It is nearly half a century since I studied qm. I wrestled with solutions to Schrodingers equation to get similar results. Now Dirac creates qm virtually out of nothing. A bit like the way something comes of nothing when that nothing is the quantum vacuum, in fact. Truly amazing.
After years of learning bits and pieces of QM from various sources, this lecture of yours and the 3 that immediately preceded it is what peeled away the final layer of the mathematics and the conventions to reveal the true understanding of the core of this incredible theory. You truly have a gift, thank you.
Excellent video. This is the first time that I have understood how deep Heisenberg's Uncertainty Principle goes. That last 1 minute in the video was worth its weight in gold. For a while, I thought this was a measurement/accuracy problem. That was killed off when I more fully understood the interaction problem. Now that has been killed off with the lack of common eigenvectors. I finally get it. You are an excellent teacher. You deserve a raise.
What have you made is just one of the best structured lectures i have ever attended. This is the way, that from now on knowledge will flow from one person to the other WITHOUT the INSTITUTIONS between. People like you Transform Internet to the most powerfull University , that is one that is everywhere and nowhere at the same time.
I think if you actually understood what your explaining then you must then Know there is a red rainbow and a white one and then all that makes perfect sense and Hermes explains it better and sums it up much better with AS ABOVE SO BELOW so why are you going on like some labottamy student
Hello sir, Your lectures are ultimate all these lectures have cleared my doubt on the quantum physics, specially lectures on bra and ket vectors, and Schrodinger equations. Thanks for shearing such informative lectures. thank you very much
thank you so so much sir! your youtube channel is an absolute treaser for me! never have come across a proffesor like who makes everything look so easy. im a huge fan of yours sir! i have listened to all of your songs! follow you on twitter! i would be so mighty pleased if you once again resume delivering such incredibly valuable lectures on physics !
That's right provided the probability for the two outcomes was the same. As I shall show in a new series on particle physics (for later in the autumn) you can have a 3 term eigenvector where the probabilities are not the same.
Wonderful, but brilliant, but fantastic. Excellent explanation, I guess I now have a better understanding of the formulation of quantum mechanics. Thank you.
I made the assumption that a) the prob of finding the electron beyond certain distances was zero (so as you say, it could only be found inside L) and b) that the electron was as likely to be at one point as any other (which of course might not be right). Psi* Psi then reps the prob of finding the electron at any given point and the sum of all probabilities must sum to 1.
Hello, Thanks for all your videos. I just have a remark at 34:05. You said for Psi: It is the linear superposition for all these points in the continuum. I though that Psi (in your formula) should be the eigenvecteur after the observation corresponding to the x eigenvalue observed by the detector (described by the operator X). Then, the linear superposition does not exist anymore in the Psi of your formula (which should be called Psi of x). Your definition is correct before the observation and the wave function collapse. Thanks.
I appreciate these videos, they are excelent, BUT what would really be great is either a PDF with problems and solutions or a recommendation for a companion book with problems and some solutions. Thanks
At approx 38:20 we drop duv/dx integral on the basis that it is zero at + - infinity, but earlier we established this integral had a value of 1 across L - whilst I understand why it is zero at infinity it is non-zero across the interval [- infinity to + infinity] : what have I misunderstood? Thanks for the videos - there is nothing to match them on the net.
Big thanks. I finally understood what the commutator is. I takes me a really long time to go through the lectures, because I have to keep replaying to catch the logic. Worth the effort, though.
I remembered how a university lecturer failed all his students in the quantum mechanics exam in a physics course at my university some 20 years ago. He said only three people in the world understood quantum mechanics (and that included him). It turned out that the reason of the 100% failure rate was because he didn't introduce the notion of what an operator was (it is a matrix) and didn't describe the notations.
Why if we are complex conjugating an operator we have to dagger it at 25:03 . Since we can simply complex conjugate it's elements without transposing the matrix.Please do reply sir.
It was not explained in the video. But for any vector: = when G is the transposed version of H. So in the case of the video next to conjugating H, the matrix have to be transposed -> Dagger
At 43:45 why did you need to have negative i added to the operator instead of adding positive i? It seems that it would still pass the Hermitian test without having the (-) sign, so why is it necessary?
At @28:56, regarding the position operator; would it be correct to say that in order for Psi to be a valid position state, Psi must be a function Psi : R --> C with the requirement that the integral of Psi^2 over R is 1 ? In other words, the space of possible position states is the collection of functions from R to C whose L2 norm evaluates to 1?
@Joanna Lada. It's because H in that situation represents a hermitian matrix. A matrix can be used as an operator but not all operators are matrices. The special type of matrix known as hermitian matrix has real elements along its diagonal and the off diagonal elements that are mirror positions of each other flip. Its diagonal elements are complex conjugates of themselves since they are real and the eigenvalues of hermitian matrices are thus always real numbers which is quite useful in solving problems in QM. So as he says at 24: 15 It's the transposed complex conjugate,, not just the complex conjugate.
At 26:00 the measurement of the particles position is treated as a similar probabilistic problem as what is derived from the photon polarisation and the electron spin experiments. Is there any experimental proof that the position of an electron is REALLY a probabilistic problem or is this generalization assumed to be true? If so, where is this assumption based on?
