Great explanation. I was looking for a graphical visualization of a qbit that made sense and read about the bloch sphere. Your description of how the qbit maps to the sphere is wonderful. It is always amazing when a complex idea can be represented by a simple visual concept.
🙏🙏🙏🙏 was very helpful man!! I have started learning quantum computing now!! First I learnt quantum mechanics then linear algebra and I tried lot of videos there wasn't hard math videos anywhere! I literally understood what actually is happening 🙏🙏🙏thanks a ton!!
Best explanation of Bloch sphere I have found. You started with the math and then that naturally led up to the sphere picture. Other videos go the reverse way which is not as transparent. If you have not already, can you please post a video on why a mixed state lies inside the sphere. Thanks again.
This video is really helpful for me, because I saw other video related to block's sphere, these all confused me. But your way of explaining is really amazing. I praised. Thanks alot............... Love from Pakistan
From all of the content I have studied so far I found this one to be the best explained. For sure this can be sure to grasp everything at once. But after a couple of reviews (and some math reminders with a pen and a piece of paper) things get clearer and clearer. Thank a lot, blessed youtuber!
@@mu-hoch-380 Have you considered by any chance writing a blog or even a paper w/ markdown? I started writing (French) notes to make things clearer in my mind. Anyway I like your approach because I’m personally fed up with those math nerds who sometimes enjoy not being understood by others.
Slight explanation: The south pole is e^{i \phi} |1> \equiv |1> since an overall phase is immaterial (p.s. I like your introduction of the equivalence symbol.)
Dude, actually speaking, I have finished almost all the playlists in Quantum computing and Quantum information available in youtube. But if you make one in English, yours will be the best. Please go for it.
There is another component that can be applied to the qubit. It looks like applying spin we can then talk about spinors. The state space of a spinor is out of phase with physical space by a factor of two. I believe this is why θ -> θ/2.
Excellent explanation. The good teaching method is to make the things as easy as possible, not for some textbooks, to make things too complicated to understand.
Thanks for such a nice video, one doubt: when I am considering a sphere in 3d eucleadian space, I feel the projection of any vector on unit sphere is cos(\theta) and not cos(\theta/2), what I am doing wrong? one question: if \theta is pi, then how \si will impact the qubit One suggestion/Request: It would be great if you can upload corrected (not autogenerated) subtitles in English
1. yes, the probability to measure a qubit in state |0> is (cos theta/2)^2 and to measure it in state |1> is (sin theta/2)^2 2. also yes :-) are you familiar with the concept of "spin" in quantum mechanics? if you realize a qubit as the spin of an electron in a magnetic field, then the vector pointing from the center of the bloch-sphere to the point that represents the quantum state of this qubit is simply the vector of the angular momentum of the electron (in the three-dimensional space) :-) when you measure the spin you either get spin up or spin down (the wavefunction collapses, when you measure!), but before a measurement, the spin can be in a superposition of the basis states |up> and |down> which means that its angular momentum vector points to any(!) direction in three-dimensional space. so for a "spin qubit" the angle phi just denotes the horizontal angel of its angular momentum vector.
Is there some reason we would want 0 and 1 to be opposite each other on the sphere? They do not look orthogonal to me when they are like that, which I do not like.
hi, thank you for your comment :-) as we have to represent a qubit on a three-dimensional "thing" (i.e, a sphere), and we have TWO "opposite" states, we want to place them as far as possible from each other. so living on a (three-dimensional) sphere, we get the biggest distance when we place |0> and |1> on the poles of the bloch sphere :-) the price we have to pay for this is, that the vectors representing the qubit states |0> and >1> have an angle of 180 degrees between them - and not 90 degrees, as we are used to when we speak about orthogonality...
Someone: What cha doin’? Me: Studying Quantum Mechanics. Someone: Wow! What’s your favourite topic? Me: Block Sphere, but Block with an ‘H’ instead of ‘K’ Someone: *Confused as hell! Block like Sphere, Sphere like Block, WTF! I knew something is not right about this subject when I heard of someone ‘s cat! 😂
How would one represent Rabi oscillations on the Bloch sphere, would it be an axis rotation of angle phi with tan(ø) = perturbation/difference of energy between both observables? If so, why?
