What happens after a quantum measurement?

[@pier_3885]

Thank you for the great video and the explanation. Real cases really help to put together all the principles we study. Well done!👍

[@ProfessorMdoesScience]

Glad you find this useful!

[@jhanzaibhumayun5782]

Just started our QM course. This is a godsend!

[@ProfessorMdoesScience]

Glad you find it helpful! :)

[@mahmoudjbely1464]

Please consider a QED playlist!

[@ProfessorMdoesScience]

Thanks for the suggestion! We are hoping to cover a range of more advanced topics after we are done with the basics of quantum mechanics.

[@meenusa8224]

Please do a video on POVM

[@ProfessorMdoesScience]

Thanks for the suggestion! Others have also asked for this, so it is on our list; however, it may take some time before we get there...

[@SergeyPopach]

So, applying the measurement operation triggers collapse of wave funciton and initiates a time evolution of the quantum state and after measurement it comes back to its original (stationary) state (when solving a eigenvalue for Hamiltonian)?

[@ProfessorMdoesScience]

Under the standard time evolution determined by the Schrödinger equation, if the system starts off in a state that isn't an energy eigenstate, then in general it will always stay as a superposition and will not become a stationary state. This will only happen if the measurement is for the energy. I hope this helps!

[@wadelamble2156]

Thanks for the videos. Incredible content! In the Schroedinger equation video, I thought you said that we solve in the energy basis and gave a general recipe for solving physical scenarios using the eigenvalue equation in the energy basis, but in this video and the infinite square well, you seem to use the position basis. Am I mistaken in equating these approaches? Are there examples of problems solved in the energy basis. Would Psi then not be a function of position? If so, how would I convert back to the position basis.

[@ProfessorMdoesScience]

Good question! I think the confusion arises because we may not have been absolutely careful in the nomenclature used, sorry! But to clarify: here we are using the position representation in the sense that we write quantum states in terms of wave functions. However, these quantum states can be any state, and specifically they can be energy eigenstates. This is precisely what we need to study time evolution: to expand our general state (written as a wave function) in terms of energy eigenstates (also written as wave functions). We do this because, as we discuss in the video on the Schrödinger equation, when we write states in terms of energy eigenstates, then the time evolution of conservative systems becomes very easy. I hope this helps!

[@wadelamble5462]

​@@ProfessorMdoesScience This is still elusive to me. Is there a way to set up an analogy to a finite-dimensional Euclidean situation? For example, if, in finding an equation for a spherically symmetric potential, we would use spherical rather than cartesian coordinates, in the quantum case above, what would the coordinate systems (bases) be that we can choose from? I guess one is "the basis whose axes are wave functions that are eigenstates of the energy operator." So our "choice" that is analogous to spherical vs cartesian coordinates is energy eigenstates vs, I don't know, some other observable's eigenstates. That makes sense to me in terms of defining different representations as different choices of wave functions. But is there a choice of bases that doesn't use wave functions at all? A couple of related questions. First, what is an actual alternative to expanding in the energy eigenstate basis that, while messy, would be physically equivalent? Second, what are the input parameters of the wave function? I see Psi(x,t). Is x here just the spatial coordinate(s)? Could this alternatively be Psi(p,t), where p is momentum? Third, and this is quite vague, sorry, is there some interesting relationship between the quantum state (the wave function) and the common notion of a state in "phase space"?

[@ProfessorMdoesScience]

@@wadelamble5462 Here are some thoughts: 1. Yes, the eigenstates of every observable define a valid basis, so the energy basis will be given by the eigenstates of the Hamiltonian, but you could also use other quantities to define a basis. As an example, the angular momentum operators are a very convenient way to define a basis when a particle moves in a central potential. 2. Yes, it is possible to write the wavefunction as psi(x,t), and equivalently as psi(p,t). And transforming between the position and momentum representations is described in this video: kzbin.info/www/bejne/aJ3VZJR3aduUeNU 3. The notion of phase space becomes central in the theory of statistical mechanics. In its quantum version, the relationship between these concepts becomes clear. We hope to do a series on statistical mechanics in the future, although for now we still have quite a few topics to cover on quantum mechanics... I hope this helps!

[@alexanderheller2039]

13:38 must be the coolest thing I've seen in quantum mechanics. It makes me think of that quote... "How do we know the moon is there when we're not looking at it?"

[@ProfessorMdoesScience]

I agree it is really cool :)

[@gnyszbr4187]

This will be a more practical question, but how do we know the amount of states our superposition can collapse into in a real example? I mean sure the energy we put in while constructing the system must at least be higher than the nth energy eigenstate's energy amount in order to have n states but then what happens when we put in a "nth state" amount of energy and superposition collapses into (n-1)th state? Isn't the energy lost here? Or is it because our way of putting energy in the first place also includes some uncertainty so there is a range including both higher results and lower results?

