Andrew Dotson How long did it take you to get to this point in your understanding of physics? My question comes from the fact that I am 27, but only recently decided to change directions in life to push towards my true passion of majoring in physics and seeing where it takes me. Anyways, just curious man. Thanks in advance.
@@K24_ej1 Romans, I can't speak for Andrew, obviously, but I can briefly tell you my experience. I decided to make a career change when I was 28. I was in a dead end job with no prospects, so I decided to start working on a Physics degree at my local University as a part time student (2 classes per semester). Bear in mind that I already had a degree (art related), so I did not need to take my General Education classes again, just the math and science courses. I kind of had to reshape my entire life. I needed to keep a full time job to pay the bills, so I applied for and finally found a full-time clerical position at the school I was attending. I was able to get a tuition discount from the school since I was an employee, use my lunch break for classes, etc. That was about 4 years ago. I'm almost 32 now. I graduate this coming May. I have a bunch of applications out right now to graduate schools to hopefully start my PhD work in the Fall. If you are dedicated, persistent, and perhaps a bit obsessive compulsive than anyone can learn the material. It's the whole "4 years of your life" thing that will stop most people. It's been a major undertaking, but worth every second for me, especially since I can finally see the light at the end of the tunnel. The material he is covering here is, truth be told, not overly complex. It's certainly not beginner material, but it's not the most complicated either. Good luck with whatever you decide.
Lyle Arnett, Jr Thank you for the reply Lyle. That definitely gives me a boost in confidence since we seem to be (were) living in the same boat. Dead end job. GE our of the way in college etc. I’ll take note of the things you said. I do appreciate it. 🙏🏻 thanks so much.
C'mon now, Schrödinger still did a very useful thing there! Just as we don't need even Special Relativity to do a lot of physics, we can still do a lot of QM with Schrödinger's Equation. It gives us quantized energy levels for confined particles, hydrogen wave functions, quantum tunneling, etc., etc. Fred
Glad to have stumbled upon your channel. Physics is really all about asking yourself why something is wrong/ right and actually calculating the answer by yourself (and of course, coming up with a hand waving argument before ).
5:06 This is why we often express vectors of the holonomic basis as ∂i , since there is really an equivalence between vectors and directional derivatives as well as how vectors and partial derivatives transform. In good old flat space this is just a curiosity, but this is crucial in Riemannian geometry where you want to define a tangent space without an embedding. The tangent space can be defined as some abstract vector space made out of directional derivative operators.
The only explanation my lecturers give us was SE doesn’t treat spacetime on equal footing, KG and Dirac does Nice to see some more detailed explanations
Nice! Another pre-maths observation: The Schrodinger equation (for a free particle) has a fixed mass value, which is also a relativistic red flag. Another decent exercise: Showing the Schrodinger equation is invariant under Galilean transformations - this also lets you see that mass defines a superselection variable in nonrelativistic QM ('Bargmann mass superselection rule').
Math is literally a language, and so is logic. Look up formal systems and formal languages for more information. English is a natural language (along with the rest of the spoken languages).
math, logic, geometrie are per definition languages. All Equations are sentences about quatitative or countable relations of individual elements of a defined universal set and so on. They all have the same structure as normal languages, they have a syntax, a semantics and a truth value and the only thing which can be true is a sentence (ok in normal language its utterances, which include context and sentence, because in normal languages there is also a context principle)
Instead of working out the coordinate transformation, one can glance briefly at the equation itself. It expresses the non-relativistic relation between total, kinetic and potential energy in the operator form required by Quantum Mechanics: E = p^2/2m + V. This is essentially different from the relation in Special Relativity. Done.
I like the initial explanation, and I love the fact that there's so much I'll get the chance to learn in some time to understand the following part. Greetings!
Extraordinarily clear explanation of the mathematics of both Lorentz transforms and how to apply it to an equation involving space and time. Absolutely fantastic. So many folks talking about these aspects of physics have never done the math. It is abundantly clear that you have, likely many times over.
My god this looks and sounds so tough. I going through high school, studying mathematics and physics and I'm about to apply for a university and try to get my masters in medical physics. Feels like I have a long way to go when I compare my high school material to this. Oh well.. I'm not afraid of hard work.
If you say that the Schrödinger equation is i d/dt [e, t> = H [e, t> it totally works for a special relativistic hamiltonian. The familiar wave equation form of the Schrödinger equation is just it's representation in Position Basis with THE non-relativistic Hamiltonian: H=p^2/2m + V(x) In the case of a free relativistic Scalar particle: H =1/2m^2 * (pI * pI + m^2) where pI denotes the I-th element of Momentum, where I=2 or 3 This is so called lightcone quantization Add a potential V(x) and you're good to go. (you then obviously have a non-free Scalar particle)
Thank you Andrew for showing the rigorous calculation . I like to point out that without going through the long functional derivative part for the 2nd order term you can just square it. (d/dx)^2 = (rhs)^2. That will give the same result and a big time saver.
