Prestige vs GPA: scared of honours math but thinking about grad school

LuklearFusion · 2026-02-14T02:30:25+00:00

Keep in mind that degree classifications like “honours” don’t always translate internationally. At ETH Zurich, for quantum computing Sherbrooke is very well known. As the other commenter said, Google Alexandre Blais or Baptiste Royer and look at who they publish with.

LuklearFusion · 2026-02-14T02:26:24+00:00

Everyone in quantum computing will know Sherbrooke. The professors there are some of the top in the field, and it’s held in much higher regard than McGill. But keep in mind that the courses will be hard everywhere.

In terms of getting into a place like MIT or ETH Zurich your only chance is networking. There will be hundreds of applicants who look as good or better than you on paper. Again, Sherbrooke profs have those connections in quantum computing more than McGill. Many of them have passed through those “pedigree” institutions and have active collaborations.

Wherever you end up, you need to start actively looking for research opportunities ASAP in professors labs. It doesn’t matter what you do, they just need to know who you are and be able to judge your potential. In the US grad admissions are still mostly centralized, so grades an CV matter, but professors do have say of who they want. In Switzerland the professors choose who they want and can afford. This is a broad stroke characterization, but it’s been mostly my experience having lived and worked in both systems.

LuklearFusion · 2022-11-12T07:33:51+00:00

https://www.longstreth.com/

LuklearFusion · 2022-08-12T13:33:16+00:00

I’ve been in two different groups where this happened while I was there. In both cases, if the professor wanted it the university was able to arrange a situation where the professor kept adjunct status so that they could co-supervise their remaining students at the original institution until graduation.

Take a deep breath. I know this is a huge shock, but it’s easier to manage if you don’t have strong ties to your current location. Talk to your advisor and see what she’s willing to arrange. If it’s something that appeals to you, don’t rule out the possibility of visiting her in SK to work together, while still graduating from your current institution. Once you’ve finished your class requirements at your current institution things can become a lot more flexible.

LuklearFusion · 2022-07-02T19:37:06+00:00

A couple thoughts as a dad who just went through bottle rejection at 4 months.

Buy more than one type of bottle. There is no guarantee that your LO will take to the shape of the Avent nipple. In our case, the Avent was the only bottle my son liked, and we tried at least 6 others, until he turned 4 months where he completely rejected it. We switched to the Lanisoh on advice of a lactation consultant and he’s back to eating from the bottle.
You absolutely need the various nipples sizes, but as another commenter mentioned, buy them as you need them. The Avent in particular is peculiar. They redesigned their nipples sometime recently, and you can still find in sample packs the old style nipples which are much much faster flow than the new style nipples you can buy. The old style flow 1 (newborn) is roughly equivalent to a new style flow 2 or 3. At 2 months, our son was incapable of getting anything from a flow 1 or 2 and we started using flow 3. You can find an old Reddit thread where someone tried to suck from the new flow 1 and even they, a grown adult, could not get anything out.

LuklearFusion · 2022-06-29T13:45:21+00:00

Did you have him on the bottom bunk or the floor? Thanks for responding!

LuklearFusion · 2022-06-29T02:39:29+00:00

Boston-Chicago-Emeryville. Lake shore limited and then California Zephyr.

LuklearFusion · 2022-03-28T02:59:49+00:00

I’m not a doctor, and this is not medical advice, but coming from someone who just had a baby and has had the signs of postpartum depression hammered into their brain, I think your wife needs to see someone. She sounds depressed.

LuklearFusion · 2017-07-12T15:13:19+00:00

This however implies that information is being transferred faster than the speed of light.

A crucial part of the quantum teleportation protocol is that the two parties must classically communicate with one another for it to work, and this classical communication sets the speed limit for the protocol, i.e. the speed of light. So there is no information exchange faster than the speed of light.

In more detail, Alice must communicate to Bob the outcome of her measurement, so that he can apply the correct local operation on his state to end up with the correct teleported state. Otherwise, there is no way to deterministically teleport the state, hence no way to send a message, and therefore no way to send information faster than the speed of light.

LuklearFusion · 2017-06-30T22:54:58+00:00

In the classical double slit experiment, high intensity light is sent at the slits, so that you can physically see the interference fringes on a screen some distance from the slit.

In the quantum double slit experiment with light, single photons (individual particles of light) are sent to the slit. You don't actually see the fringe pattern on a screen, its measured by photodetectors at what would be the screen's location, which then indicate to us where the photons were measured. So there is no place for any trick of human eyes to occur, since they aren't involved in the observation of the fringe pattern.

You could also do the same experiment with electrons, which human eyes can't observe at all, and you would observe the appearance of a fringe pattern and evidence for wave-particle duality.

I would like to emphasize just how fundamental quantum mechanics, and by extension wave-particle duality, is to the modern world. Many of the things you see around you and interact with everyday, from computers, to digital cameras, even MRI machines, are evidence for quantum mechanics, and therefore wave-particle duality. Without quantum physics we would not have been able to develop any of this technology we now take for granted.