I can't understand why pick r_alpha as cos (theta/2) and r_betta as sin (theta/2), can i pick arbitrary combination considering r_a^2 + r_b^2 = 1, but we use this in specific cause they live in the first cuadrant?
They are very much orthogonal! However, observing spin in x or y direction destroys the state into a superposition as those observations (corresponding to theta = pi/2, phi = 0 for x or pi/2 for y) arent eigenstates on the pauli z-matrix Hence your next observation of zspin might change as pauli matrices do not commute/arent compatible/ do not share a common base of eigenvectors
e^(ix) is only 1 for x = n 2Pi (where n is any integer) but |e^(ix)| = 1. so you multiply the state with a number which has the absolute value of 1, you can do this, as it doesen’t change the physics of the qubit.
A way to think about is this: Recall that the r_\alpha and r_\beta are positive real numbers, and so if we visualize them on the unit circle, they only occupy the first quadrant, i.e. \theta \in [0,\pi/2]. However, we want to map this interval to that of the unit sphere which goes from [0,\pi], and so we want define the angle in the unit circle as \theta/2 so that it stretches out to the entirety of the range of angles in the unit sphere.
Thanks alot, but could you, please, tell me why the range of thêta/2 is from 0 to Pi/2 (quad of circle) and not 2*Pi (all the circle). Also another question how to define Phi (spherical angle) for a point on the axis....thanks again sir
first of all: r_alpha and r_beta are both positive (!) numbers, so the point (r_alpha, r_beta) must be in the first quadrant and therefor theta/2 goes from 0 to pi/2 (and not further :-) question 2: you mean phi for the states |0> and |1> ? here the value of phi is irrelevant as one angle describes completly the states. you could pick any angle, but lets see what would that means, for example: theta = 0 and phi = 0 means state |0> theta = 0 and phi = pi/2 means state i|0> theta = 0 and phi = pi means state -|0> theta = 0 and phi = 3pi/2 means state -i|0> BUT: all this state are absolutly the same! so as i said: with theta = 0 we know exactly the state of the qubit, no further knowledge (here phi) is necessary :-)
@@mu-hoch-380 Thanks, but about Phi I am wondering about what is the answer for the question: what is the spherical coordinates for the point with cartesian coordinates (0, 0, z) ? Thanks again
for a point (0,0,z) Phi ist simply undefined! you can look up here for more information about spherical coordinates: en.m.wikipedia.org/wiki/Spherical_coordinate_system
taken from wikipedia: „… if θ is 0° or 180° (elevation is 90° or −90°) then the azimuth angle is arbitrary; and if r is zero, both azimuth and inclination/elevation are arbitrary. To make the coordinates unique, one can use the convention that in these cases the arbitrary coordinates are zero.“
@@sayanjitb please look at 6:03 - there you see how phi and theta can be calculated: phi is the difference of the phases of alpha and beta. in your case alpha and beta are REAL numbers, so this difference is 0, and therefore phi is 0. theta is 2 arccos(r_alpha), where r_alpha ist the absolute value of alpha (i hope you are familiar with writing a complex number in exponential form?). as in our case r_alpha = 1/sqrt(2) we get: theta = 2 arccos(1/sqrt(2)) = 90° (or pi/2) that's it: the state 1/sqrt(2) (|0> + |1>) is at the equator of the blochsphere, as theta = 90°, exactly at that place where the positive x-axis intersects the blochsphere. I post a picture of the blochsphere with some typical states below this post - I hope I could help you 😊
@@mu-hoch-380 thank you so much for your invaluable assistance over online teaching. What you have answered is for pure spin state. What if I take mixed state into account? Don't i have to deal with density matrix? How can I represent that in Bloch sphere representation? TIA
Amazing explanation! Only your Psi angle on your drawn Bloch sphere seems to be wrong? Looks like it is working on the same axis as the theta? The final moving model makes it very clear though!