I'm hoping I've fully understood what you are asking: As for the electron, look up 'electron diffraction'. This wave behaviour of the electron means that in the time between observations, the possible positions of the electron get more spread out, leading to the probabilities. The probabilistic nature of position is not just limited to the electron though, for that, see 'wave-particle duality' and the de Broglie hypothesis. Hope this helps.
I'm not sure I understand ~25 mins in. Since *, inspecting H shows it's defined as [H]_ij=[H*]_ji, so if complex conjugated gives [H*]_ij=[H**]_ji=[H]_ji, then [H-dagger]_ji=[H]ji, so by the definitions the dagger is required for it all to fit?
Thank for excellent lecture. In derivation of Heisenberg's Uncertainty, for two operators \bar{x} \bar{p} acting on the state, you said that the operator \bar{x} simply returns x. Thus, "\bar{x} \bar{p} | \psi" reduces to x \bar{p} |\psi. In this case the position operator should act on the state \bar{p}|\psi. As a result, \bar{x} \bar{p} | \psi ot= x \bar{p}| \psi. Is this right?
The integral is indeed one. You are referring to UV|_(-inf)^(+inf). This asks the probability to find an electron at infinity and at minus infinity. Those probBilitiesare zero, but th UV|_(-inf)^(inf) term alone doesn't compute the probability to find electron anywhere.
The integral of ΨΨ* from -inf to inf is 1. If we are saying that ΨΨ*(inf) and ΨΨ*(-inf) are both 0 and that we know the electron is "almost certainly" somewhere in the room, then d(ΨΨ*) must be zero "almost everywhere". The integral of zero almost everywhere is zero. That's the best I can make of it, anyway
On seventh thought, it's simpler than that. What goes up must come down. In order for the value to start at zero, go up to some nonzero maximum, and back down to zero, it has to travel the same amount of distance in both directions. This means the derivative has to have just as much positiveness as negativeness. This cancels out in the integral.
Could the uncertainty we have between position and momentum represent the uncertainty we have in everyday life? We can set up an experiment in classical mechanics and within the isolated reference frame of the experiment we can predict the outcome every time. This is because the experiment is based on the laws of physics and may have been done many times before. But there will always be the uncertainty that something outside the isolated reference frame of the experiment will distort the outcome. For the nature of the universe is not based of isolated reference frames it is based on an interactive process that in this theory can be represented at the smallest scale by the mathematics of quantum mechanics in the form of Heisenberg’s Uncertainty Principle ∆×∆pᵪ≥h/4π.
This is a good introduction to representing momentum operators . Most beginners can benefit from this video. However, prior knowledge of Lie Groups is useful here.
Sometimes you can consider the way the ancients put it; "chieng' mane osudo kar chakruok". In the Nile valley, stellar trajectories were meant to start at x(0). This place was called "ka tchehen" in Metu Neter(Madhumne Ter). Thus x(t) = Q(t, x(0)). And what determined the trajectory was surprisingly known by their system of ODEs dy/dx = F (x(t)). I tell you, man. Nut ma ba ka (Ng'utmbaka)
The basis vectors (xi, xj, xk, ..., xn) are orthonormal - i.e. orthogonal and of unit length. An inner (dot) product of two vectors A & B is given by: ||A|| ||B|| cos(theta) where theta is the angle between the vectors. Because A and B are defined to be orthogonal: theta = 90degrees = pi/2 radians Which means the cosine of theta always equals 0: cos(90 degrees) = cos(pi/2 radians) = 0 the two basis vectors point in the same direction, in which case theta = 0 cos(theta) = cos(0) = 1 ||A|| ||B|| cos(theta) = ||A|| ||B|| if and only if A & B are colinear A further restriction is placed on A and B by making them of unit length - if they are colinear AND of unit length, they must be identical vectors, hence A = B (or xi = xj in your example)
If it's possible to have more than two probability terms in the summation, and the sum of the terms must equal one, shouldn't the number under the root of each probability be equal to the number of terms in the summation? For example, 1/root2 would be for two terms, 1/root3 for three terms, etc.? Sorry if this sounds like a stupid question.
Since ∫ΨΨ*dx=1. Just wondering if you were to graph this so the y axis was ΨΨ* and the x axis was just one spatial dimension for where an electron could be. Wouldn't it not be some normally distributed curve? I mean as you get further away the probability (ΨΨ*) would get lesser but this is the same value on the other side as you got further. So take x=5, ΨΨ* would be the same as x=-5.
+AlchemistOfNirnroot Yes you can plot the probability amplitude, Ψ, but it is more complicated than a normal distribution. For example, many textbooks will calculate and plot Ψ in one dimension just as you say, for a square potential well or a simple harmonic oscillator potential and other simple shapes within which a particle can be confined. Because Ψ takes the form of sines and cosines you get a probability distribution like sin^2 or cos^2 (depending on boundary conditions) which means there are peaks and troughs where the particle can be found when it's measured, or never found at all. If you look at the Ψ distributions for the electron position in a hydrogen atom (pretty complicated but still possible to write down the solution analytically without resorting to solving numerically on a computer) then you get all sorts of weird shapes in 3 dimensional space depending on the energy and angular momentum of the electron state - it is not simply spherical. The wavefunction is described beautifully by spherical harmonics and laguerre and legendre polynomial functions.
