00:02 Comparing pure and mixed quantum states 01:40 Quantum measurements lead to state collapses and probabilities of eigenvalues. 03:22 Quantum states can be represented as statistical mixtures of states 05:06 Quantum mechanics and classical mechanics give the same measurement results. 06:40 Measurement of pure vs. mixed quantum states 08:29 Superposition states require interference term for accurate predictions. 10:13 Quantum systems have pure and mixed states. 11:47 Mixed quantum states contain partial information and use probabilities to describe the system.
Something that changed the way I looked at superposition was noting the fact that a superposition is as much a characteristic of the state as of the basis (and/or observable) - you can have a superposition in a given basis but not in another, for the same state: the state "itself" just kind of /is/.
Great point! This insight has tremendous consequences in quantum mechanics, one of the most famous being the uncertainty principle. You can find our take on this here: th-cam.com/video/pfjQtyLBBHw/w-d-xo.html
I'm guessing this is really important in dealing with error correction in quantum computation? Because it looks like the error correction for some 'computational target', itself needs error correction - iterate until you're as sure of the computational outcome, as you need? Please indicate if I'm talking sense - or not!
This video, together with the videos on the density operator, are indeed central to all aspects of quantum information. We may get to a series on these topics in the future, which will allow us to explore these questions in more detail!
At 12:13, you say that for mixed states, we don't know the actual state but rather a mixture of states with respective probabilities. My question is how do we get this probability distribution? There are infinite states. When we don't know about the system, how can we tell that it may exist in |psi_1> to |psi_n> with some associated probabilities?
Your question is a very important one and is essentially the topic of "statistical mechanics". As an example, statistical mechanics tells us that, if the system is at thermal equilibrium, then the corresponding distribution is described by the canonical ensemble. We do hope to start a series on statistical mechanics in the future, but we first want to finish with quantum mechanics.
At 7:40 you have the prefactor of 2Re. You say that the sum of a number plus its complex conjugate results in twice its real part. This makes sense to me. Does this mean that we only take the real number part of the term inside the brackets { ... }, and this is signified by the Re prefactor? Why is the complex conjugate * symbol removed from the c_1 and terms, but not the c_2 and term? Will that symbol be removed once we apply the Re prefactor, and the * is left for clarity in the proof? Or am I missing something about this?
Yes, "Re" means that we take the real number part of the term inside brackets. In more detail, let's consider the starting multiplication: (c*_1 * + c*_2 *) (c_1 + c_2 ) If we multiply this through, we get four terms (this is the step I skipped in the video): c*_1c_1* + c*_2c_2 * + c*_1c_2* + c*_2c_1* The first two terms give: |c_1|^2||^2 + |c_2|^2||^2 And the last two terms are the complex conjugate of each other, let's call the first one z* and the second one z. We then use z*+z = 2Re(z), so we end up with the term you refer to: 2Re(c*_2c_1*) where the conjugate symbol "*" is with the c_2 and the . We could, of course, have chosen z*+z = 2Re(z*), and then the conjugate symbol would be with the c_1 and the . Both should give the same answer. Once you actually evaluate the real part of the term in brackets, then you'll end up with a real number. I hope this helps!
@@ProfessorMdoesScience thank you so much this really helps! Also, thank you in general for making these videos and taking the time to respond to questions I and others post. It is very helpful and It is very appreciated. I'm jealous of your students for having such a wonderful teacher haha.
Thanks for the suggestion! We hope to get there eventually, but in the meantime, we very briefly discussed entangled states in the video on tensor products: th-cam.com/video/kz3206S2B6Q/w-d-xo.html
I think it’s easier to define a pure state and a mixed state in the following way. A pure state is just a state where we can define a fixed phase between the states. For example, in the superposition pure state there is interference term that carries a phase for both u1 and u2. The same can’t be said about mixed states. The mixed state doesn’t have a fixed phase. Mixed states are just classical statistical ensembles. I hope this helps someone. 😊
The copies of states you talked about in statistical mixture of states doesn't contradict the no-cloning theorem that states that we cannot have two copies of the same state?