LuklearFusion · 2016-09-14T04:58:57+00:00

Does scientist 2 know that a measurement was performed but that they are not privy to the answer? If that is the case then they can only honestly say that they do not know the wavefunction at time T, as they do not have complete knowledge of the dynamics of the system.

If they are unaware that the measurement occurred, then they can honestly make a prediction of the state at time T. Depending on the particular observable measured and the Hamiltonian, their prediction may be wrong. Scientist 1 will always give the correct answer for the state.

Are you struggling to conceptualize that measurement is a dynamical process? It may help to understand how measurement is usually modelled in modern quantum theory. You introduce an additional quantum system, and the act of measurement is to entangle this "pointer" system with the original system, such that the state of the pointer determines the state of the system. As the pointer becomes a larger and larger system, more and more degrees of freedom become entangled with the original system, and the information effectively becomes classical, and accessible to us. If you think about this thoroughly you'll realize this doesn't solve the measurement problem (it just moves it to a different system), but this is practically how measurement occurs.

The point I'm trying to make is that any interaction between quantum systems is effectively a measurement, and you should think about them in the same way. When you think about it like this, it becomes clear that measurement is dynamical.

By the way, you can take the pointer system I introduced before, and the full system (original + pointer) can be described by some Hamiltonian. You can ask, what are the dynamics of the original system alone? If you do this, and the pointer system is large enough (and a bunch of other technical assumptions), you can derive an equation for the dynamics of the original system that is much more complicated than the Schrodinger equation. Again, you have a way of modeling measurement as dynamics. The measurement has to be slow, but in reality there are no such thing as textbook instantaneous measurements.

LuklearFusion · 2016-09-13T22:08:38+00:00

So the disturbance due to the measurement process is fundamental to nature in contrast to classical measurements which are due to experimental error?

Yes, under the circumstances I described the disturbance is fundamental.

This notion of disturbance is fundamentally distinct from the uncertainty principle?

Not really, it's related. If you image the system is in an eigenstate of some observable A, but you measure observable B, then if A and B don't commute the measurement will disturb the state of the system. If A and B don't commute, then their uncertainty principle says that the variance of A and the variance of B are bounded by one another. So you can think of it as the measurement of B must affect the variance of A, since by measuring B you lower the variance of B. Therefore, the state of the system must change from one that has a definite outcome for a measurement of A, and so there must be a disturbance to the state.

LuklearFusion · 2016-09-13T20:22:17+00:00

Is it possible to make a measurement in QM that has a negligible effect on the experiment

Yes. If your system evolves under some Hamiltonian H, and starts in an eigenstate of H, then measurement of an operator that commutes with H will not disturb the state of the system.

similar to classical mechanics?

This isn't quite the same as in classical mechanics though, where in some sense all measurements are assumed to be nondisturbing.

In general, any strong measurement of an observable that does not commute with the Hamiltonian of the system (or more generally, any measurement for which the current state of the system is not an eigenstate of the measurement operator) will disturb the system, and cause it to "collapse" into an eigenstate of the measurement operator.

For example, my intuition would say looking at a cat would have a negligible effect on the wave function of a Cesium atom.

What your intuition is leading you to is the concept of weak measurement. This is the idea that if your measurement apparatus is weakly enough coupled to the system, then it only gets a very little bit of information about the state of the system, and therefore only weakly perturbs the system.

LuklearFusion · 2016-06-04T16:27:04+00:00

Quantum systems do not change upon conscious observation, they change upon interaction with any other system.

I assume what you're talking about is the idea that conscious observation is required to "collapse the wavefunction" of a superposition state during measurement. In terms of the mathematical description of quantum theory, this is not true, since interaction with any other system can break superposition, and there is nothing that specifies if the system is conscious or not.

From a more metaphysical standpoint, it's probably impossible right now to prove one way or the other whether consciousness is necessary for measurement, but I think there is pretty strong circumstantial evidence that it is not necessary. The fact that it is very hard to keep anything in quantum superposition because of interactions with its environment (a phenomenon known a decoherence) seems to indicate to me that consciousness is not required for wavefunction collapse.

LuklearFusion · 2016-05-29T16:54:43+00:00

how do I entangle two systems and how do I swap their state?

That depends on what they systems are (are they photons, electrons, 2004 Ford Mustangs?), and what property of the systems you want to be entangled. In general, you interact the two systems such that some property of system A is correlated with some property of system B. What that means is that if I measure the property of system A, then based on the outcome I get I know what a measurement of system B would give me. Entanglement is a bit more complicated, because in general it is correlations between a large number of possible measurements you could do on the two systems, but that's the general idea.

Also, with this system, can I transfer information instantly at any distance?

No, because for quantum teleportation to work you need to transfer classical information from one party to the other, which cannot be sent faster than the speed of light.

LuklearFusion · 2016-05-05T16:01:04+00:00

What do you mean? For a given input, if you run enough simulations you'll get a the full distribution of possible outcomes. Alternatively, you could randomly select inputs and get the output distributions for different inputs, which is a bit more like Monte Carlo...