@@nickxjohnson ngl my maths has had a boost since - still can't do the spherical harmonics but have seen very good sources on legendre polynomials. Just wondering, what are the prereqs for spherical harmonics? I can do [some] linear algebra and vector calculus and some basic complex analysis (not really there but on the way, hopefully by next academic year) btw.
@@AlchemistOfNirnroot Just some calculus and complex numbers and you're there! They're very interesting because, even though they do crop up as the spatial solutions to the Schroedinger equation for a spherically symmetric potential (the hydrogen atom), they are useful all over physics particularly in the study of gravity and electromagnetism. Hope that helps.
@@nickxjohnson Yh, I've done MT1 and MT2 now from my Physics course and it covers elementart calc, complex numbers, matrices, vector calc, fourier series and transforms and it's really helped with this series. I've now made notes on it now; prob was dumb attempting this in GCSE but hey at least I learned I need to learn more to learn this :D Electromagnetism is hard, but manageable - we only use integrals where symmetry may be used (think this is standard for all 1st year modules).
0:31:00 You can't have a uniform probability distribution over an infinite set if the probabilities must add up to 1. In fact, the area Psi*PsiL is not the sum of the probabilities of all points. So, how can it make sense to say that the electron has the same probability to be found at any point? Wouldn't it be more accurate to say that it has the same probability to be found in any interval within L? Or what am I missing?
You're right, it is probability density rather than probability, so probability to be found in a certain interval divided throught the width of that interval.
is it more accurate to say that ' a quantum system is not allowed to have exact position and momentum at the same time' than 'the position and momentum of a quantum system cannot be measured at the same time'? The key is, is it really relevant to the measurement?
Hi DrPhysicsA, how did the two sigmas at 07:10 merge into one sigma, is it the general rule or a special case? So when Sigma i is multiplied by Sigma J, you just write Sigma i,j and multiply the insides? Thank you!
It's just a notational trick. Instead of writing two sigmas, we write one and everyone just assumes we mean all possible combinations of values of i and j. In the end people tend to drop the sigmas entirely because you get tired of writing them all the time
27:20. A single real number erects a Hilbert space. Here each possible valueof x acts as an orthogonal basis vector. This makes sense to me, but is not obvious at all. This transition deserves a paragraph of explanation.
You're saying the Heisenberg Uncertainty Principle (HUP) comes from the fact that the x and p operator don't share Eigenvectors. But isn't that itself derived from the HUP. In other words, you're explaining x with y but postulating that y depends on x. I am still struggling to understand the fundamental reason for the UHP. Can you help?
An other way to look at it is by establishing that something with a well defined momentum is characterized by a sine wave, which by definition has no location. As soon as you limit this wave to a certain range of positions, it will not be a pure sine anymore but a sum (or rather an integral) of an infinite number of waves with different frequencies, so representing different momentums. Hence the momentum will not be defined anymore, but will also be in a range. It's not so special. If you have a musical note of finite duration, it can never consist of just one frequency, it's spectrum is spread out as well. You can hear this with short bass notes. It's principally impossible to assign them an exact frequency. The effect is a "boom bass", an amorph drumlike sound with no clear frequency.
Thank you DrPhysicsA, I love listening to your lectures. They are very clear and understandable. Are there any good problems that goes along with these lectures? Going from here to Shankar is a bit of a leap. I wonder if there are good "in between" materials.
Because spatial frequency in x an y directions are independent. One can have a spread out sine in x direction (momentum) and a spatially non spread out pulse in y direction (location).
Just take the limit of the sum of interval width times times value of Psi in that interval. Making the interval smaller and smaller this sum will arbitrarily close approach a certain value. That's what integration is about. "Infinite" is not a number, it's a limit. And then the sum of interval times Psi can also be shown to approach limit, namely one.
@@wiztech6563 That would never give you the probability of finding the system at a particular point. My point is that you can have a probability density function but not a probability mass function. And you seem to agree since the operation you suggest involves a probability density function.
The part that confused me: You didn't define a position operator, and you calculated the commutator [x,p] by replacing the x position operator with x. But I understood x to be an eigenvalue of the operator rather than the operator itself.
The video contains a lot of useful information. Still working my way through it. Let me make a prior suggestion more precise. 0:31:40 "The sum of the probabilities of all the different points must add up to 1". Impossible for infinitely many points, as here. Maybe, we shouldn't speak of the probability of finding the particle at a point but within an interval. This is a density probability distribution; its being uniform means all equally long subintervals have equal probability, not all points.