you mean the angle phi (not psi)? for this angle you have to project the position of the qubit onto the x-y plane. then phi is the angle between the positive x-axis and the vector pointing from the origin to the projected qubit in the x-y plane :-) so theta "works" on the z-axis and phi on the x-axis...
hi k r 😀 no, ist is not, because this time we go from the north pole of the bloch sphere, where theta is 0, to the south pole, where theta is pi. these locations represent different states: |0> on the north and |1> on the south pole.
@@DrDeuteron the range for θ is 0 ≤ θ ≤ π, so θ = 2 π is not possible. The north pole of the bloch sphere is at θ = 0, the south pole at θ = π. for θ = π the system is in the state |1> (not -|0> )
are you familiar with the concept of writing a complex number in the polar form a = r_a * e ^ (i phi_a) where r ist he absolute value of a and phi_a ist the phase or "argument" of z? phi is simply the difference between the phases oft the complex numbers a and b: phi = phi_b - phi_a (if this difference is negativ you normally add 2 Pi).
oh i see… here we just write the complex numbers alpha and beta in the so called „Euler form“ - here the wikipedia article where you can see how it works :-) en.m.wikipedia.org/wiki/Euler%27s_formula
@@jimsagubigula7337 actually... not in the near future, i have just too many other things to do... (i started to work on a video about the deutsch-algorithm, but since 6 month i didn't find the time to finish it)
the angle that describes the relation between r_alpha and r_beta lies between 0 and pi/2. we could define ANY parameter for this angle, but as you see from 6:02 we want that the parameter lies between 0 and pi, so that we can interprete the theta as the azimuthal angle of a location on a sphere.
@@mu-hoch-380 Thanks alot, but could you, please, tell me why the range of thêta/2 is from 0 to Pi/2 (quad of circle) and not 2*Pi (all the circle). Also another question how to define Phi (spherical angle) for a point on the axis....thanks again sir
@@SkanderTALEBHACINE I think it's because all of theta represents what we see as the vertical axis, and phi defines the entire horizontal axis. If theta could go up to 2pi the line would wrap back around the circle and indicate a value already represented by the range with just pi.
@@SkanderTALEBHACINE theta/2 is equal to sininv(r_alpha) and cosinv(r_beta), which have the ranges [-pi/2, pi/2] and [0, pi] respectively. If you take their intersection, since theta must satisfy both, you'll get the range [0, pi/2].
One of the best explanation of Bloch Sphere on RUclips. Highly recommended for those who are new to Quantum Computing.
I am writing an essay about quantum computers and have a problem with visualizing qubits in bloch sphere, your video just helped me a lot. Thanks
thank you for your comment - i would like to read your essay :-)
This uploader deserves more subscribers. Come on, everybody
thank you :-)
Done
Very rigorous. Definitely a well prepared and good explanation of the Bloch sphere representation. Well done.
Great explanation. I was looking for a graphical visualization of a qbit that made sense and read about the bloch sphere. Your description of how the qbit maps to the sphere is wonderful. It is always amazing when a complex idea can be represented by a simple visual concept.
A very clear explanation. Thank you.
Really well explained! Thank you.
I feel like I need to pay you for this wonderful explanation
🙏🙏🙏🙏 was very helpful man!! I have started learning quantum computing now!! First I learnt quantum mechanics then linear algebra and I tried lot of videos there wasn't hard math videos anywhere! I literally understood what actually is happening 🙏🙏🙏thanks a ton!!
While sitting in class, wondering what is this. Finally, i understand. Thanks to you.
Amazing explanation. Best one found yet. Go on with this kind of tutorials.
Yeah pls
You should do more videos (on QM?). Fine style of presenting and explaining!
thank you for your encouraging words :-) actually I really think about doing a whole YT series on quantum computing...
@@mu-hoch-380 Seems like a good idea to me. Wish you good luck!
