A few people have asked in the comments how we get the matrix representation at 34:41. I gave this as a reply to one such comment, but I'll re-post it in a separate comment for greater visibility: Think about the deterministic operation where we set Y equal to X. The action on standard basis vectors looks like this: 00 -> 00, 01 -> 00, 10 -> 11, 11 -> 11. In words, the second bit gets replaced by the first. The matrix representation of this deterministic operation is the first one on the right-hand side of the equation (not including the factor of 1/2). In a similar way, setting X equal to Y gives the second matrix on the right-hand side (again, not including the factor of 1/2). If we perform one of the two, each with probability 1/2, we average the matrices to get the matrix on the left-hand side.
I guess you also considered the previous slide and thought "columns". But actually it is not the case. The symmetricity of the previous example misled us. If we consider the definition at the 23. page of the first lecture slides and explicitly write state trasitions and superpose them we actually obtain that very matrices of page 30 of the lecture 2 👍@@Gnefitisis
Although I am already familiar with Q information Q - ML and Python, your explanation is so good, that certain "gaps" I had, your lecture, filled them happily. I am now "inspired" ; for lesson 2. There is always something to learn. Knowledge has many surfaces where the sunlight shines and glows differently. Thanks. E
This is literally as good as many uni lectures if not even better than when I was still a student - amazing job on this one aswell; can't wait to work with it and refer this video series to other people trying to get into the field. Thank you!
At 31:55 is probably a small mistake on the slides in the sence: "1. The probability that a measurement of Y yields an outcome a \in \Sigma is...". I believe there should be "1. The probability that a measurement of Y yields an outcome b \in \Gamma". Otherwise, totally amazing lesson :)
Dear Professor, excellent presentation. May I use your presentation for my classes? I would like to prepare slides based on the material covered in your lectures and possibly supplement the lecture with my own examples. Best regards,
I’m a physician by profession, with a passion for physics. I was struggling for the last three months to understand when a composite quantum state is said to be entangled. Your discussion on tensor products, and when probabilities in a compound system can or can’t be written as products of the probabilities of single systems, came as a revelation. I was wondering if composite quantum systems have analogs of basis vectors in single systems, and then you started talking about Bell states. So now I know. Thank you! You’re a wonderful teacher!
If you love physics (or mathematics) be warned that there's a serious flaw in this presentation. The motivation behind it seems to be to make quantum theory look like classical probability theory and conseqently be more easily absorbed. That's a laudable aim because in fact quantum theory /is/ just [an algebraic generalisation of classical, Kolmogorovian] probability[1] and it should be possible to make it easy[2], but this isn't the way to do it. Unfortunately, its focus on superficial, calculational, similarities comes at the expense of conceptual understanding and an inapt distinction between quantum theory and (classical) probability theory is made[3]. [2] I don't know if there is a good introduction at the same (elementary) level that isn't problematic in the ways that this course is but there is a wonderful book by Ingemar Bengtsson and Karol Życzkowski, Geometry of Quantum States (An Introduction to Quantum Entanglement). [1] See e.g. arXiv:quant-ph/0601158 [3] Notably facilitated by a horrible abuse of Dirac notation whereby the convex set of (classical) probabilty states is represented by "bra"s.
@@Verlamian I accept your criticism and am sorry to hear you don't like my approach, but I stand by it. I'm sure there are other resources for those who prefer a different approach, and welcome anyone to offer suggestions in these comments.
@@John.Watrous Well I hope you'll reconsider that stance because your approach is excellent apart from that one serious but very easily fixed flaw. Classical (commutative) and quantum (noncommutative) probability do fit into the same algebraic formalism after all (arXiv:quant-ph/0601158)* and a - flawless! - course covering the finite dimensional part of it, as yours so very nearly is, would be a major contribution. * NB. And all that forbidding operator algebra theory isn't needed for the basic ideas and in the finite dimensional context - Tom Banks once wrote an article for a guest post on Sean Carroll's blog demonstrating this. I can provide a link to it if you're interested.
We'll see quantum and probabilistic states fitting very nicely into a single algebraic framework when we discuss the general formulation of quantum information in Unit 3. For now it's about keeping things simple and focusing on the basics.