Not understanding it is completely normal - it is deeply weird & IMO there are things no one can explain fully other than saying 'a wizard did it' eg Why do some electrons teleport through walls whilst other identical electrons do not? Why does mass curve space? Why do fast things have an increase in mass (or momentum) which means they can't go faster than c? Why does empty space has some resistance to formation of EM fields (known as permittivity & permeability) IMO permittivity slows light down because an electric field moves electron clouds & these electron clouds want to be around protons so they resist being moved i.e. they are like springs (& in a vacuum this happens with virtual particles) & springs take a bit of time to compress. Bigger permittivity = bigger springs = bigger time loss compressing the springs = lower speed. & permeability slows down light because highly permeable materials eg iron bend magnetic field lines a lot (unlike air which barely bends them) so permeabililty is an amplifier of magnetism & the excellent LC circuit hydraulic analogy of inductor = heavy paddle wheel means that light's magnetic field = a form of momentum. & an object with momentum takes longer to slow down than an object without momentum. In What is Light? DrPhysicsA expertly derives |E|/|M| = c so bigger magnetic field = slower c
These lectures are great, but they would be better if you could be sure the audience understood linear algebra. Better still if they knew about Lie Groups, Lie Algebras and representation theory.
Why do we need Lie groups and algebras here? Lecture presents all necessary concepts and there are no aces up a sleeve. It would be nice to add Hilbert and Hausdorff spaces, but that would bloat a lecture. Anyone interested in foundations of this formalism should start with secular equation in linear algebra.
+sndik16, Pretty much. Sigma_i (a) * Sigma_j (b) is really just a complicated way to write (a1 + a2 + ... + ai)(b1 + b2 + ... + bj). If you use the Distribution Property, and Sigma notation, you'll find you get Sigma_i (a) * Sigma_j (b) = Sigma_i (a*Sigma_j(b)). Once you understand that, the only thing left is to learn that Sigma_ij means i and j are both independent variables, so you have to go from 1 to i and 1 to j independently, so at some point, you'll have every combination of i and j, so Sigma_i (a) * Sigma_j (b) = Sigma_i (a*Sigma_j (b)) = Sigma_ij (ab), because all the terms will be identical.
i'm not going to say that QM is a complete waste of time....all this math struggle and weird witchy notation must have some use somewhere (beyond just generating a paycheck) .... but all of this mental jumping through hoops like a trained dog will not grow one grain of wheat or produce one gallon of gasoline or make one pound of wool which are all the things we need to live and survive.....QM is analogous to Tchaikovsky's The Nutcracker - all very good entertainment for czars and clever elite feudal russian oligarchs but you can't eat it or wear it or use it to keep the rain out ......anybody can learn QM if they want to memorize all the quirky notation and bizarre tortured math...in much the same way anybody can learn japanese if they want to memorize the kanji and train their vocal muscles to do somersaults...but at least learning japanese has some useful value in the mundane world..................yah yah yah QM gave us the transistor and the atomic bomb both of which have turned into curses upon mankind
PUH SYE sounds so pretentious and forced, especially when he pronounces it as the more common SYE every time his concentration is elsewhere. really really annoying.
It is nearly half a century since I studied qm. I wrestled with solutions to Schrodingers equation to get similar results. Now Dirac creates qm virtually out of nothing. A bit like the way something comes of nothing when that nothing is the quantum vacuum, in fact.
Truly amazing.
After years of learning bits and pieces of QM from various sources, this lecture of yours and the 3 that immediately preceded it is what peeled away the final layer of the mathematics and the conventions to reveal the true understanding of the core of this incredible theory.
You truly have a gift, thank you.
Excellent video. This is the first time that I have understood how deep Heisenberg's Uncertainty Principle goes. That last 1 minute in the video was worth its weight in gold. For a while, I thought this was a measurement/accuracy problem. That was killed off when I more fully understood the interaction problem. Now that has been killed off with the lack of common eigenvectors. I finally get it. You are an excellent teacher. You deserve a raise.
What have you made is just one of the best structured lectures i have ever attended. This is the way, that from now on knowledge will flow from one person to the other WITHOUT the INSTITUTIONS between. People like you Transform Internet to the most powerfull University , that is one that is everywhere and nowhere at the same time.
51:44
'If the system is in a state, that's the state it's in'
-- Dr.PhysicsA
What a brilliantly simple way of deriving the Heisenberg uncertainty principle from the mathematical formalism!
but can you cook on it
I think if you actually understood what your explaining then you must then Know there is a red rainbow and a white one and then all that makes perfect sense and Hermes explains it better and sums it up much better with AS ABOVE SO BELOW so why are you going on like some labottamy student
@@christemp1724 ???
Hello sir, Your lectures are ultimate all these lectures have cleared my doubt on the quantum physics, specially lectures on bra and ket vectors, and Schrodinger equations. Thanks for shearing such informative lectures. thank you very much
thank you so so much sir! your youtube channel is an absolute treaser for me! never have come across a proffesor like who makes everything look so easy. im a huge fan of yours sir! i have listened to all of your songs! follow you on twitter! i would be so mighty pleased if you once again resume delivering such incredibly valuable lectures on physics !
Good comment, but perhaps you mean: absolute treasure?
Thank you for these wonderfully clear and challenging videos. You have done a great service by making and sharing them.
That's right provided the probability for the two outcomes was the same. As I shall show in a new series on particle physics (for later in the autumn) you can have a 3 term eigenvector where the probabilities are not the same.
