Best Bloch Sphere video ever! I was always confused how |0> and |1> were orthogonal states but somehow "on the same line" in the Bloch Sphere, and your walkthrough of the math & how it assigns this antipodal outcome was super clear.
You and not only you are puzzled, because they do not correctly represent the angle of the theta in the figure. Instead of “theta” in the figure, “theta / 2” should be signed, as in the formula. Then the vector I1> will be along the x-axis and thus orthogonal to the vector I 0>. Theta angle should change, as well as phi angle, from 0 to 2 pi. With theta equal to 2pi, the vector -I0> will, as expected, be depicted opposite to the z-axis (down). Moreover, the vector I 0> will return to its original position with theta equal to 4 pi, which actually happens with spin 1/2.
there is good explanation at the 7:00, it is kind of counterintuitive if we consider it in a 3d space (block sphere). In the 2d complex vector space, they are orthogonal.
@@karabaskruger That is a good suggestion. To paraphrase, if you are using theta/2 instead of theta, then for |1> to be along |x>, theta/2 must equal pi/2, i.e. theta = pi. Moreover, theta must range from 0 to 2pi so theta/2 can range from 0 to pi. Though the reader should keep in mind that this is an equivalent set-up from the original one -- differing only in the use of symbols; that is, instead of using 0
A bit late, but making a measurement on a quantum state involves taking an inner product with a complex conjugate of some state. Here for example you could calculate the probabilities of measuring the |0> by taking a product < 0 | Psi >. The conjugate
@@kubafrank96 your answer is bit ambiguous, overall phase is irrelevant as when measurement is performed the qubit collapse into one of the eigenstates of the measurement operator. Consider operator X whose eigenstates are |+> and |-> and probabilities that state |a> collapse into |+> or |-> are and respectively ( notice |+> are involved the probabilities remain unaffected by the overall phase. Also the state after measurement is (|+>)/sqrt( ) or (|->)/sqrt( ) based on weather |a> collapses into |+> or |-> (notice here the overall phase is still present and has not vanished which is totally correct).
We just rotate the whole thing by angle phi_0. Anyway when we represent a complex plane we need two linearly independent vectors. So we can assume alpha as oriented along a new reference axis which rotates the beta by (phi1-phi0)
So we have linear combinations of |0> and |1> represented as points on the surface of a 3-dimensional sphere, but where is the connection to our 3-D space. The Bloch sphere lives in some 3-D spin space, but again, what is the connection to our space? Does the state corresponding to +y represent a spin with a +1 component in the y-direction? And so on for all points? In other words, does the direction a vector in the Bloch sphere coincide with a spin pointing in the same direction in ordinary space?
Lenny D. sez: At 4 minutes for clarity to explain disappearance of the arbitrary phase ( e^iphi) when measurement occurs should show it is multiplied by ( e^ - i phi ) , the product of these being unity regardless of the value of phi
At 8:05 when he calculates which state corresponds to the -|z> on the Bloch sphere he doesn't mention the value of Φ. So my question is why is he omitting that? Shouldn't it be that -|z> corresponds to exp(iΦ)|1> for any value 0
theta is the angle between the vector and z-axis. When the vector on the x-axis, it will be orthogonal to the z-axis. Then, the theta will be 90 degree and it equals pi/2.
Is it because when we make a measurement, the probability is determined by the amplitude squared and when you take e^i(theta) and square it, you always get 1?
На мой арифметически простой взгляд, Сфера Блоха - это *не* физический объект. Можно сказать, что это условное вспомогательное мнемоническое представление о характере взаимодействия физических объектов. Природа не оперирует подобными трансцендентными представлениями. Поэтому на их основе невозможно строить логически правильные умозаключения о практической реализации этих представлений. 09.09.2024.
На интуитивном уровне предполагаю, что Природа оперирует квантовыми процессами в первую очередь в соответствии с симметричными кристаллографическими соотношениями. 13.09.2024.