In our statistical mixture we are creating copies of known states (say |u_1> and |u_2>), what we don't know is which one we'll pick when we perform a measurement. I think this is somewhat different to hte no-cloning theorem about making copies of unknown states. I hope this helps!
Both of your lectures are excellent! I would like to ask a question. What is the difference between state vector norm and operator norm in Quantum Mechanics?
State vector norms are the "lengths" of the vectors in the (complex) vector space in which we describe quantum states. We typically work with normalized state vectors, as then the interpretation of quantities such as probability of measurement outcome become very simple. The operator norm is used less frequenty in quantum mechanics. In general, the norm of a Hermitian operator is equal to the magnitude of the largest (positive or negative) eigenvalue in the operator spectrum. However, in quantum mechanics many important operators have an unbound eigenvalue spectrum, so the operator norm doesn't provide much insight. I hope this helps!
Very helpful, thaank you! Could you please give me an example of the Stern-Gerlach experiment to get an intuition? When do we have a mixed state and when do we have a pure state? What is the difference between state selection and preparation?
Thank you so much! I've been struggling with this when reading about density functional theory. Would it be possible to do a video series on the Hartree-Fock method and Kohn-Sham equations?
Thanks for the suggestion! We are hoping to cover topics such as density functional theory in the future, but we first want to finish the basics of quantum mechanics, so it may still take some time for us to get there...
for interference pattern ,we prepare pure state by double slit so interference happen and if we try to measure after the slit before the screen then we convert it to mixed state so interference disppear is this right ?
I am not sure I completely follow your question. But in general, the double slit experiment illustrates interference, which is a phenomenon that occurs even in pure states!
double slit experiment setup that I mean is as follow 1-source then double slit then screen ( interference happen ) 2-source then double slit then detector then screen (no interference) in the first setup we start with mixed state (as there no complete information) then convert to pure state(complete information) by the two slit so interference pattern appear in the second setup detector(incomplete information) convert pure state again to mixed state so no interference happen.
If your source is generating a "classical" statistical mixture of states, then it should be described as a mixed state. I think you are then trying to then use the double slit to somehow select a pure state out of this initial mixed state. The standard double slit setup will not do this, but you could select a pure state using something like a polarizer. I hope this helps!
Thanks for the suggestion, and we do hope to include a video on entanglement. In the meantime, we (very briefly) mention entanglement in this video: th-cam.com/video/kz3206S2B6Q/w-d-xo.html I hope this helps!
@@ProfessorMdoesScience Because I see there is superposition in a pure state. This lead me to relate different orbital superposition. Now I think their difference is the hybridization arises from the superposition of wave functions.
00:02 Comparing pure and mixed quantum states
01:40 Quantum measurements lead to state collapses and probabilities of eigenvalues.
03:22 Quantum states can be represented as statistical mixtures of states
05:06 Quantum mechanics and classical mechanics give the same measurement results.
06:40 Measurement of pure vs. mixed quantum states
08:29 Superposition states require interference term for accurate predictions.
10:13 Quantum systems have pure and mixed states.
11:47 Mixed quantum states contain partial information and use probabilities to describe the system.
Something that changed the way I looked at superposition was noting the fact that a superposition is as much a characteristic of the state as of the basis (and/or observable) - you can have a superposition in a given basis but not in another, for the same state: the state "itself" just kind of /is/.
Great point! This insight has tremendous consequences in quantum mechanics, one of the most famous being the uncertainty principle. You can find our take on this here:
th-cam.com/video/pfjQtyLBBHw/w-d-xo.html
Came here after the recommendation by #Parth G
Good Explanation
Loved it👌👌👌🙏♥️
Thanks for watching, and glad you liked it! :)
me too!
@@evilotis01 hope you liked it!
@@ProfessorMdoesScience very much!
Nice explanation! Density operator for mixed states coming soon, I hope?
Thanks. Just published actually :) Here is the link:
th-cam.com/video/bg0SMY40q8Q/w-d-xo.html
Found this video very useful, These are the type of videos that will help me get through my masters. Thank you!!