The internet + DrPhysicsA· = A wonderful time
That has gone 6 years ago😰😰😰
Yet another superb presentation. Really well explained. A pleasure to follow. Thank you DrPhysicsA. All your videos are excellent!
Love these videos, my school needs teachers like you.
Wonderful, but brilliant, but fantastic. Excellent explanation, I guess I now have a better understanding of the formulation of quantum mechanics. Thank you.
You're simply amazing, thank you infinitely.
I made the assumption that a) the prob of finding the electron beyond certain distances was zero (so as you say, it could only be found inside L) and b) that the electron was as likely to be at one point as any other (which of course might not be right). Psi* Psi then reps the prob of finding the electron at any given point and the sum of all probabilities must sum to 1.
That was truly mind blowing!
Many thanks for your lectures men. They are very nice prepare.
Brilliantly delivered. I'm hooked.
Hello,
Thanks for all your videos.
I just have a remark at 34:05.
You said for Psi: It is the linear superposition for all these points in the continuum.
I though that Psi (in your formula) should be the eigenvecteur after the observation
corresponding to the x eigenvalue observed by the detector (described by the operator X). Then, the linear superposition does not exist anymore in the Psi of your formula (which should be called Psi of x).
Your definition is correct before the observation and the wave function collapse.
Thanks.
Your videos are so wonderful. Thank you for providing such helpful information!
I appreciate these videos, they are excelent, BUT what would really be great is either a PDF with problems and solutions or a recommendation for a companion book with problems and some solutions. Thanks
yes. Did you find any book elsewhere?
At approx 38:20 we drop duv/dx integral on the basis that it is zero at + - infinity, but earlier we established this integral had a value of 1 across L - whilst I understand why it is zero at infinity it is non-zero across the interval [- infinity to + infinity] : what have I misunderstood? Thanks for the videos - there is nothing to match them on the net.
Let's consider this beast here... :D
Big thanks. I finally understood what the commutator is.
I takes me a really long time to go through the lectures, because I have to keep replaying to catch the logic. Worth the effort, though.
I remembered how a university lecturer failed all his students in the quantum mechanics exam in a physics course at my university some 20 years ago. He said only three people in the world understood quantum mechanics (and that included him). It turned out that the reason of the 100% failure rate was because he didn't introduce the notion of what an operator was (it is a matrix) and didn't describe the notations.
A matrix among others.....
Could anyone explain why at 25:03 when the complex conjugate is taken of we have to take the 'dagger' of H and not just complex conjugate it?
I like your videos and you at the same time. I always knew I was a commutable fella.
Is there an error at 5:58 of making b j complex conjugated? If you complex configured both you're adding an additional minus.
I was hoping you still had plans to do more on this subject like multiple kets |010> or quantum computing concepts.
Why if we are complex conjugating an operator we have to dagger it at 25:03 . Since we can simply complex conjugate it's elements without transposing the matrix.Please do reply sir.
It was not explained in the video. But for any vector: = when G is the transposed version of H. So in the case of the video next to conjugating H, the matrix have to be transposed -> Dagger
3:43 a "P" appeared on the whiteboard, now I am scared
At 43:45 why did you need to have negative i added to the operator instead of adding positive i? It seems that it would still pass the Hermitian test without having the (-) sign, so why is it necessary?
at that case the wavefunction would be e^-ikx, not e^ikx.
At @28:56, regarding the position operator; would it be correct to say that in order for Psi to be a valid position state, Psi must be a function Psi : R --> C with the requirement that the integral of Psi^2 over R is 1 ? In other words, the space of possible position states is the collection of functions from R to C whose L2 norm evaluates to 1?
Then @29:55, you assert that the wave functions (Psi) of a laboratory electron must have compact support. Yes?
@Joanna Lada. It's because H in that situation represents a hermitian matrix. A matrix can be used as an operator but not all operators are matrices. The special type of matrix known as hermitian matrix has real elements along its diagonal and the off diagonal elements that are mirror positions of each other flip. Its diagonal elements are complex conjugates of themselves since they are real and the eigenvalues of hermitian matrices are thus always real numbers which is quite useful in solving problems in QM. So as he says at 24: 15 It's the transposed complex conjugate,, not just the complex conjugate.
At 26:00 the measurement of the particles position is treated as a similar probabilistic problem as what is derived from the photon polarisation and the electron spin experiments. Is there any experimental proof that the position of an electron is REALLY a probabilistic problem or is this generalization assumed to be true? If so, where is this assumption based on?
I'm hoping I've fully understood what you are asking: As for the electron, look up 'electron diffraction'. This wave behaviour of the electron means that in the time between observations, the possible positions of the electron get more spread out, leading to the probabilities. The probabilistic nature of position is not just limited to the electron though, for that, see 'wave-particle duality' and the de Broglie hypothesis. Hope this helps.
I'm not sure I understand ~25 mins in.
Since *, inspecting H shows it's defined as [H]_ij=[H*]_ji, so if complex conjugated gives [H*]_ij=[H**]_ji=[H]_ji, then [H-dagger]_ji=[H]ji, so by the definitions the dagger is required for it all to fit?
