Errata: 1) At 2:35, the South Pole corresponds to theta = pi. Therefore, "psi(pi/2, phi)" → "psi(pi, phi)". 2) At 3:44, in the last line, we forgot the exp(i phi). This is necessary in order to write everything as psi again.
Thankyou so much, one can know the extent of your knowledge simply by how you define things, im only starting with this and watched dozens of videos (literally) and none mentioned explicitly that points can exist inside the Block Sphere (ie not on surface)
At 3:44 last line, why can we remove the e^(i*phi) when substitution the sin(v/2) identity? Shouldn't we leave the e^(i*phi) inside so that it is equal to the state=cos(v/2)|0>+e^(i*phi)*sin(v/2)|1> instead of just state=cos(v/2)|0>+sin(v/2)|1> ? Thank you for your wonderful videos!
For the last one, in order to calculate the probability for the random state is found in spin upstate in the positive X axis is given by the inner product of those two states right?
The probability that a truly random state (a state you know nothing of) will be either up or down (along any axis) is 1/2. If you want to calculate the transition probability, you need to compute the inner product of |initial> and |final>.
This is the most general ansatz to solve the equation for alpha and beta. Since we're dealing with absolute squares, the phase of the numbers is not uniquely determined, hence we need the exp(...) factors.
Due to our recent poll, the next videos will be about quantum mechanics, and after that nuclear/particle physics. We haven’t decided yet what comes after that!
Thanks!!! I understood everything except one part, when you plug (pi,phi) into the general state, I got exp(i phi)|1> which isn't the same as the |1> state. Did I make a mistake with that?
This exp(i phi) is an unobservable phase, you it doesn’t matter what phi is. Furthermore, at the north/south pole of a sphere, phi loses its meaning (= there is no difference between theta = 0, phi = 0 and theta = 0, phi = pi, etc).
@@joshuapasa4229 for polar representation of the complex number we use this form: \a = |a|*exp(\i \phi) . Why do you think exponent should be a general function of \theta and \phi?
We definitely want to cover more quantum information topics! Currently, our supporters on Patreon have voted on QM and nuclear/particle physics, so maybe we‘ll do some QI after that :D
Hi I have a few questions: 1. Can we measure a qubit (single qubit, without entanglement)? 2. Actually, what is the output of measurement of a qubit? Is it 1 bit ? How about 2 qubits that entangled? Is it 2 bits? 3. Actually, what's the reason a qubit cannot be copied? Is it because its state uncertain? Thank you. Hopefully someone can help. Because I already read number of articles, but I still cannot get the theory.
1) Yes, performing a measurement on a qubit works the same way as measurement in quantum mechanics. 2) The result of measuring a qubit is an eigenstate of the operator that's used for the measurement. It could be something simple like |1>, or a linear combination like (|1>+|0>)/sqrt②. 3) You might want to watch our video on the No-Cloning theorem: ruclips.net/video/ydh7Xxh-8lg/видео.html
@@PrettyMuchPhysics Thank you for your explanation. 1. What I understand is the outcome of a single qubit measurement is one classical bit, while for 2 qubits that are entangled will give two classical bits (same value, if we measure particle A the result is |1> automatically we know the other particle also hold same value)? Does my understanding correct? 2. Yes, I already watched the no-cloning theorem from your channel. But I still couldn't understand. Thank you.
It depends on how exactly they are entangled. For instance, if you have the state |10> + |01> (where I skipped the normalization) then if you measure „1“ in the first sub system, then the other has to be a „0“ (due to the first ket state). And if the first sub system turns out to be a „0“, then we are dealing with the second ket and we get a „1“ in the second sub system.
@@PrettyMuchPhysics Ok tq.. I Is it possible to suggest me book or article or anything that I can refer to, for better understanding? Because I thought, measurement of two entangled qubits will make the two systems hold same value after the measurement. After measurement the superposition collapse to the measured state... Is the entanglement between particles also destroy?
If the entangled state is instead, |00> + |11>, then you would get the same state in both systems. This is why it depends on how exactly the states are entangled. If you perform a measurement, the entanglement is gone. For more information on entanglement, I‘d suggest a quantum mechanics book, like the one by Griffiths!
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.
@@PrettyMuchPhysics yeah delta potential ok I got it I don't need perturbation theory but is it perturbed state?and can I solve it with perturbation theory?
*God I needed this. Doing this course on Coherence and Quantum entanglement. This came just at the right moment.*
Glad you liked it, Quahntasy! :D
Errata:
1) At 2:35, the South Pole corresponds to theta = pi. Therefore, "psi(pi/2, phi)" → "psi(pi, phi)".
2) At 3:44, in the last line, we forgot the exp(i phi). This is necessary in order to write everything as psi again.
Thankyou so much, one can know the extent of your knowledge simply by how you define things, im only starting with this and watched dozens of videos (literally) and none mentioned explicitly that points can exist inside the Block Sphere (ie not on surface)
Bro, your channel is saving my carreer.
🤩
At 3:44 last line, why can we remove the e^(i*phi) when substitution the sin(v/2) identity? Shouldn't we leave the e^(i*phi) inside so that it is equal to the state=cos(v/2)|0>+e^(i*phi)*sin(v/2)|1> instead of just state=cos(v/2)|0>+sin(v/2)|1> ? Thank you for your wonderful videos!
