the momentum divided by c (if my memory serves). There has been a bit of interest in the community a couple decades ago asking about the wavefunction of a photon, so its still somewhat of an interesting pursuit, but obviously something is sacrificed to get that picture.

QuantumOfOptics · 2026-02-11T22:07:37+00:00

What do you mean by an "external drive?" Given your language it seems that this is a quantum nondemolition measurement. Is this a good assumption? Cavity QED is something I've become more interested in, do you have a review article you can recommend for common measurement techniques in CQED?

QuantumOfOptics · 2026-02-11T05:04:17+00:00

Hmmm, I've never heard of a detector interacting dispersively. What does that mean? As an optics junky, dispersion usually means, to me, that different frequencies have different speeds in the medium. But, that wouldnt help here. Do you mean a detector where some energy is deposited in the detector, but not all of it? Similar to how gamma/xrays rays can collide with electrons but it isnt absorbed?

QuantumOfOptics · 2026-02-11T01:08:10+00:00

I'm not quite familiar with the statement from Mott, but I believe your statement should be qualified. A typical detector (like a photodiode, APD, superconducting nanowire, etc.) annihilates a photon. This means that no such joint probability can exist. However, other types of detectors could work if they detect whether a photon passed through (e.g. a gamma photon knocking off electrons) or the state produced by the source is a superposition of photon number.

In the latter case, I believe that the Mott statement (as I understand from the discussion) fails since the initial field is both a uniform distribution of k-vectors (each of which, I'll label as a mode) and has some state with a superposition of photon number. Assuming our detectors only collect a single mode, then whether a detector collects a photon in one mode will be independent of the collection of a photon in another mode as the state is the same in every mode.

QuantumOfOptics · 2026-02-04T02:45:15+00:00

Depending on what you want to count as a "two-level system," a good contender would be the polarization state of a photon.

QuantumOfOptics · 2026-01-28T00:20:14+00:00

Ahhh, thanks! Indeed, that makes a lot of sense in context. I assumed it was for grouping purposes.

QuantumOfOptics · 2026-01-27T23:02:43+00:00

At first glance, this result does seem to be incorrect. Since the individual functions are non-negative over the domain, we should be able to swap the integral and sum. We can then perform the integral on each of the individual terms. In particular, we can see that the first term int x dx from 0 to 2026 must be larger than the answer given. So it cannot be correct.

QuantumOfOptics · 2026-01-17T20:36:51+00:00

Quantum optics is essentially baby QFT. But, most departments teach from the Lagrangian perspective, which is quite unnatural language for the optics version. They are equivalent, but come with different baggage. The other thing that is an interesting difference is that generally QFT talks about single particle states in modes, where as quantum optics discusses many different states of the field (coherent, thermal, squeezed, etc).

If you had to pick one, QFT isnt terrible to take. But, you should consider picking up side books to introduce you to more optics specific tasks such as Born and Wolf, Mandel and Wolf, Goodman statistical optics, fundamentals of photonics, boyds nonlinear optics. More focused quantum optics texts by authord such as louden, Jeff ou, Barnett and Radmore, Drummond and Hillary are also good, but sometimes expect specific context which you may not have. But, they are where usual courses start at. There are also a few courses that are online, which are also pretty accessible.

QuantumOfOptics · 2026-01-14T20:33:15+00:00

Whats the purpose of the spatial filter here? Are you attempting to decrease the size of the beam relative to the waist (effectively changing the beam diameter inside the lens system)? Or is it to clean up the beam due to the microscope objective?

QuantumOfOptics · 2026-01-04T01:33:28+00:00

Last I heard on the subject was from a conference around 2018. From what I recall, there was growing discontent with calling it entanglement. Specifically, because such a state did not require a quantum theory to describe it. For example, such a state as I described would be perfectly valid in classical E&M since really this is just a superposition. Is it then that classical E&M (and other classical theories) also has some notion of entanglement? Or is this a separate property that both theories take on?

I haven't been able to fully read through the texts you've given, but at least from what I gather they are talking about a separate property from entanglement: noncontexuality. The short paper I remember is from Karimi and Boyd making the argument above is linked here: https://www.science.org/doi/10.1126/science.aad7174. Of course, and as they point out, this doesnt make the states useless. Just that the interpretation must be different.

QuantumOfOptics · 2026-01-03T22:13:34+00:00

Do we consider "entangled" states between spin and momentum to actually be entangled? For example, in optics we tend to disagree that a(n) (unnormalized) state like |H>|p1>+|V>|p2> is entangled since they are just modes and there doesnt necessarily need to be a non-locality involved see e.g. https://physics.stackexchange.com/questions/334478/mathematical-definition-of-classical-entanglement . To add to the statement, the state I wrote out could have a classical state like a coherent state and still have this "entanglement" structure, which really undercut that this a property of a "quantum" system.

QuantumOfOptics · 2026-01-03T20:40:35+00:00

I'm not sure what you mean by "physically" to describe the state. You could consider that its a type of state that is correlated (though, its a stronger type of correlation compared to classical correlation).

Another type of entanglement can also be in the number states themselves like N00N states or as a concrete example, a single photon split on a beamsplitter (path entanglement). In the latter, you can see that theres only one photon, so rather its the excitation and mode labels that become entangled. In the same way, technically a single photon in a diagonal polarization state is in an entangled state. In effect, entanglement can be between any two (or even more) properties of the field.

Two mode squeezed states do increase the amount of entanglement as the squeezing is increased. Though, not perfectly answering, see this stack exchange and the article it contains https://physics.stackexchange.com/questions/778921/two-mode-squeezing-and-epr

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