[@ProfessorMdoesScience]

This is a very interesting question! Energy conservation still works in quantum mechanics, but to reconcile it with the act of taking a measurement, you need to consider the *total* system, which includes your quantum particle *and* the measuring apparatus. I hope this helps!

[@gnyszbr4187]

@@ProfessorMdoesScience Got it. But what about the "how do we know the amount of states our superposition can collapse into in a real example" part? What is the parameter (if there is any) for that? Or is there always a chance to collapse even into most extreme state like 9999. energy eigenstate?

[@ProfessorMdoesScience]

@@gnyszbr4187 This will depend on the state you start with. For example, if your state is an eigenstate of the property you are trying to measure, the state can only "collapse" into itself, so only one option. The most general states will have contributions from every eigenstate spanning the state space, and you could collapse into any of those (with the corresponding probability given by the expansion coefficient squared of that state). I hope this helps!

[@saulberardo5826]

Fantastic!

[@ProfessorMdoesScience]

Glad you like it!

[@georgerevell5643]

How is this powerful video not more popular? People scared a little undergrad maths!?

[@ProfessorMdoesScience]

Thanks for your support! :)

[@abdelrhmanyousif7657]

Thanks it is very helpful to me!!!!

[@ProfessorMdoesScience]

Glad to be helpful! :)

[@khoanguyen5321]

If i take a stationary state and the wave function collpased at x_o, does it also evolves to wave function with constant probability as the end of the video? Or it evolves back to the same stationary state?

[@ProfessorMdoesScience]

The wave function will not evolve back to the same stationary state, in general it will go to a constant probability. When we measure the position of the particle, we radically change the state of the system, and right after the measurement there is no "memory" of what the state was beforehand. We explore measurements in more detail in these two videos: kzbin.info/www/bejne/q2K1ZJ6IjM1km80 kzbin.info/www/bejne/pZWvqIiOgL5jgNU I hope this helps!

[@khoanguyen5321]

@@ProfessorMdoesScience Thank you, I had watched the other two videos before this one and other videos on your youtube channel; it helps me a lot !!!. The thing that I am still concerned about is energy conservation. If the state is energy eigenstate with E = E_o and I measure position. Then, the wave function will become superposition of all energy eigenstate E_n with coefficient C_n. After that, I measure energy again and assume that we get energy eigenstate with E=E_2 which is different than E_o that we started with. How is the energy is conserved in this case? I know that the measurement involves the photon interaction. I think that somehow photon is the one that gives energy to the system, but I am not sure how explicitly. When wave function is in the superposition of energy eigenstate, does particle is "entangle" with the photon that is also in the superposition of energy eigenstate?

[@ProfessorMdoesScience]

@@khoanguyen5321 You are correct that to have energy conservation at the measurement step we need to consider the energy of the larger system+measuring device. Describing this whole process accurately can be incredibly complex, but roughly speaking, the higher the accuracy you need for your measurement the higher the energy of photon you need, and therefore the more energy that can be transfered to the system. I hope this helps!

[@senkum80]

You are the best!!!!

[@ProfessorMdoesScience]

Glad you like this! :)

[@workerpowernow]

great video

[@ProfessorMdoesScience]

Thanks for watching!

[@XinyanXiang]

Then, how do we prepare a quantum state before the measurement?

[@ProfessorMdoesScience]

This is an excellent question, which has a very difficult answer. There is no general procedure to prepare a given quantum state, and active research is going into this for specific cases. A strategy that is often used is to "filter" out states from a mixture of initial states, for example as done in the Stern-Gerlach experiment. I hope this helps!

[@XinyanXiang]

@@ProfessorMdoesScience Thank you so much!

[@vancuongpham699]

Thanks for your video, I am a starter, may I ask you some questions about this one, 1. if particle is at x0 and and the function of particle is Delta (x-x0), so what I understand here that the variance of x = 0, at least at the mathematical theory. It means if I take a measurement at that time, we can collect the absolute value without any error? Can you correct me? 2. and the second inquiry is about that you take a set of the eigen function of energy as a set of the eigen function of position, I confuse this a lot, because as far I know, they have their own eigen function (I understand eigen function and associated wave function are the same?) for every eigen value, maybe this a silly question, 3. The 3rd question is if x0 is eigen value, so what is eigen function (out of the above set of eigen function) for this eigen value. 4. And the 4th question, I saw in your video the E (j=1, k=2), but actually, I remember, it just take only 1 function after the measurement, is it right? I hope your clearing to help me more, Thanks again for your clip,