On Bell states derivation. I hope what I'm writing here is meaninful to you, at last. I noticed that the two Bell states |phi_plus> and |phi_minus> appear to be very similar to the result of Hadamard Operator H acting on the two base states |0> and |1>, but substituting them with |00> and |11>. A similar argument can be expressed on |psi_plus> and |psi_minus>. So, one can think a bigger "Hadamard operator", say B, to be applied to the four base states |00>, |01>, |10> and |11> and use it to get the four Bell states from them. Starting from the fact that: H = (X + Z) / sqrt(2)
I exploited some block-matrix B adding these two tensor products and normalizing the same way: B := ( X § X + Z § 1 ) / sqrt(2)
here "§" stands for "tensor-product-operator" and "1" stands for identity matrix. The definition of B, amazingly for me, looks similar to the above equation of H ~ X+Z. Expanding the tensor product using block-matrices as intermediate results, so that B looks like { 0 X } { 1 0 } { 1 X } { X 0 } { 0 -1 } { X -1 } ________ + ________ = ________ sqrt(2) sqrt(2) sqrt(2) and expanding completely: { 0 0 0 1 } { 1 0 0 0 } { 1 0 0 1 } { 0 0 1 0 } { 0 1 0 0 } { 0 1 1 0 } { 0 1 0 0 } { 0 0 -1 0 } { 0 1 -1 0 } { 1 0 0 0 } { 0 0 0 -1 } { 1 0 0 -1 } ______________ + _____________ = ______________ sqrt(2) sqrt(2) sqrt(2) The columns of matrix B are the coefficients of the four Bell states, so that: B |00> = |phi_plus> B |11> = |phi_minus> B |01> = |psi_plus> B |10> = |psi_minus>
1:04:50 your theoretical quantum calculations bother me because they assume physics is not local, so you assume relativity is false. I'm not sure we can design a realistic technological device with such assumptions. We could wonder if these calcultations have enough solid experimental grounding. Relativity could be false, Physic could be non local... no causality anymore🤔
Thanks for the great presentation. I am not clear at 1:03:26 when you say that for controlled-Not with Y as control, you swap the order of tensor products. I would appreciate help (may be, it has been taken up earlier and I am not able to locate it). Thanks
Hello sir, thank you for your efforts, it is a great video. I came across N00N status on the internet. I learn it makes precise phase measurement when optical interferometer is used in quantum optics. Can such a situation be used in quantum algorithms? If so, where is it usually used?
Awesome lecture, the whole series. Around 53:37, small misspelling in Measurement/example, the coefficient in front of |10> has to be with the sign "-" and the coefficient in front of |11> has to be with the sign "+", or similar to restore the normalization condition to the unity of the mixed state.
I don't see an error here. We can change the coefficients as you suggest to get a different example, but it's a unit vector - and hence a valid quantum state vector - either way.
Can you give some hints on how to get matrix in example on 34:56? in general, how to translate a statement, such as in 34:56 example, to a matrix representation of an operation? thanks
Hi! I'm following along and find this series to be very good, however I'm struggling with the example matrix U at 57:20. To test my knowledge I'm trying to decompose it into a matrix A tensor B acting on X and Y respectively. If I'm correct A should be a 3x3 matrix and B should be a 2x2 matrix. So far, so good. However, I cannot for the life of my figure out how (for example) u_41 can be 0. If my understanding is correct u_41 = a_21 * b_21. This implies that either a_21 or b_21 is zero. If we assume a_21 is 0, then that should also imply that u_31 is 0 as we can rewrite it as a_21 * b_11, but it isn't. If we assume b_21 is 0, then that should also imply that u_21 is 0 as we can rewrite is as a_11 * b_21, but it isn't. The same sort of logic can be applied to all 0-valued entries in U as far as I can tell. So I seem to arrive at the contradiction that A and B must have all non-zero entries that somehow multiply to 0 once they are tensored. So either there's an error or (much more likely) a gap in my understanding. Any clarification would be greatly appreciated!
My working hypothesis is that this must in fact be an entanglement-generating gate thus trying to decompose it won't work, as it cannot be applied independently to XY. Any rebuttal or hints as to how I can prove it would be nice! :)
I think you have already figured it out, but if any of the comment readers get puzzled (from my understanding): Key point: If X and Y undergoes A and B operations, total operation can be written by tensoring them. But it's reverse is not always true. The matrix was just a 6x6 unitary matrix to just illustrate an example of unitary matrices which means operations on a system. This is a valid operation for that system. But it is not necessary for the matrix, to be an operation for any two bit systems, to come from tensor producting individual operations on X and Y. Any unitary matrix can represent an operation. We will be using tensor for the reverse case. If we know that A and B are operations on X and Y, then to find the total operation, tensor-product is the solution. But decomposing to tensor isn't necessary for it to be a valid operation. And yes (I think from my understanding), your @BeepBoop's calculation is the procedure to prove your hypothesis.