Интересно, чт0, глядя на Сферу Блоха, рассказывает продвинутый Искусственный Интеллект (AI) о технической и технологической возможности / невозможности создания полноценно работающего квантового компьютера? 26.09.2024.
The “Bloch Sphere” in this figure is the most idiotic invention I have ever seen. It is called "do not believe your eyes." If you see the theta angle, then this is the 2 times smallest angle. But at the same time, if you see the angle phi, then this is the angle phi ...
its an accurate model of the constraints on a vector in 3 dimensional space. it only requires theta over two because allowing it to project the entire domain would be redundant (the angle phi represents the qubits XY component entirely) the one angle in the third dimension doesnt need to project a plane rather an angle
Best Bloch Sphere video ever!
I was always confused how |0> and |1> were orthogonal states but somehow "on the same line" in the Bloch Sphere, and your walkthrough of the math & how it assigns this antipodal outcome was super clear.
You and not only you are puzzled, because they do not correctly represent the angle of the theta in the figure. Instead of “theta” in the figure, “theta / 2” should be signed, as in the formula. Then the vector I1> will be along the x-axis and thus orthogonal to the vector I 0>. Theta angle should change, as well as phi angle, from 0 to 2 pi. With theta equal to 2pi, the vector -I0> will, as expected, be depicted opposite to the z-axis (down). Moreover, the vector I 0> will return to its original position with theta equal to 4 pi, which actually happens with spin 1/2.
there is good explanation at the 7:00, it is kind of counterintuitive if we consider it in a 3d space (block sphere). In the 2d complex vector space, they are orthogonal.
can someone tell, why in a 2d complex vector space they are orthogonal?
@@karabaskruger That is a good suggestion. To paraphrase, if you are using theta/2 instead of theta, then for |1> to be along |x>, theta/2 must equal pi/2, i.e. theta = pi. Moreover, theta must range from 0 to 2pi so theta/2 can range from 0 to pi. Though the reader should keep in mind that this is an equivalent set-up from the original one -- differing only in the use of symbols; that is, instead of using 0
One of the underrated channel for Quantum
Clear and concise! Thank you!
Great video! I spent so much time trying to understand the Bloch Sphere and this video really helped!
Best Bloch Sphere video ever!
it is very very clear. Thanks for the explanation
This video is highly underrated.
Beautiful explanation... thank you
Very clear explanation, thank you very much.
@3:42 Could you explain why the overall phase is physically irrelevant? Thank you.
A bit late, but making a measurement on a quantum state involves taking an inner product with a complex conjugate of some state. Here for example you could calculate the probabilities of measuring the |0> by taking a product < 0 | Psi >. The conjugate
@@kubafrank96 your answer is bit ambiguous, overall phase is irrelevant as when measurement is performed the qubit collapse into one of the eigenstates of the measurement operator. Consider operator X whose eigenstates are |+> and |-> and probabilities that state |a> collapse into |+> or |-> are and respectively ( notice |+> are involved the probabilities remain unaffected by the overall phase. Also the state after measurement is (|+>)/sqrt( ) or (|->)/sqrt( ) based on weather |a> collapses into |+> or |-> (notice here the overall phase is still present and has not vanished which is totally correct).
We just rotate the whole thing by angle phi_0. Anyway when we represent a complex plane we need two linearly independent vectors. So we can assume alpha as oriented along a new reference axis which rotates the beta by (phi1-phi0)
So we have linear combinations of |0> and |1> represented as points on the surface of a 3-dimensional sphere, but where is the connection to our 3-D space. The Bloch sphere lives in some 3-D spin space, but again, what is the connection to our space? Does the state corresponding to +y represent a spin with a +1 component in the y-direction? And so on for all points? In other words, does the direction a vector in the Bloch sphere coincide with a spin pointing in the same direction in ordinary space?