Best explanation of Bloch sphere I have found. You started with the math and then that naturally led up to the sphere picture. Other videos go the reverse way which is not as transparent. If you have not already, can you please post a video on why a mixed state lies inside the sphere. Thanks again.
the beeeeeeeeeeeeeeest video explaining the bloch sphere on the internet ..thank u soooooo much
This is the best explanation I've seen on the Bloch Sphere, thank you!
yesssssss
you saved my grade in quantum mechanics, thank you
wow - this alone would be reason enough to do this video, thanx for the feedback :-)
Excellent explanation
Such a good video, I understood much more than in two lectures at university about quantum computing. Thanks!! :)
It was a very nice explanation, sir. Yours is the only place I could understand. Thank you
Good job! Very nice explanation in combination with a calming voice :)
thank you - made my day :-)
This video is really helpful for me, because I saw other video related to block's sphere, these all confused me. But your way of explaining is really amazing. I praised. Thanks alot............... Love from Pakistan
Thank you @mu-hoch-3. I like how you explained the Probability of finding the qubit and the superposition states with the use of the Bloch Sphere. !
thank you for your nice comment :-)
seriously great job, all other videos on this are super confusing, you deserve +10k subscribers
wow... thank you!! +10k is difficult to reach with only one video on youtube... ;-)
From all of the content I have studied so far I found this one to be the best explained.
For sure this can be sure to grasp everything at once. But after a couple of reviews (and some math reminders with a pen and a piece of paper) things get clearer and clearer.
Thank a lot, blessed youtuber!
thank you so much! :-) it encourages me to do more qc videos - if I just had more time.....
@@mu-hoch-380 Have you considered by any chance writing a blog or even a paper w/ markdown? I started writing (French) notes to make things clearer in my mind. Anyway I like your approach because I’m personally fed up with those math nerds who sometimes enjoy not being understood by others.
@@geoffreyroyer452 yes, i do... plan is to start a blog in january 2021 :-)
very nicely done, appreciate the presentation. :)
Slight explanation: The south pole is e^{i \phi} |1> \equiv |1> since an overall phase is immaterial (p.s. I like your introduction of the equivalence symbol.)
thank you explained very beautifully
very clear explanation ! Thanks !!
Damn, this is a really informative video
very good i kinda get it now (a bit more) thanks
Very well explained. Cleared a lot of doubts. Thank you so much
Thank you, well explained.!!
Digvijay Singh Rawat thank you 😃
Simple and to the point. Keep it up!
Dude, actually speaking,
I have finished almost all the playlists in Quantum computing and Quantum information available in youtube.
But if you make one in English, yours will be the best.
Please go for it.
Excellent explanation! Thanks a lot!!
Well explained.
My apprrciations.
There is another component that can be applied to the qubit. It looks like applying spin we can then talk about spinors. The state space of a spinor is out of phase with physical space by a factor of two. I believe this is why θ -> θ/2.
Please make more videos in quantum computing, will be really really great 😊 🌸
Excellent explanation. The good teaching method is to make the things as easy as possible, not for some textbooks, to make things too complicated to understand.
Just what I needed. Danke!
Please keep posting about Quantum Computing
nicely explained. thanks a lot.
Sehr gute Erklärung! Die meisten, die ich zuvor gesehen habe, haben mich verwirrt.
Superbly clear
great explenation, i was struggling in that part in a course, thanks!
so well-explained! thank you.
Really nice explanation of the maths behind the Bloch sphere, made my day ^_^
Bro, it is precisely what I wanted to know!
so clear. thank you
Thank you sir.
Thanks for such a nice video,
one doubt: when I am considering a sphere in 3d eucleadian space, I feel the projection of any vector on unit sphere is cos(\theta) and not cos(\theta/2), what I am doing wrong?
one question: if \theta is pi, then how \si will impact the qubit
One suggestion/Request: It would be great if you can upload corrected (not autogenerated) subtitles in English
so theta indicates the probability of the qubit being in one of the two states, right? Is there any intuition about what phi denotes in this case?