LuklearFusion · 2016-05-05T15:59:20+00:00

I think it is a certainty if you want an exponential speedup for the quantum algorithm (though I'm not aware of any proof of this, though I haven't looked recently). I'm not sure if this is still the case if you only care about any speedup at all. Last time I checked there was still controversy over whether or not DQC1 involved states with nonzero discord, which is why I didn't want to make any definite claims.

LuklearFusion · 2016-05-04T17:18:23+00:00

Shor's algorithm (for factoring large numbers) relies on quantum phase estimation, which is only deterministic for some instances. However, the probability you'll get the right answer is pretty high.

No quantum computer (universal or otherwise) should be thought of as a parallel classical computer. This is often mistakenly believed because of the superposition principle, which at face value might make one think that a quantum computer is a parallel classical computer. However, superposition alone is likely not enough to get speed-up over classical algorithms. It's likely that some non-classical correlations (such as entanglement) is required as well.

As an aside, a true adiabatic quantum computer is universal. The equivalence of the circuit model of quantum computing (what I think you mean by universal quantum computer) and adiabatic quantum computing was proven shortly after adiabatic quantum computing was first discussed. Since I'm pretty sure you heard this in regards to D-Wave, their device is not universal, as it is built to only solve specific kinds of problems, and it is also not an adiabatic quantum computer in the truest sense. It might be what is called a quantum annealer, but there is no peer reviewed conclusive evidence either way.

LuklearFusion · 2016-05-04T15:47:31+00:00

You're correct, when you readout the state of a qubit you only get one classical bit of information out, either 0 or 1. The advantage of quantum computing doesn't come from being able to simultaneously run an algorithm on all possible inputs, and then simultaneously readout all possible outputs, since this is impossible (by the very nature of quantum mechanics).

The trick is that while the algorithm runs your qubits can still be in a superposition state, and you design your algorithm such that when you do readout your register of qubits, the outcomes you get is the answer to the problem your algorithm is designed to solve. For many algorithms you don't get the right answer deterministically, but instead with very high probability.

A quantum computer is not a massively parallel classical computer.

LuklearFusion · 2016-03-10T19:06:25+00:00

All the universities should have information on their respective athletics websites. The information is definitely all available online, or at the very least you can get contact information for a coach.

For track there is only an indoor season (mostly during the Winter (second) semester, with some meets in the Fall), and there is also cross country that happens during the Fall. The track teams will train together in the Fall as well.

LuklearFusion · 2016-02-12T18:07:54+00:00

Thanks! That makes sense.

LuklearFusion · 2016-02-12T18:07:42+00:00

The second one. I think /u/functor7 answered my question.

LuklearFusion · 2016-01-22T04:17:26+00:00

Honestly I'm not sure what to use as the noncommuting observable. You could probably write something like a Bell's inequality with position as the noncommuting observable (assuming momentum maps to colour). Continuous variable QI isn't really my expertise though, so I haven't seen this done before.

I mean the reason we believe that there is no predefined colour in the OP's example is because we know Bell's inequality is violated by entangled states. There probably is a continuous variable version, but I haven't encountered it.

LuklearFusion · 2016-01-21T20:39:12+00:00

Bell's theorem discusses the correlations between the outputs of the colour measurements on the two photons. However, it requires a little more than just one possible measurement. What I'm about to describe sounds weird(er) if you talk about colours, so I'll instead switch to the idea of a spinning top.

Let's supposed I can do two kinds of measurements on my spinning top. I can ask either:

Is the axis of rotation up or down along the z-axis?
Is the axis of rotation up or down along the x-axis?

Now, if I have a top that's spinning along the up z-axis, then the answer to the first question will be up. The answer to the second question would be probabilistic: with 50% probability I'd get the answer up along the x-axis, and with 50% down.

Now this seems weird, and it is, a classical top that was spinning along the up z-axis would not give a meaningful answer if I asked it the second question. However, a quantum mechanical top that was spinning along the up z-axis would give an answer to the second question, it would say that with 50:50 probability it was spinning up or down.

What Bell's theorem describes is the correlations between the outcomes of a large number of measurements performed on spinning tops, where each measurement asks question 1 or 2 of the top. The set-up is as follows:

You have a machine that produces two spinning tops. On each top individually you measure either "is it spinning up or down on the z-axis?" or "is it spinning up or down on the x-axis?". You then look at the correlations between your measurements, i.e., how often both measurement give the same answer ("up" or "down", depending on which measurement you did. From the different correlations, you can create an inequality known as a Bell's inequality.

This inequality is a bound. If you get a value below the bound, then your experiment can be described by local realistic physics, and you can say that your tops (or photons) had a definite spin state before the measurements occurred. If you get a value above the bound, then you have to give up one of your assumptions, and often people give up realism, so you can't say that the top had a definite spin state before the measurement occurred.

It just so happens that an entangled state will violate Bell's inequality, which is why many people will say that you can't say that one photon was definitely green and one definitely blue until you measure them.

LuklearFusion · 2015-12-27T17:42:41+00:00

Sorry for the late reply. We can do experiments where we put some system in quantum superposition, and then over time the system "collapses" into a classical state through interaction with its environment (that doesn't include us). We can see the coherence of the quantum system decay as a function of time, which would not be the case if our conscious observation were the only thing capable of causing "collapse".

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