I think the matrix is actually not unitary. I think the first three entries of the second row from the bottom should be instead swapped with the zeros in the last row? Just my feeling from checking a few entries... But maybe I got it mixed up - its been a while since I last did matrix multiplication ;-)
Hi there! I have a question at 52:50 . You said earlier that tensor products in quantum state represents independence of individual states. But when we measure a state X from compound system XY, the quantum state of Y changes. Doesn't it mean that the independence has collapsed along with the measurement of individual state X?
As far as I can tell independence in that case is used to refer to unentangled 2-qbit systems, meaning the individual qbits carry information about the overall system when viewed independently. That means that measuring the X qbit independently of Y will give you some amount of information about the overall system, then measuring Y will give you the rest of the information. In contrast: a (maximally) entangled system means that measuring parts of the system will cause decoherence of the system. Thus measuring X, then measuring Y will not give you information about the overall state of XY as it existed in the entangled state. So a 2-qbit system that is completely de-composable into a product state of 2 single qbit systems is independent in an information-carrying sense, but not necessarily in the probabilistic sense. On the other hand a 2-qbit system that cannot be written as a product state must be maximally entangled meaning that measuring X and Y independently in fact provides 0 information of XY, as information about the system can only be gleaned when considering a joint measurement of XY
At 35:00 the matrix representation on an operation on bits (X,Y) was provided. I am unable to understand how those 2 matrices were calculated. The textbook lesson also did not elaborate on how to calculate these matrices. Any help would be appreciated. Thank you.
Think about the deterministic operation where we set Y equal to X. The action on standard basis vectors looks like this: 00 --> 00, 01 --> 00, 10 --> 11, 11 --> 11. In words, the second bit gets replaced by the first. The matrix representation of this deterministic operation is the first one on the right-hand side of the equation (not including the factor of 1/2). In a similar way, setting X equal to Y gives the second matrix on the right-hand side (again, not including the factor of 1/2). If we perform one of the two, each with probability 1/2, we average the matrices to get the matrix on the left-hand side.
These sort of videos piss me off, because they will explain obvious things to an arrogant level... but will skip to explain the concept of a permutation matrix. In short, realize that the rows of the column vector for these qubits is represented so: [ 00 ] = 0 [ 01 ] = 1 [ 10 ] = 2 [ 11 ] = 3 which, is just a top-down writing of 2-bits with all 2^n = 4 states. Now, because this is a CNOT, it's say that the most significant bit (X,Y) (most sig, least sig) is the control bit. Meaning, if it (X) is 1 (exists) then NOT the next bit (Y). So you can describe a permutation function like so: [ 00 01 10 11 ]
Thank you for your interest in this series. We would love to be able to tell you exactly when each video in this series will be released, but the reality is that we don't know yet. We're a small team, and creating these videos takes time... we're releasing them as they are created. What I can tell you is that Lesson 3 will be released in January and Lesson 4 will be released in February, which will complete the first unit of the series. You can expect lessons from the second unit to start appearing sometime in the spring and to continue through the year. I'm afraid that the completion of the entire series is still a long ways off, and I can't offer a meaningful estimate on how long it will take at this time. We will provide future updates on the timeline as we are able.
A few people have asked in the comments how we get the matrix representation at 34:41. I gave this as a reply to one such comment, but I'll re-post it in a separate comment for greater visibility:
Think about the deterministic operation where we set Y equal to X. The action on standard basis vectors looks like this: 00 -> 00, 01 -> 00, 10 -> 11, 11 -> 11. In words, the second bit gets replaced by the first. The matrix representation of this deterministic operation is the first one on the right-hand side of the equation (not including the factor of 1/2). In a similar way, setting X equal to Y gives the second matrix on the right-hand side (again, not including the factor of 1/2). If we perform one of the two, each with probability 1/2, we average the matrices to get the matrix on the left-hand side.
Why are the resulting matrices the transpose of what I would think they are?
For example: for the first one on the right side: 00 00, Input 01 -> 11 .
I just started studying these concepts and I wonder this as well. Could you figure it out ?@@Gnefitisis
I guess you also considered the previous slide and thought "columns". But actually it is not the case. The symmetricity of the previous example misled us. If we consider the definition at the 23. page of the first lecture slides and explicitly write state trasitions and superpose them we actually obtain that very matrices of page 30 of the lecture 2 👍@@Gnefitisis
@@SukruOzan Thanks! I also just stared studying this too. It's mostly just notation and terminology.