1. yes, the probability to measure a qubit in state |0> is (cos theta/2)^2 and to measure it in state |1> is (sin theta/2)^2
2. also yes :-) are you familiar with the concept of "spin" in quantum mechanics? if you realize a qubit as the spin of an electron in a magnetic field, then the vector pointing from the center of the bloch-sphere to the point that represents the quantum state of this qubit is simply the vector of the angular momentum of the electron (in the three-dimensional space) :-)
when you measure the spin you either get spin up or spin down (the wavefunction collapses, when you measure!), but before a measurement, the spin can be in a superposition of the basis states |up> and |down> which means that its angular momentum vector points to any(!) direction in three-dimensional space.
so for a "spin qubit" the angle phi just denotes the horizontal angel of its angular momentum vector.
thank you so much sir.
Is there some reason we would want 0 and 1 to be opposite each other on the sphere? They do not look orthogonal to me when they are like that, which I do not like.
hi, thank you for your comment :-) as we have to represent a qubit on a three-dimensional "thing" (i.e, a sphere), and we have TWO "opposite" states, we want to place them as far as possible from each other. so living on a (three-dimensional) sphere, we get the biggest distance when we place |0> and |1> on the poles of the bloch sphere :-) the price we have to pay for this is, that the vectors representing the qubit states |0> and >1> have an angle of 180 degrees between them - and not 90 degrees, as we are used to when we speak about orthogonality...
totally makes sense to me, best videos relevant
Very clearly explained!!!! Thanks a lot!
man, this is beautiful
How can you go back to calculate the original angles of alpha and beta
For any general -qubit state, is it possible to visualize the state on Bloch spheres?
hi, I don‘t quite understand.. - is your question how to calculate theta and phi when you know alpha and beta?
thank you. you are a genius
Brilliant thanks.
This was fantastic! Thank you for such a clear video :)
Someone:
What cha doin’?
Me: Studying Quantum Mechanics.
Someone: Wow! What’s your favourite topic?
Me: Block Sphere, but Block with an ‘H’ instead of ‘K’
Someone: *Confused as hell! Block like Sphere, Sphere like Block, WTF!
I knew something is not right about this subject when I heard of someone ‘s cat!
😂
It’s not pronounced Block tho, it’s German. As he pronounces it.
Great explanation. Thank you!
How would one represent Rabi oscillations on the Bloch sphere, would it be an axis rotation of angle phi with tan(ø) = perturbation/difference of energy between both observables?
If so, why?
Thank you friend.
I can't understand why pick r_alpha as cos (theta/2) and r_betta as sin (theta/2), can i pick arbitrary combination considering r_a^2 + r_b^2 = 1, but we use this in specific cause they live in the first cuadrant?
Great explanation. Until now, I was wondering why |0> and |1> where not orthogonal!
They are very much orthogonal!
However, observing spin in x or y direction destroys the state into a superposition as those observations (corresponding to theta = pi/2, phi = 0 for x or pi/2 for y) arent eigenstates on the pauli z-matrix
Hence your next observation of zspin might change as pauli matrices do not commute/arent compatible/ do not share a common base of eigenvectors
@@butwhoasked1821 I meant in plain 3D.
@@rebluecrow ooh ok mb. That would be the reason of the theta/2 as he explained, sorry
Wonderful… thankyou.
Very well explained!
Great explanation.... the only thing not clear is 'the trick' ... why is e^(-i φα) = 1 ?
e^(ix) is only 1 for x = n 2Pi (where n is any integer) but |e^(ix)| = 1.
so you multiply the state with a number which has the absolute value of 1, you can do this, as it doesen’t change the physics of the qubit.
I still don't get why we take theta/2 while calculating r_alpha and r_beta. Can someone help.