Of course! We accidentally omitted the exp(i phi) in the last line! :0 Thanks for letting us know! I'll write it in a pinned comment!
One more point, why were you keep writing states in terms of bra vectors instead of kets?
For the last one, in order to calculate the probability for the random state is found in spin upstate in the positive X axis is given by the inner product of those two states right?
The probability that a truly random state (a state you know nothing of) will be either up or down (along any axis) is 1/2. If you want to calculate the transition probability, you need to compute the inner product of |initial> and |final>.
@@PrettyMuchPhysics right right , thank you so much
@ 1:01 States that lie opposite to each other on the Bloch Sphere are only orthogonal in Hilbert Space.
Great vid!
Thank you very much! :D
Very clarifying and coherent
Thanks :D
Where could I learn about that phase factor at 1:52. I understood the sine and cosine but why add a factor of e^i?
e^it is representation of a complex number by Euler's formula. A and B are complex numbers so this is a way of representing them.
This is the most general ansatz to solve the equation for alpha and beta. Since we're dealing with absolute squares, the phase of the numbers is not uniquely determined, hence we need the exp(...) factors.
could you make video about quantum information? Is it in your tudo list?
Due to our recent poll, the next videos will be about quantum mechanics, and after that nuclear/particle physics. We haven’t decided yet what comes after that!
I learned and and forgot all these cool things. Now am trying to do the same again.
That's great, I hope our videos are helpful!
Thanks!!! I understood everything except one part, when you plug (pi,phi) into the general state, I got exp(i phi)|1> which isn't the same as the |1> state. Did I make a mistake with that?
This exp(i phi) is an unobservable phase, you it doesn’t matter what phi is. Furthermore, at the north/south pole of a sphere, phi loses its meaning (= there is no difference between theta = 0, phi = 0 and theta = 0, phi = pi, etc).
@@PrettyMuchPhysics Why is the choice exp(i\phi) instead of exp(i f(\phi,\theta)), where f is a general function?
@@joshuapasa4229 for polar representation of the complex number we use this form: \a = |a|*exp(\i \phi) . Why do you think exponent should be a general function of \theta and \phi?
For any general n-qubit state, is it possible to visualize the state on n Bloch spheres?
Please go for a mini course on QIT!
We definitely want to cover more quantum information topics! Currently, our supporters on Patreon have voted on QM and nuclear/particle physics, so maybe we‘ll do some QI after that :D
@@PrettyMuchPhysics Sounds awesome! Will check out the patreon btw
🤩
really good
Thank you very much!
Hi I have a few questions:
1. Can we measure a qubit (single qubit, without entanglement)?
2. Actually, what is the output of measurement of a qubit?
Is it 1 bit ?
How about 2 qubits that entangled? Is it 2 bits?
3. Actually, what's the reason a qubit cannot be copied? Is it because its state uncertain?
Thank you.
Hopefully someone can help. Because I already read number of articles, but I still cannot get the theory.
1) Yes, performing a measurement on a qubit works the same way as measurement in quantum mechanics.
2) The result of measuring a qubit is an eigenstate of the operator that's used for the measurement. It could be something simple like |1>, or a linear combination like (|1>+|0>)/sqrt②.
3) You might want to watch our video on the No-Cloning theorem: ruclips.net/video/ydh7Xxh-8lg/видео.html
@@PrettyMuchPhysics
Thank you for your explanation.
1. What I understand is the outcome of a single qubit measurement is one classical bit, while for 2 qubits that are entangled will give two classical bits (same value, if we measure particle A the result is |1> automatically we know the other particle also hold same value)? Does my understanding correct?
2. Yes, I already watched the no-cloning theorem from your channel. But I still couldn't understand.
Thank you.
It depends on how exactly they are entangled. For instance, if you have the state
|10> + |01>
(where I skipped the normalization) then if you measure „1“ in the first sub system, then the other has to be a „0“ (due to the first ket state). And if the first sub system turns out to be a „0“, then we are dealing with the second ket and we get a „1“ in the second sub system.
@@PrettyMuchPhysics
Ok tq.. I
Is it possible to suggest me book or article or anything that I can refer to, for better understanding?
Because I thought, measurement of two entangled qubits will make the two systems hold same value after the measurement.
After measurement the superposition collapse to the measured state... Is the entanglement between particles also destroy?
If the entangled state is instead,
|00> + |11>,
then you would get the same state in both systems. This is why it depends on how exactly the states are entangled.
If you perform a measurement, the entanglement is gone.
For more information on entanglement, I‘d suggest a quantum mechanics book, like the one by Griffiths!
Awesome
Thanks!
Typo at 2:35
Yes, pi/2 should be pi!
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.
Needlessly complicated explanation
Me too😅
Ow
😮
@@PrettyMuchPhysics hey can you tell me dirac potential in quantum well is a perturbations??
Do you mean a delta potential? If yes, then you don’t need perturbation theory to solve the corresponding Schrödinger equation!
@@PrettyMuchPhysics yeah delta potential ok I got it I don't need perturbation theory but is it perturbed state?and can I solve it with perturbation theory?