Although I am already familiar with Q information Q - ML and Python, your explanation is so good, that certain "gaps" I had, your lecture, filled them happily. I am now "inspired" ; for lesson 2. There is always something to learn. Knowledge has many surfaces where the sunlight shines and glows differently. Thanks. E
Wow, I had no idea Matt Leblanc was into quantum computing now.
This is literally as good as many uni lectures if not even better than when I was still a student - amazing job on this one aswell; can't wait to work with it and refer this video series to other people trying to get into the field. Thank you!
This is great, thank you for these videos!
At 31:55 is probably a small mistake on the slides in the sence: "1. The probability that a measurement of Y yields an outcome a \in \Sigma is...".
I believe there should be "1. The probability that a measurement of Y yields an outcome b \in \Gamma".
Otherwise, totally amazing lesson :)
Yes, you are correct. Good catch!
Thanks for the video series. Looking forward to gates & circuits coming up next!
Dear Professor, excellent presentation. May I use your presentation for my classes? I would like to prepare slides based on the material covered in your lectures and possibly supplement the lecture with my own examples.
Best regards,
Yes, you are most welcome to use the videos in your classes.
the simplicity and correctness and relevance of the examples is striking
good to know what those symbols are called now after reading through a bout 8 books on this. haha
At 1:01:54 why does swapping phi- not change the state to -(phi-); wouldn’t the action from psi- apply to that one too?
I’m a physician by profession, with a passion for physics. I was struggling for the last three months to understand when a composite quantum state is said to be entangled. Your discussion on tensor products, and when probabilities in a compound system can or can’t be written as products of the probabilities of single systems, came as a revelation.
I was wondering if composite quantum systems have analogs of basis vectors in single systems, and then you started talking about Bell states. So now I know.
Thank you! You’re a wonderful teacher!
If you love physics (or mathematics) be warned that there's a serious flaw in this presentation. The motivation behind it seems to be to make quantum theory look like classical probability theory and conseqently be more easily absorbed. That's a laudable aim because in fact quantum theory /is/ just [an algebraic generalisation of classical, Kolmogorovian] probability[1] and it should be possible to make it easy[2], but this isn't the way to do it. Unfortunately, its focus on superficial, calculational, similarities comes at the expense of conceptual understanding and an inapt distinction between quantum theory and (classical) probability theory is made[3].
[2] I don't know if there is a good introduction at the same (elementary) level that isn't problematic in the ways that this course is but there is a wonderful book by Ingemar Bengtsson and Karol Życzkowski, Geometry of Quantum States (An Introduction to Quantum Entanglement).
[1] See e.g. arXiv:quant-ph/0601158
[3] Notably facilitated by a horrible abuse of Dirac notation whereby the convex set of (classical) probabilty states is represented by "bra"s.
@@Verlamian I accept your criticism and am sorry to hear you don't like my approach, but I stand by it. I'm sure there are other resources for those who prefer a different approach, and welcome anyone to offer suggestions in these comments.
@@John.Watrous Well I hope you'll reconsider that stance because your approach is excellent apart from that one serious but very easily fixed flaw. Classical (commutative) and quantum (noncommutative) probability do fit into the same algebraic formalism after all (arXiv:quant-ph/0601158)* and a - flawless! - course covering the finite dimensional part of it, as yours so very nearly is, would be a major contribution.
* NB. And all that forbidding operator algebra theory isn't needed for the basic ideas and in the finite dimensional context - Tom Banks once wrote an article for a guest post on Sean Carroll's blog demonstrating this. I can provide a link to it if you're interested.
We'll see quantum and probabilistic states fitting very nicely into a single algebraic framework when we discuss the general formulation of quantum information in Unit 3. For now it's about keeping things simple and focusing on the basics.
I didnt get the reference of " the whole is greater than the sum of the parts" said at 11:51
On Bell states derivation.
I hope what I'm writing here is meaninful to you, at last.
I noticed that the two Bell states |phi_plus> and |phi_minus> appear to be very similar to the result of Hadamard Operator H acting on the two base states |0> and |1>, but substituting them with |00> and |11>.
A similar argument can be expressed on |psi_plus> and |psi_minus>.
So, one can think a bigger "Hadamard operator", say B, to be applied to the four base states |00>, |01>, |10> and |11> and use it to get the four Bell states from them.