A way to think about is this: Recall that the r_\alpha and r_\beta are positive real numbers, and so if we visualize them on the unit circle, they only occupy the first quadrant, i.e. \theta \in [0,\pi/2]. However, we want to map this interval to that of the unit sphere which goes from [0,\pi], and so we want define the angle in the unit circle as \theta/2 so that it stretches out to the entirety of the range of angles in the unit sphere.
Thanks!
marvellous
Thanks alot, but could you, please, tell me why the range of thêta/2 is from 0 to Pi/2 (quad of circle) and not 2*Pi (all the circle). Also another question how to define Phi (spherical angle) for a point on the axis....thanks again sir
first of all: r_alpha and r_beta are both positive (!) numbers, so the point (r_alpha, r_beta) must be in the first quadrant and therefor theta/2 goes from 0 to pi/2 (and not further :-)
question 2: you mean phi for the states |0> and |1> ? here the value of phi is irrelevant as one angle describes completly the states. you could pick any angle, but lets see what would that means, for example:
theta = 0 and phi = 0 means state |0>
theta = 0 and phi = pi/2 means state i|0>
theta = 0 and phi = pi means state -|0>
theta = 0 and phi = 3pi/2 means state -i|0>
BUT: all this state are absolutly the same! so as i said: with theta = 0 we know exactly the state of the qubit, no further knowledge (here phi) is necessary :-)
@@mu-hoch-380 Thanks, but about Phi I am wondering about what is the answer for the question: what is the spherical coordinates for the point with cartesian coordinates (0, 0, z) ? Thanks again
for a point (0,0,z) Phi ist simply undefined!
you can look up here for more information about spherical coordinates: en.m.wikipedia.org/wiki/Spherical_coordinate_system
taken from wikipedia: „… if θ is 0° or 180° (elevation is 90° or −90°) then the azimuth angle is arbitrary; and if r is zero, both azimuth and inclination/elevation are arbitrary. To make the coordinates unique, one can use the convention that in these cases the arbitrary coordinates are zero.“
Dear sir, Using this representation we are able to represent pure states. But can we represent mixed states as well?
sure! give me a mixed state, and I will calculate theta and phi for you :-)
@@mu-hoch-380 Okay lets work out for :-- \pho = (1/sqrt(2)) (|0>
@@sayanjitb please look at 6:03 - there you see how phi and theta can be calculated:
phi is the difference of the phases of alpha and beta. in your case alpha and beta are REAL numbers, so this difference is 0, and therefore phi is 0.
theta is 2 arccos(r_alpha), where r_alpha ist the absolute value of alpha (i hope you are familiar with writing a complex number in exponential form?). as in our case r_alpha = 1/sqrt(2) we get: theta = 2 arccos(1/sqrt(2)) = 90° (or pi/2)
that's it: the state 1/sqrt(2) (|0> + |1>) is at the equator of the blochsphere, as theta = 90°, exactly at that place where the positive x-axis intersects the blochsphere.
I post a picture of the blochsphere with some typical states below this post - I hope I could help you 😊
imgur.com/a/rWeIxeM
@@mu-hoch-380 thank you so much for your invaluable assistance over online teaching.
What you have answered is for pure spin state. What if I take mixed state into account? Don't i have to deal with density matrix? How can I represent that in Bloch sphere representation?
TIA
keep making more
Well explained sir
I thank you very much, good sir!
Amazing explanation! Only your Psi angle on your drawn Bloch sphere seems to be wrong? Looks like it is working on the same axis as the theta? The final moving model makes it very clear though!
you mean the angle phi (not psi)? for this angle you have to project the position of the qubit onto the x-y plane. then phi is the angle between the positive x-axis and the vector pointing from the origin to the projected qubit in the x-y plane :-)
so theta "works" on the z-axis and phi on the x-axis...
@@mu-hoch-380 Completely correct! I meant Phi my bad
I understand too guys
@ 4:43 Here is the second "trick" in this video.
I thought in the Title says @ SIMPLY EXPLAINED@
It is simple explained. If you do not find it simple, then you don’t have the basics of complex numbers, trigonometry…etc.