[@ProfessorMdoesScience]

Here are some thoughts: 1. If the particle is described by the wave function psi(x), then then probability of finding the particle in the range (x,x+dx) is given by |psi(x)|^2dx. Indeed, if the state of the system is psi(x)=delta(x-x0), then the particle will be found at x0. However, note that this state is physically impossible, we use it only as a mathematically convenient device. A true state will always have some spread around x0. 2. I am not sure I completely follow your point. Any observable (e.g. position or energy) will have its own set of eigenstates, which will, in general be different from each other. You can represent a general quantum state in terms of any set of eigenstates, and while the maths will be different depending on the set you choose, the physics will be the same. 3. The position eigenfunction associated with eigenvalue x0 is a delta function centered on that point, delta(x-x0). 4. After an energy measurement, the state of the system will be the corresponding energy eigenfunction. However, one can create quantum states that are superpositions of energy eigenstates, and in that example we look at the simplest possible such case, one in which the state is made of the superposition of only two energy eigenstates. Overall, I would recommend that you take a look at our playlist on the postulates of quantum mechanics, which builds all the above ideas (representations, measurements, etc) from the very basics: kzbin.info/aero/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb I hope this helps!

[@vancuongpham699]

@@ProfessorMdoesScience Thanks your feedback, it is meaningful a lot. For the 2nd question, I mean we have to expand the eigenstate of position operator in term of energy eigenstate. I don't know why? it is right for another operator? The others, I am clear, thanks,

[@vancuongpham699]

Oh, I get the answer for my confused point on your feedback already, thank you very much

[@vancuongpham699]

@@ProfessorMdoesScience hi man, I really don't understand why after measurement, the system will fall into their eigenfunction. My argument is as the following, please correct me. 1. For position measurement, as you said" if the state of the system is psi(x)=delta(x-x0), then the particle will be found at x0. However, note that this state is physically impossible". so measuring position of the systems as x0, I think it physically impossible too. why we not say measurement of position of the system as x0 with some variance, it is easier! 2. For energy measurement, I really don't know why the system after the measurement become this eigenstate, but not that eigenstate. And I know the measurement change the quantum state, but why they have to fall into the eigenstate? Please help me more, I really hit the wall

[@ProfessorMdoesScience]

For your first point, you are correct that whenever we have a continuous variable like position, we should more precisely say that our measurement outcome falls within some range of possible values. For your second point, the answer is that this is a postulate of quantum mechanics: when a measurement gives an eigenvalue as an outcome, the state immediately after the measurement is the associated eigenstate. So there is nothing to "understand", it is an assumption of quantum mechanics on which we build the rest of the theory. We have a few videos on measurements, which I recommend you check out for a more complete answer, starting with this one: kzbin.info/www/bejne/q2K1ZJ6IjM1km80 I hope this helps!

[@TheWingEmpire]

Awesome

[@ProfessorMdoesScience]

Thanks! :)

[@geraldpellegrini2782]

Summation of Heisenberg's time dependant matrix elements seems to solve this "mind blowing" problem. See my emails to you on July 31 9:30am, and August 14 8:45am.

[@geraldpellegrini2782]

Summation of Heisenberg's time dependent matrix elements solves this "mind blowing" problem. See emails to you on July 31 9:30 am and August 14 8:45 am. Why no reply??

[@geraldpellegrini2782]

If the particle in a box is isolated it is forced to be in an energy eigenstate. Contrary to Ehrenfest's theorem the classical correspondence is NOT achieved for this isolated system! According to the Ehrenfest equation of motion the expectation value of particle positions is stationary for this eigenstate, which is not the classical correspondence motion!!

[@amaljeevk3950]

❤

[@geraldpellegrini2782]

Why does no one see this a proof the standard use of a single state 'wave function' as a complete description of the particle motion is WRONG!

[@geraldpellegrini2782]

This video should demonstrate this "mind blowing" approach to the particle's motion is completely unrealistic and wrong! Especially in light of the fact that summation of Heisenberg's time dependant matrix elements for a given energy eigenstate gives the classical motion in the classical limit.

[@geraldpellegrini2782]

What happens after you measure the energy?? All probability distributions are time independent and all motion stops, correct? That's even more "mind blowing"

[@geraldpellegrini2782]

Why do you believe this rediculous conclusion? There comes a time when you have to make sense!!

[@geraldpellegrini2782]

Does everyone actually believe this rediculous conclusion?

[@schmetterling4477]

There is no such thing as an "after a measurement". A measurement is irreversible, i.e. it stays around forever.