Thank for excellent lecture. In derivation of Heisenberg's Uncertainty, for two operators \bar{x} \bar{p} acting on the state, you said that the operator \bar{x} simply returns x. Thus, "\bar{x} \bar{p} | \psi" reduces to x \bar{p} |\psi.
In this case the position operator should act on the state \bar{p}|\psi. As a result,
\bar{x} \bar{p} | \psi
ot= x \bar{p}| \psi. Is this right?
at 49:16 Why can't we take a derivative with respect to p and get the position operator in the correct form: x_hat = +i h_bar d/dp
at 38:42, why is integral of d(ΨΨ*)/dx from -infinity to infinity not 1 but 0?
The integral is indeed one. You are referring to UV|_(-inf)^(+inf). This asks the probability to find an electron at infinity and at minus infinity. Those probBilitiesare zero, but th UV|_(-inf)^(inf) term alone doesn't compute the probability to find electron anywhere.
The integral of ΨΨ* from -inf to inf is 1. If we are saying that ΨΨ*(inf) and ΨΨ*(-inf) are both 0 and that we know the electron is "almost certainly" somewhere in the room, then d(ΨΨ*) must be zero "almost everywhere". The integral of zero almost everywhere is zero. That's the best I can make of it, anyway
On seventh thought, it's simpler than that. What goes up must come down. In order for the value to start at zero, go up to some nonzero maximum, and back down to zero, it has to travel the same amount of distance in both directions. This means the derivative has to have just as much positiveness as negativeness. This cancels out in the integral.
thank you dr physics
46:43 "In this context, k of course is 2pi/lamb". Can anyone explain why ?
Largely definitional. k is defined as 2pi/lambda.
+157 239n "k" is called the wave number because it tells you the number of waves in a full(2 pi) circle
But sir you chose the letter k arbitrarily . I think it was just the eigenvalue when we acted k(operator) on psi
BRILLIANT!
Thanks!
Could the uncertainty we have between position and momentum represent the uncertainty we have in everyday life? We can set up an experiment in classical mechanics and within the isolated reference frame of the experiment we can predict the outcome every time. This is because the experiment is based on the laws of physics and may have been done many times before. But there will always be the uncertainty that something outside the isolated reference frame of the experiment will distort the outcome. For the nature of the universe is not based of isolated reference frames it is based on an interactive process that in this theory can be represented at the smallest scale by the mathematics of quantum mechanics in the form of Heisenberg’s Uncertainty Principle ∆×∆pᵪ≥h/4π.
The 'P' in 'Psi' is finally getting the credit it deserves 😁
Very helpful. Thanks!
Better explanation of the outer product as an identity operator than my 600+ page textbook offered. Any textbooks in the cards?
This is a good introduction to representing momentum operators . Most beginners can benefit from this video. However, prior knowledge of Lie Groups is useful here.
What is a Lie group? A group of all "move from (0,0) to (x,y)"?
Sometimes you can consider the way the ancients put it; "chieng' mane osudo kar chakruok". In the Nile valley, stellar trajectories were meant to start at x(0). This place was called "ka tchehen" in Metu Neter(Madhumne Ter). Thus x(t) = Q(t, x(0)). And what determined the trajectory was surprisingly known by their system of ODEs dy/dx = F (x(t)). I tell you, man. Nut ma ba ka (Ng'utmbaka)
I didn't watch all of th others, actually only the first, but why is the inproduct only one when Xi and Xj are the same? around 12:30
The basis vectors (xi, xj, xk, ..., xn) are orthonormal - i.e. orthogonal and of unit length. An inner (dot) product of two vectors A & B is given by:
||A|| ||B|| cos(theta)
where theta is the angle between the vectors.
Because A and B are defined to be orthogonal:
theta = 90degrees = pi/2 radians
Which means the cosine of theta always equals 0:
cos(90 degrees) = cos(pi/2 radians) = 0
the two basis vectors point in the same direction, in which case theta = 0
cos(theta) = cos(0) = 1
||A|| ||B|| cos(theta) = ||A|| ||B|| if and only if A & B are colinear
A further restriction is placed on A and B by making them of unit length - if they are colinear AND of unit length, they must be identical vectors, hence A = B (or xi = xj in your example)
Len Dacruz thanks, man!
Thank you sir for this video
If it's possible to have more than two probability terms in the summation, and the sum of the terms must equal one, shouldn't the number under the root of each probability be equal to the number of terms in the summation? For example, 1/root2 would be for two terms, 1/root3 for three terms, etc.? Sorry if this sounds like a stupid question.
Since ∫ΨΨ*dx=1. Just wondering if you were to graph this so the y axis was ΨΨ* and the x axis was just one spatial dimension for where an electron could be. Wouldn't it not be some normally distributed curve? I mean as you get further away the probability (ΨΨ*) would get lesser but this is the same value on the other side as you got further. So take x=5, ΨΨ* would be the same as x=-5.