The answer to your question is given in Lecture 15 -4.
thanks, and what is the state 1/sqrt2 |0> + i/sqrt2 |1> representing?
|Y>
When thêta equals zero , phi is arbitrary? Am I right?
very clear, thank you!
Please make the representation of a concrete state in the bloch sphere.
What courses would you reccomend for someone who wants to understand this stuff better?
MIT quantum mechanics, susskind quantum mechanics, but you may need some classical physics to understand what they are talking about.
Lenny D. sez: At 4 minutes for clarity to explain disappearance of the arbitrary phase ( e^iphi) when measurement occurs should show it is multiplied by ( e^ - i phi ) , the product of these being unity regardless of the value of phi
yes, this is due to fact that probability distribution would always be same
Thanks,very helpful!
great
While normalising the coefficients of the wave function, why is it (r0)^2 + (r1)^2 = 1 and not (r0)^2 + (r1*e^iφ)^2 = 1?
Got it. I just forgot that r1 is itself the magnitude of r1(e^iφ).
thanks
At 8:05 when he calculates which state corresponds to the -|z> on the Bloch sphere he doesn't mention the value of Φ. So my question is why is he omitting that? Shouldn't it be that -|z> corresponds to exp(iΦ)|1> for any value 0
I think the phase factor is irrelevant. Geometrically, any value of Φ you plug in will get you to the -|z> pt.
Nice one.
9:10 how is theta in x dir pi/2? Does it start 0 at z and then pi/2 anti clockwise from there?
theta is the angle between the vector and z-axis. When the vector on the x-axis, it will be orthogonal to the z-axis. Then, the theta will be 90 degree and it equals pi/2.
@@sayici so would it be -π/2 if you co clockwise from positive z to negative z?
@@nityadevaraj7554 There is no starting point. If you are at the negative z, it would be |1>.
@@sayici sorry it was a typo, meant to ask if theta would be -pi/2 if you went from positive z to negative X...
@@nityadevaraj7554 Yes, in that case theta would be -pi/2 and the overall state become |-> state.
I know this is often repeated, but why is it that we can drop the overall phase?
Is it because when we make a measurement, the probability is determined by the amplitude squared and when you take e^i(theta) and square it, you always get 1?
So gracefully explained Bloch sphere!!! lol
На мой арифметически простой взгляд, Сфера Блоха - это *не* физический объект. Можно сказать, что это условное вспомогательное мнемоническое представление о характере взаимодействия физических объектов. Природа не оперирует подобными трансцендентными представлениями. Поэтому на их основе невозможно строить логически правильные умозаключения о практической реализации этих представлений.
09.09.2024.
На интуитивном уровне предполагаю, что Природа оперирует квантовыми процессами в первую очередь в соответствии с симметричными кристаллографическими соотношениями.
13.09.2024.
Интересно, чт0, глядя на Сферу Блоха, рассказывает продвинутый Искусственный Интеллект (AI) о технической и технологической возможности / невозможности создания полноценно работающего квантового компьютера?
26.09.2024.
Wherever you look , an Indian is there...don't know, should be proud of or hate it.🤣😅🤣🤣
@ 3:38 This is so poor! Easy to show why this is so. @ 4:30 But he never comes back to this point explicitly.
why e power i phi sub zero is not considered and it is dropped?
some one explain pls.
The “Bloch Sphere” in this figure is the most idiotic invention I have ever seen. It is called "do not believe your eyes." If you see the theta angle, then this is the 2 times smallest angle. But at the same time, if you see the angle phi, then this is the angle phi ...
its an accurate model of the constraints on a vector in 3 dimensional space. it only requires theta over two because allowing it to project the entire domain would be redundant (the angle phi represents the qubits XY component entirely) the one angle in the third dimension doesnt need to project a plane rather an angle
@@andrewegge14 don't ask this guy how he finds where he is positioned on the globe. He might be one of those flat earth people.
Triggered by sphere