Glad to hear we'll be useful! What Masters are you doing, and where, if we may ask?
@@ProfessorMdoesScience I am Specializing in Quantum computing and Quantum information from SPPU, India.
Great explanation - way better than a textbook!
Glad you like it! :)
This is the best quantum mechanics! Thanks! I love them!
Glad you like them! :)
I'm guessing this is really important in dealing with error correction in quantum computation? Because it looks like the error correction for some 'computational target', itself needs error correction - iterate until you're as sure of the computational outcome, as you need? Please indicate if I'm talking sense - or not!
This video, together with the videos on the density operator, are indeed central to all aspects of quantum information. We may get to a series on these topics in the future, which will allow us to explore these questions in more detail!
Good topic to cover, thanks!
Glad you like it!
At 12:13, you say that for mixed states, we don't know the actual state but rather a mixture of states with respective probabilities. My question is how do we get this probability distribution? There are infinite states. When we don't know about the system, how can we tell that it may exist in |psi_1> to |psi_n> with some associated probabilities?
Your question is a very important one and is essentially the topic of "statistical mechanics". As an example, statistical mechanics tells us that, if the system is at thermal equilibrium, then the corresponding distribution is described by the canonical ensemble. We do hope to start a series on statistical mechanics in the future, but we first want to finish with quantum mechanics.
At 7:40 you have the prefactor of 2Re. You say that the sum of a number plus its complex conjugate results in twice its real part. This makes sense to me. Does this mean that we only take the real number part of the term inside the brackets { ... }, and this is signified by the Re prefactor?
Why is the complex conjugate * symbol removed from the c_1 and terms, but not the c_2 and term? Will that symbol be removed once we apply the Re prefactor, and the * is left for clarity in the proof?
Or am I missing something about this?
Yes, "Re" means that we take the real number part of the term inside brackets. In more detail, let's consider the starting multiplication:
(c*_1 * + c*_2 *) (c_1 + c_2 )
If we multiply this through, we get four terms (this is the step I skipped in the video):
c*_1c_1* + c*_2c_2 * + c*_1c_2* + c*_2c_1*
The first two terms give:
|c_1|^2||^2 + |c_2|^2||^2
And the last two terms are the complex conjugate of each other, let's call the first one z* and the second one z. We then use z*+z = 2Re(z), so we end up with the term you refer to:
2Re(c*_2c_1*)
where the conjugate symbol "*" is with the c_2 and the . We could, of course, have chosen z*+z = 2Re(z*), and then the conjugate symbol would be with the c_1 and the . Both should give the same answer.
Once you actually evaluate the real part of the term in brackets, then you'll end up with a real number. I hope this helps!
@@ProfessorMdoesScience thank you so much this really helps!
Also, thank you in general for making these videos and taking the time to respond to questions I and others post. It is very helpful and It is very appreciated. I'm jealous of your students for having such a wonderful teacher haha.
I am so glad that I found your channel. Nice video and extremely useful content :DDD
Glad you like them! :)
Good one! Can you do a video about bell states and entanglement?
Thanks for the suggestion! We hope to get there eventually, but in the meantime, we very briefly discussed entangled states in the video on tensor products:
th-cam.com/video/kz3206S2B6Q/w-d-xo.html
Thanku for the clear and crisp explanation.. looking for more of this ;)
Glad you like it! :)
Best explaination ever! I finally understand this topic 😍
Glad to hear this! :)
Amazing ! Loved the explanation
Glad you liked it! :)
I think it’s easier to define a pure state and a mixed state in the following way. A pure state is just a state where we can define a fixed phase between the states. For example, in the superposition pure state there is interference term that carries a phase for both u1 and u2. The same can’t be said about mixed states. The mixed state doesn’t have a fixed phase. Mixed states are just classical statistical ensembles. I hope this helps someone. 😊
Thanks for your insights!
Thanks for this awesome video,the best explanation that i could possibly get,thank you so much!!!!
Glad it helped!!