Isn't zeta being pi the same as zeta being 0, as both the vectors will be overlapping ? ( Applying the same reasoning as used for phi at 5:33 )
hi k r 😀 no, ist is not, because this time we go from the north pole of the bloch sphere, where theta is 0, to the south pole, where theta is pi. these locations represent different states: |0> on the north and |1> on the south pole.
@@mu-hoch-380 thanks for the clarification! The video was really helpful👌
@@DrDeuteron the range for θ is 0 ≤ θ ≤ π, so θ = 2 π is not possible. The north pole of the bloch sphere is at θ = 0, the south pole at θ = π. for θ = π the system is in the state |1> (not -|0> )
@@mu-hoch-380 right, anything past pi is a phase factor, imma delete that.
given alpha and beta, how do you compute phi? (... assuming I have also found theta by doing 2*arccos(sqrt(alpha)) ... )
are you familiar with the concept of writing a complex number in the polar form a = r_a * e ^ (i phi_a) where r ist he absolute value of a and phi_a ist the phase or "argument" of z? phi is simply the difference between the phases oft the complex numbers a and b: phi = phi_b - phi_a (if this difference is negativ you normally add 2 Pi).
@@mu-hoch-380 and how do you find phi_b and phi_a ?
@@pablobiedma watch this :-) ruclips.net/video/a5IQIGuFdCM/видео.html
Good video but where does the second line come from?
what do mean by „second line“?
@@mu-hoch-380 2nd line from the top with ket psi = r(alpha) e^(iPhialpha) + r(beta) e^(iPhibeta)
oh i see… here we just write the complex numbers alpha and beta in the so called „Euler form“ - here the wikipedia article where you can see how it works :-)
en.m.wikipedia.org/wiki/Euler%27s_formula
@@mu-hoch-380 cheers mate
well done!
Looks like bogoliubov transform for fermions?
What happens in a mixed state?
mixed states are located INSIDE the bloch sphere - but that would be subject of a whole new video…
@@mu-hoch-380 Hmm, pretty interesting. Do you intend to do anything like that in the near future?
@@jimsagubigula7337 actually... not in the near future, i have just too many other things to do... (i started to work on a video about the deutsch-algorithm, but since 6 month i didn't find the time to finish it)
@@mu-hoch-380 OK, I understand
amazing!
not to be confused with the Baloch sphere which sadly doesn't exist
why theta/2 ? can you explain that?
the angle that describes the relation between r_alpha and r_beta lies between 0 and pi/2. we could define ANY parameter for this angle, but as you see from 6:02 we want that the parameter lies between 0 and pi, so that we can interprete the theta as the azimuthal angle of a location on a sphere.
Sehr gut!
Dankeschön :-)
Great.
Why the range of thêta is Pi and not 4*Pi? Thanks
theta/2 has the range [0, pi/2], so theta has the range [0, pi].
@@mu-hoch-380 Thanks alot, but could you, please, tell me why the range of thêta/2 is from 0 to Pi/2 (quad of circle) and not 2*Pi (all the circle). Also another question how to define Phi (spherical angle) for a point on the axis....thanks again sir
@@SkanderTALEBHACINE I think it's because all of theta represents what we see as the vertical axis, and phi defines the entire horizontal axis. If theta could go up to 2pi the line would wrap back around the circle and indicate a value already represented by the range with just pi.
@@SkanderTALEBHACINE theta/2 is equal to sininv(r_alpha) and cosinv(r_beta), which have the ranges [-pi/2, pi/2] and [0, pi] respectively. If you take their intersection, since theta must satisfy both, you'll get the range [0, pi/2].
@@amritabhatia9910 Thanks but what do you mean with cosinv ? sininv ? also r_alpha and r_beta ? ( excuse me it is also 6 months ago!) , thanks again
epic. thank you
👏👏👏👏👏👏
Wishing I found this earlier
why u took $\theta / 2$ ?
Hi, I answerd this before - see what I replied in this thread to „Manuel Aguirre“ :-)