Schroedinger started with relativistic equation, the calculations following this approach were inconsistent with experimental results of that time, so he did a non-relativistic approximation, which is exactly the Schroedinger equation we know. When the experiments became more precise, the need for relativistic approach became obvious.
I was always confused when I heard that the Schrödinger equation was non-relativistic, when I knew you could just sub in a relativistic Hamiltonian and it just works (being identical to the Dirac equation); I didn't realise that 'the Schrödinger equation' and 'the general Schrödinger equation' were considered distinct, because I'd always only been taught the latter i.e. id/dt=H. I guess this makes sense now.
With that particular notation it has two nice uses: it tells you in a millisecond that the equation is Lorentz invariant and makes finding the Green's function extremely simple. It's a lot easier to remember the following: 1) No free Lorentz indices means Lorentz invariance 2) box +m^2 -> 1/(w^2-p^2-m^2) in momentum space. This is just a Fourier transform. Why write more down then you have to? General relativity can get tedious enough to work with using index notation/ differential forms. It would be an absolute nightmare without it.
I am proud of myself that I understood almost 96% of what was presented. I am a Physics major with minor in Mathematics. I still have more work to do. Kudos for the brilliant explanation.
Me too...🎉. I am really proud of myself. I did have to rewind on some parts. But I got it. Thank you Andrew. One thing I did not realize until now was how we could apply all the mathematical physics stuff into every derivation and understand that these are correlated. I am so inspired.🎉🎉🎉
There was a mistake in your proof. You assumed that the wavefunction is a scalar -- it is not. Let's take psy to be a Gaussian which, at it's peak, has the value p. In a primed frame the value at the peak cannot be the same as p because if you Lorentz contract a function that preserves probability, the peak will be larger. Taking this into account does not save the Schrodinger's equation from being non-relativistic, but people should be aware of this argument.
It's worth noting that not only is the Schrodinger restricted to the non-relativistic domain, but it is also restricted to the description of electrons in a spin eigenstate! The Pauli equation is a non-relativistic approximation of the Dirac equation that reduces to the Schrodinger equation when in an eigenstate of spin.
@@AndrewDotsonvideos formulations in applications is algorithms. I am not smart enough to follow formulations, always. Takes me long time to get to the applications, then guess the algorithms and then look at the formulations and definitions of the universes and multiverses. The attempted long term goals of the algorithms is to change the definitions.
Water can't be wet. Wetness is a property of something with water on it. If water has water on it there is just more water, not wet water. You inherently need a second object in which to frame the state of wetness. So a bowl with water in it could be described as wet. But a globe of water floating in the vacuum of space is just water.
Hi Andrew! Regarding the Schrodinger work and the process he used to find his equation it may be useful to read his article on Physical Review n.6 Vol.28 December, 1926. In pragraph n.10 he explains why he didn’t include a relativistic analisis.
I think you can directly square the partial x differentiation operator? It'll be easier this way, since we don't need to apply the definiation of it again.
On the same line of thought you will see that gallelian transformation doesn't keep SE invariant. But actually it does. While proving such invariance we can't simply take psi going to psi prime.
Hey Andrew. Here's my main confusion: it looks like the Schrodinger Equation isn't even invariant under Galilean transformations! In particular, looking at the final equation you wrote, taking the limits as v/c approaches zero and gamma approaches 1, there still remains the rightmost term of the right side of the equation. Any ideas on why this lingering term appears?
Ohhh I've figured it out. That "lingering" term actually needs to be there! (tldr: the Schrodinger equation is in fact invariant under Galilean transformations.) In taking the partial with respect to t', we're taking x' to be constant, not x. But in verifying the validity of the transformation, we have to start with the non-transformed Schrodinger equation, which originally uses psi(x,t) not psi(x',t'). In taking the partial of psi(x,t) with respect to t' (keeping x' constant, not x!), there is an extra term v(delPsi/delx) which is exactly what is cancelled out by the "lingering" term.
Watching this I'm so proud of you bro, you understand this shit like it's second nature. And I know that took crazy effort. Will meet you at the top someday
I've got a real question. Love your vids, bringing some real physics on youtube, that's juste pure kindness and it is quiet satisfying to watch :) But in your video you said multiple time that "Schrödinger's eq. leads to wave equation" I'm not quite sure, this is what De Broglie said "stationary wave" but in fact all the Schrödinger's eq. solution's aren't wave, the solution are just "stationary state" without the proper form of a wave eq. Anyway that's what I've been told. I would to hear you on that one :) !
but lorentz transform is only for the special relativity right? should we work in functional form for the coordinate transforms, it would be more general i suppose.