Great work!
Thanks for watching!!
Thank you for this video, your explanation is very clear.
Glad you like it! :)
Oh my god parth g suggested this and this is amazing 🤩
Glad you found us! :)
Thank you so very much for such lucid explanation.
Glad you like it! :)
The copies of states you talked about in statistical mixture of states doesn't contradict the no-cloning theorem that states that we cannot have two copies of the same state?
In our statistical mixture we are creating copies of known states (say |u_1> and |u_2>), what we don't know is which one we'll pick when we perform a measurement. I think this is somewhat different to hte no-cloning theorem about making copies of unknown states. I hope this helps!
very nice explanation. Thanks
Glad you like it! :)
Both of your lectures are excellent! I would like to ask a question. What is the difference between state vector norm and operator norm in Quantum Mechanics?
State vector norms are the "lengths" of the vectors in the (complex) vector space in which we describe quantum states. We typically work with normalized state vectors, as then the interpretation of quantities such as probability of measurement outcome become very simple. The operator norm is used less frequenty in quantum mechanics. In general, the norm of a Hermitian operator is equal to the magnitude of the largest (positive or negative) eigenvalue in the operator spectrum. However, in quantum mechanics many important operators have an unbound eigenvalue spectrum, so the operator norm doesn't provide much insight. I hope this helps!
@@ProfessorMdoesScience Very good! Thank you very much, and once again, your lessons are great!
Very helpful, thaank you! Could you please give me an example of the Stern-Gerlach experiment to get an intuition? When do we have a mixed state and when do we have a pure state? What is the difference between state selection and preparation?
Thank you so much! I've been struggling with this when reading about density functional theory. Would it be possible to do a video series on the Hartree-Fock method and Kohn-Sham equations?
Thanks for the suggestion! We are hoping to cover topics such as density functional theory in the future, but we first want to finish the basics of quantum mechanics, so it may still take some time for us to get there...
Sir please make
A video regarding polarization of photon and helicity concept.
Thanks for your wonderful videos🙏
Thanks for the suggestion, and glad you like the videos!
Hey, thanks a lot for the nice video. Just wondering, if a pure state could also be an entangled state?
Yes! We briefly discuss this in this video: th-cam.com/video/kz3206S2B6Q/w-d-xo.html
I hope this helps!
for interference pattern ,we prepare pure state by double slit so interference happen
and if we try to measure after the slit before the screen then we convert it to mixed state so interference disppear
is this right ?
I am not sure I completely follow your question. But in general, the double slit experiment illustrates interference, which is a phenomenon that occurs even in pure states!
double slit experiment setup that I mean is as follow
1-source then double slit then screen ( interference happen )
2-source then double slit then detector then screen (no interference)
in the first setup we start with mixed state (as there no complete information) then convert to pure state(complete information) by the two slit so interference pattern appear
in the second setup detector(incomplete information) convert pure state again to mixed state so no interference happen.
If your source is generating a "classical" statistical mixture of states, then it should be described as a mixed state. I think you are then trying to then use the double slit to somehow select a pure state out of this initial mixed state. The standard double slit setup will not do this, but you could select a pure state using something like a polarizer. I hope this helps!
Sir can u make vedio about entanglement please
Thanks for the suggestion, and we do hope to include a video on entanglement. In the meantime, we (very briefly) mention entanglement in this video: th-cam.com/video/kz3206S2B6Q/w-d-xo.html
I hope this helps!
Ty!!! Great Video!:)
Glad you like it!
Are pure states hybridization?
Hybridization is a different concept from that of pure states; how do you think they are related?
@@ProfessorMdoesScience Because I
see there is superposition in a pure state. This lead me to relate different orbital superposition. Now I think their difference is the hybridization arises from the superposition of wave functions.
Awesome
Glad you like it!
great
Glad you like it!
Vicey annualized arable topic cover nice o1
Thanks for watching!
Criticas respecto de las desigualdades tipo Bell: th-cam.com/video/ZVbj93nqL4Y/w-d-xo.html
Gracias por el link!