Starting from the fact that:
H = (X + Z) / sqrt(2)
I exploited some block-matrix B adding these two tensor products and normalizing the same way:
B := ( X § X + Z § 1 ) / sqrt(2)
here "§" stands for "tensor-product-operator" and "1" stands for identity matrix.
The definition of B, amazingly for me, looks similar to the above equation of H ~ X+Z.
Expanding the tensor product using block-matrices as intermediate results, so that B looks like
{ 0 X } { 1 0 } { 1 X }
{ X 0 } { 0 -1 } { X -1 }
________ + ________ = ________
sqrt(2) sqrt(2) sqrt(2)
and expanding completely:
{ 0 0 0 1 } { 1 0 0 0 } { 1 0 0 1 }
{ 0 0 1 0 } { 0 1 0 0 } { 0 1 1 0 }
{ 0 1 0 0 } { 0 0 -1 0 } { 0 1 -1 0 }
{ 1 0 0 0 } { 0 0 0 -1 } { 1 0 0 -1 }
______________ + _____________ = ______________
sqrt(2) sqrt(2) sqrt(2)
The columns of matrix B are the coefficients of the four Bell states, so that:
B |00> = |phi_plus>
B |11> = |phi_minus>
B |01> = |psi_plus>
B |10> = |psi_minus>
_Matteo
I found this cool! Thanks.
suggestions for material/textbooks/exercise books to practice these questions?
Обясните пожалуйста понятнее, слишком сложно
Мой мозг взорвался
1:04:50 your theoretical quantum calculations bother me because they assume physics is not local, so you assume relativity is false. I'm not sure we can design a realistic technological device with such assumptions. We could wonder if these calcultations have enough solid experimental grounding. Relativity could be false, Physic could be non local... no causality anymore🤔
Ничего не понятно
Thanks for the great presentation.
I am not clear at 1:03:26 when you say that for controlled-Not with Y as control, you swap the order of tensor products. I would appreciate help (may be, it has been taken up earlier and I am not able to locate it). Thanks
Hello sir, thank you for your efforts, it is a great video.
I came across N00N status on the internet. I learn it makes precise phase measurement when optical interferometer is used in quantum optics. Can such a situation be used in quantum algorithms? If so, where is it usually used?
Awesome lecture, the whole series. Around 53:37, small misspelling in Measurement/example, the coefficient in front of |10> has to be with the sign "-" and the coefficient in front of |11> has to be with the sign "+", or similar to restore the normalization condition to the unity of the mixed state.
I don't see an error here. We can change the coefficients as you suggest to get a different example, but it's a unit vector - and hence a valid quantum state vector - either way.
I see now, you were right. Thanks.
Could you make subtitles available in Barasil's Portuguese language for all videos?
Epic
Can you give some hints on how to get matrix in example on 34:56? in general, how to translate a statement, such as in 34:56 example, to a matrix representation of an operation? thanks
amazing examples and instruction
what is a qubit? and How can we find how many qubits do we have?
In 1:01:16, the SWAP of phi- is negative phi- similar to the case of the psi-minor. Is the wording in the video wrong?
No, performing the swap operation on phi-minus is just phi-minus, not negative phi-minus.
Hi! I'm following along and find this series to be very good, however I'm struggling with the example matrix U at 57:20. To test my knowledge I'm trying to decompose it into a matrix A tensor B acting on X and Y respectively. If I'm correct A should be a 3x3 matrix and B should be a 2x2 matrix.
So far, so good.
However, I cannot for the life of my figure out how (for example) u_41 can be 0.
If my understanding is correct u_41 = a_21 * b_21. This implies that either a_21 or b_21 is zero.
If we assume a_21 is 0, then that should also imply that u_31 is 0 as we can rewrite it as a_21 * b_11, but it isn't.
If we assume b_21 is 0, then that should also imply that u_21 is 0 as we can rewrite is as a_11 * b_21, but it isn't.
The same sort of logic can be applied to all 0-valued entries in U as far as I can tell. So I seem to arrive at the contradiction that A and B must have all non-zero entries that somehow multiply to 0 once they are tensored. So either there's an error or (much more likely) a gap in my understanding. Any clarification would be greatly appreciated!
My working hypothesis is that this must in fact be an entanglement-generating gate thus trying to decompose it won't work, as it cannot be applied independently to XY. Any rebuttal or hints as to how I can prove it would be nice! :)
I think you have already figured it out, but if any of the comment readers get puzzled (from my understanding):
Key point: If X and Y undergoes A and B operations, total operation can be written by tensoring them. But it's reverse is not always true.