+AlchemistOfNirnroot Yes you can plot the probability amplitude, Ψ, but it is more complicated than a normal distribution. For example, many textbooks will calculate and plot Ψ in one dimension just as you say, for a square potential well or a simple harmonic oscillator potential and other simple shapes within which a particle can be confined. Because Ψ takes the form of sines and cosines you get a probability distribution like sin^2 or cos^2 (depending on boundary conditions) which means there are peaks and troughs where the particle can be found when it's measured, or never found at all. If you look at the Ψ distributions for the electron position in a hydrogen atom (pretty complicated but still possible to write down the solution analytically without resorting to solving numerically on a computer) then you get all sorts of weird shapes in 3 dimensional space depending on the energy and angular momentum of the electron state - it is not simply spherical. The wavefunction is described beautifully by spherical harmonics and laguerre and legendre polynomial functions.
@@nickxjohnson ngl my maths has had a boost since - still can't do the spherical harmonics but have seen very good sources on legendre polynomials. Just wondering, what are the prereqs for spherical harmonics? I can do [some] linear algebra and vector calculus and some basic complex analysis (not really there but on the way, hopefully by next academic year) btw.
@@AlchemistOfNirnroot Just some calculus and complex numbers and you're there! They're very interesting because, even though they do crop up as the spatial solutions to the Schroedinger equation for a spherically symmetric potential (the hydrogen atom), they are useful all over physics particularly in the study of gravity and electromagnetism. Hope that helps.
@@nickxjohnson Yh, I've done MT1 and MT2 now from my Physics course and it covers elementart calc, complex numbers, matrices, vector calc, fourier series and transforms and it's really helped with this series. I've now made notes on it now; prob was dumb attempting this in GCSE but hey at least I learned I need to learn more to learn this :D Electromagnetism is hard, but manageable - we only use integrals where symmetry may be used (think this is standard for all 1st year modules).
0:31:00 You can't have a uniform probability distribution over an infinite set if the probabilities must add up to 1. In fact, the area Psi*PsiL is not the sum of the probabilities of all points. So, how can it make sense to say that the electron has the same probability to be found at any point? Wouldn't it be more accurate to say that it has the same probability to be found in any interval within L? Or what am I missing?
You're right, it is probability density rather than probability, so probability to be found in a certain interval divided throught the width of that interval.
@24:20 I think you want to not just complex conjugate it* but you want to conjugate transpose it ⟨ϴ|H|Ψ⟩† (aka Hermitian conjugate)
At 55:00, you say " And the commutator will not be zero". You mean "will be zero", right? DrPhysicsA
Oh, got it now. Nevermind.
is it more accurate to say that ' a quantum system is not allowed to have exact position and momentum at the same time' than 'the position and momentum of a quantum system cannot be measured at the same time'? The key is, is it really relevant to the measurement?
Hi DrPhysicsA, how did the two sigmas at 07:10 merge into one sigma, is it the general rule or a special case?
So when Sigma i is multiplied by Sigma J, you just write Sigma i,j and multiply the insides?
Thank you!
It's just a notational trick. Instead of writing two sigmas, we write one and everyone just assumes we mean all possible combinations of values of i and j.
In the end people tend to drop the sigmas entirely because you get tired of writing them all the time
Sir thank u very much
very good, thanks!
Thanks!!!
27:20. A single real number erects a Hilbert space.
Here each possible valueof x acts as an orthogonal basis vector. This makes sense to me, but is not obvious at all. This transition deserves a paragraph of explanation.
I am working my way through these as a keen amateur
I am working throuth this as someone who did QM modelling of organic molecules for years.
lovely!
You're saying the Heisenberg Uncertainty Principle (HUP) comes from the fact that the x and p operator don't share Eigenvectors. But isn't that itself derived from the HUP. In other words, you're explaining x with y but postulating that y depends on x. I am still struggling to understand the fundamental reason for the UHP. Can you help?
An other way to look at it is by establishing that something with a well defined momentum is characterized by a sine wave, which by definition has no location. As soon as you limit this wave to a certain range of positions, it will not be a pure sine anymore but a sum (or rather an integral) of an infinite number of waves with different frequencies, so representing different momentums. Hence the momentum will not be defined anymore, but will also be in a range. It's not so special. If you have a musical note of finite duration, it can never consist of just one frequency, it's spectrum is spread out as well. You can hear this with short bass notes. It's principally impossible to assign them an exact frequency. The effect is a "boom bass", an amorph drumlike sound with no clear frequency.
Thank you DrPhysicsA, I love listening to your lectures. They are very clear and understandable. Are there any good problems that goes along with these lectures? Going from here to Shankar is a bit of a leap. I wonder if there are good "in between" materials.
at the end of this video, why if does not apply to the measurement of the x and y coordinate of momentum?
Because spatial frequency in x an y directions are independent. One can have a spread out sine in x direction (momentum) and a spatially non spread out pulse in y direction (location).
brillant
There are uncountably many points in any interval. What number Psi*Psi times the cardinality of the continuum equals 1?
Just take the limit of the sum of interval width times times value of Psi in that interval. Making the interval smaller and smaller this sum will arbitrarily close approach a certain value. That's what integration is about. "Infinite" is not a number, it's a limit. And then the sum of interval times Psi can also be shown to approach limit, namely one.
@@wiztech6563 That would never give you the probability of finding the system at a particular point. My point is that you can have a probability density function but not a probability mass function. And you seem to agree since the operation you suggest involves a probability density function.
The part that confused me: You didn't define a position operator, and you calculated the commutator [x,p] by replacing the x position operator with x. But I understood x to be an eigenvalue of the operator rather than the operator itself.
TY very much
très bien
I barely know what a nuclear transformation is . What am I doing here ??
The video contains a lot of useful information. Still working my way through it. Let me make a prior suggestion more precise.
0:31:40 "The sum of the probabilities of all the different points must add up to 1".
Impossible for infinitely many points, as here.
Maybe, we shouldn't speak of the probability of finding the particle at a point but within an interval. This is a density probability distribution; its being uniform means all equally long subintervals have equal probability, not all points.
Envy you Armstrong.
drphysics = beast
***** Argonians should overthrow Tamriel!
@AlphaArgonian How ya feeling about TESVI? Not too sure after Fallout 4/76.
basis vectors are not by definition orthonormal, nor are they orthogonal, basis vectors by definition are only linearly independent.
sorry nor are they "by definition" orthogonal.
You can take any set of basis vectors and turn it into a set of orthonormal basis vectors fairly easily, so I wouldn't worry too much about that
R.I.P DrPhysiscsA
Is it weird I like listening to quantum physics but yet understand 0% of it lol?
Not understanding it is completely normal - it is deeply weird & IMO there are things no one can explain fully other than saying 'a wizard did it' eg
Why do some electrons teleport through walls whilst other identical electrons do not?
Why does mass curve space?
Why do fast things have an increase in mass (or momentum) which means they can't go faster than c?
Why does empty space has some resistance to formation of EM fields (known as permittivity & permeability)
IMO permittivity slows light down because an electric field moves electron clouds & these electron clouds want to be around protons so they resist being moved i.e. they are like springs (& in a vacuum this happens with virtual particles) & springs take a bit of time to compress. Bigger permittivity = bigger springs = bigger time loss compressing the springs = lower speed.
& permeability slows down light because highly permeable materials eg iron bend magnetic field lines a lot (unlike air which barely bends them) so permeabililty is an amplifier of magnetism & the excellent LC circuit hydraulic analogy of inductor = heavy paddle wheel means that light's magnetic field = a form of momentum. & an object with momentum takes longer to slow down than an object without momentum.
In What is Light? DrPhysicsA expertly derives |E|/|M| = c so bigger magnetic field = slower c
These lectures are great, but they would be better if you could be sure the audience understood linear algebra. Better still if they knew about Lie Groups, Lie Algebras and representation theory.
Why do we need Lie groups and algebras here? Lecture presents all necessary concepts and there are no aces up a sleeve. It would be nice to add Hilbert and Hausdorff spaces, but that would bloat a lecture. Anyone interested in foundations of this formalism should start with secular equation in linear algebra.
+sndik16, Pretty much. Sigma_i (a) * Sigma_j (b) is really just a complicated way to write (a1 + a2 + ... + ai)(b1 + b2 + ... + bj). If you use the Distribution Property, and Sigma notation, you'll find you get Sigma_i (a) * Sigma_j (b) = Sigma_i (a*Sigma_j(b)). Once you understand that, the only thing left is to learn that Sigma_ij means i and j are both independent variables, so you have to go from 1 to i and 1 to j independently, so at some point, you'll have every combination of i and j, so Sigma_i (a) * Sigma_j (b) = Sigma_i (a*Sigma_j (b)) = Sigma_ij (ab), because all the terms will be identical.
i lolled when you said bj, still a good video though
Note to Self: Finish Later
5:34 hah bj
First!
hahah he said 'bj'
i'm not going to say that QM is a complete waste of time....all this math struggle and weird witchy notation must have some use somewhere (beyond just generating a paycheck) .... but all of this mental jumping through hoops like a trained dog will not grow one grain of wheat or produce one gallon of gasoline or make one pound of wool which are all the things we need to live and survive.....QM is analogous to Tchaikovsky's The Nutcracker - all very good entertainment for czars and clever elite feudal russian oligarchs but you can't eat it or wear it or use it to keep the rain out ......anybody can learn QM if they want to memorize all the quirky notation and bizarre tortured math...in much the same way anybody can learn japanese if they want to memorize the kanji and train their vocal muscles to do somersaults...but at least learning japanese has some useful value in the mundane world..................yah yah yah QM gave us the transistor and the atomic bomb both of which have turned into curses upon mankind
if you know all this stuff why don't you try to create some theories with it
+Anthony Shuey like quantum mechanics XD?
+Anthony Shuey Yeah sounds easy as fuck.Just create a theory, I do that every day.
Nice videos, but you tell everything a hundred times..it's annoying.
PUH SYE sounds so pretentious and forced, especially when he pronounces it as the more common SYE every time his concentration is elsewhere. really really annoying.
meditate on love & u won't have this problem
nonsensical theory...how many Nobel prizes in vain were given for this delusion?...and everything is happening in the imaginary world
Sounds like someone didn't quite understand the theory. :)
BRILLIANT!