The matrix was just a 6x6 unitary matrix to just illustrate an example of unitary matrices which means operations on a system. This is a valid operation for that system. But it is not necessary for the matrix, to be an operation for any two bit systems, to come from tensor producting individual operations on X and Y. Any unitary matrix can represent an operation. We will be using tensor for the reverse case. If we know that A and B are operations on X and Y, then to find the total operation, tensor-product is the solution. But decomposing to tensor isn't necessary for it to be a valid operation.
And yes (I think from my understanding), your @BeepBoop's calculation is the procedure to prove your hypothesis.
I think the matrix is actually not unitary. I think the first three entries of the second row from the bottom should be instead swapped with the zeros in the last row? Just my feeling from checking a few entries... But maybe I got it mixed up - its been a while since I last did matrix multiplication ;-)
Hi there! I have a question at 52:50 . You said earlier that tensor products in quantum state represents independence of individual states. But when we measure a state X from compound system XY, the quantum state of Y changes. Doesn't it mean that the independence has collapsed along with the measurement of individual state X?
As far as I can tell independence in that case is used to refer to unentangled 2-qbit systems, meaning the individual qbits carry information about the overall system when viewed independently. That means that measuring the X qbit independently of Y will give you some amount of information about the overall system, then measuring Y will give you the rest of the information. In contrast: a (maximally) entangled system means that measuring parts of the system will cause decoherence of the system. Thus measuring X, then measuring Y will not give you information about the overall state of XY as it existed in the entangled state.
So a 2-qbit system that is completely de-composable into a product state of 2 single qbit systems is independent in an information-carrying sense, but not necessarily in the probabilistic sense. On the other hand a 2-qbit system that cannot be written as a product state must be maximally entangled meaning that measuring X and Y independently in fact provides 0 information of XY, as information about the system can only be gleaned when considering a joint measurement of XY
@@BeeepBo0op Ok understood. Thx
At 35:00 the matrix representation on an operation on bits (X,Y) was provided. I am unable to understand how those 2 matrices were calculated. The textbook lesson also did not elaborate on how to calculate these matrices. Any help would be appreciated. Thank you.
Think about the deterministic operation where we set Y equal to X. The action on standard basis vectors looks like this: 00 --> 00, 01 --> 00, 10 --> 11, 11 --> 11. In words, the second bit gets replaced by the first. The matrix representation of this deterministic operation is the first one on the right-hand side of the equation (not including the factor of 1/2). In a similar way, setting X equal to Y gives the second matrix on the right-hand side (again, not including the factor of 1/2). If we perform one of the two, each with probability 1/2, we average the matrices to get the matrix on the left-hand side.
@@John.Watrous Hi, Thanks for that clarification. It makes sense now. Can't wait for the next lesson!
Thanks!
34:16 how do we get the matrix representation?
These sort of videos piss me off, because they will explain obvious things to an arrogant level... but will skip to explain the concept of a permutation matrix.
In short, realize that the rows of the column vector for these qubits is represented so:
[ 00 ] = 0
[ 01 ] = 1
[ 10 ] = 2
[ 11 ] = 3
which, is just a top-down writing of 2-bits with all 2^n = 4 states.
Now, because this is a CNOT, it's say that the most significant bit (X,Y) (most sig, least sig) is the control bit. Meaning, if it (X) is 1 (exists) then NOT the next bit (Y).
So you can describe a permutation function like so:
[ 00 01 10 11 ]
@@Gnefitisis THANK U ^-^
Awesome video
when are you going to release the next lecture?
We are going to release all videos in a unit one month apart, with a short break between units
@@qiskit then how long will it take to release the whole course?
In how many days will we see the lesson 3? It would be really helpful if it’s soon! :)
@@tobiasthaller1312 it's sad to know that, they don't have any responsive admin :(
Thank you for your interest in this series.
We would love to be able to tell you exactly when each video in this series will be released, but the reality is that we don't know yet. We're a small team, and creating these videos takes time... we're releasing them as they are created.
What I can tell you is that Lesson 3 will be released in January and Lesson 4 will be released in February, which will complete the first unit of the series. You can expect lessons from the second unit to start appearing sometime in the spring and to continue through the year.
I'm afraid that the completion of the entire series is still a long ways off, and I can't offer a meaningful estimate on how long it will take at this time. We will provide future updates on the timeline as we are able.