Which brings up a question. Where is the line between relativistic and non relativistic. For example up to what limit does Schrodinger equation work? ie, what is the fastest speed an electron can go and Schrodinger still works?
What role would velocity "v" play if your final result? I know that v isn't well defined in QM, and that the Schrodinger equation doesn't use velocity. So how can you combine the v from the Lorentz transform with the Schrodinger equation?
The 'v' is the velocity of the moving frame of the observer, with respect to the frame of the previous observer who was previously viewing the system and describing it via the Schrodinger equation. Particle velocity is, indeed, ill-defined in QM, but the velocity of the frame is well-defined. If Schrodinger equation were a relativistic equation, it should have been of the same form, as viewed from any non-accelerating frame, which is the first principle of relativity. Now, from the second principle of relativity (speed of light is same as viewed from any non-accelerating frame), we obtain that the correct transformation law between the space-time coordinates is the Lorentz transformation. So, Schrodinger equation should have been invariant under these transformations if it wanted to pass the test of being relativistic. The whole point of the exercise was that it fails to do so. P.S. - I apologise if I am not able to explain it properly. In that I hope someone can do it better than me and to your satisfaction. All the best.
I have a feeling there are three types of people who watch these videos: 1. The people who actually understand 2. The people who don’t fully understand but want to do physics in the future 3. The people who don’t understand and don’t know why they are here
I don't understand why the chain rule is used. My understanding is that it's used to solve composite functions, but i only did Calc one so this level of math is a bit beyond me...
Lovely, thanks. Can someone help me with one question? At the end of the video, Andrew replaces V(x) with V(x') and also Psi(x,t) with Psi(x' , t' ). Why did that happen? Given the Lorentz transformation relating x, x', t and t' we can show x = r (x' + vt' ) . Where r= gamma but I can't write greek symbols. Then, it would seem, V(x) is replaced by V( r(x' + vt') ) and not V(x' ). Similarly Psi (x, t) = Psi (r(x' + vt') , r(t' + (v/c^2)x' ) and not just Psi (x' , t' ). Is it essential that the wave function and potential function are (separately) invariant functions if we were only interested in checking that the Schrodinger wave equation as a whole is preserved under a Lorentz transformation?
Completed this entire major in physics, but still not much info is taught about equations being general relativity or special relativity besides the obvious such as Maxwell's equations and the E=mc2 equations. But I think Schrodinger's equation describes only the wave and simply gives a probability for where the particles is but does not tell the speed of the particle, just its energy and mass are involved. There is particle wave duality and Schrodinger equation focuses on the wave part.
Great video Andrew! One question though: Since v(x,t) = \frac{ \partial x}{ \partial t }, should't you also differentiate the v^2 terms when you Lorentz transform? Cheers, Henrik
No, he doesn't differentiate them, because v is constant, so that makes the whole γ factor a constant. Because v is the velocity of the moving inertial reference frame S'. If v was a function of t that would mean, that S' is not inertial, because it would have acceleration.
Another thing i stumbled across is, that when you take the stationary SE and lets say you have an elektron which 1. is in no force field and therefore has no potential energy And 2. has no kinetic energy. Then the resulting energy should be m*c^2, but the stationary SE says its 0. Is this correct??
I'm wondering if we should interpret this as the Schrodinger Hamiltonian is invalid at high energies, or that the energy is observer-frame dependent...Watching your Klein Gordon derivation next, thanks for the great videos.
Question:If entanglement is a defining aspect of QM,why would you expect relativistic terms in equations describing it. Isn't that essentially why folks like the whole " The universe is a hologram " idea? Doesn't entanglement mean there is no spacetime at the am level????
I am glad to see that some American youth are not destroying their brains with music and drugs but are doing something constructive, productive and illuminating.
Schrödinger was looking after a relativistic equation, like Klein-Gordon, but couldn't find one. Then he let down his work and went on vacation. On return, he deemed that a non relativistic equation was "good enough." And even today, we use the Schrödinger equation because it encompasses all of the features of quantum mechanics, the Dirac equation never really worked correctly, and it is still good enough.
Shouldn't you consider a multiplicative phase onto the wavefunction? When showing the Schrödinger equation is invariant under guage transformations for an electromagnetic potential we pick up such a factor in the wavefunction!
How the transformation coefficients can be figured out just like as if lorrentz transformation is an orthogonal transformation? I only know that these coefficients can be represented in derivative form if the transformation between two coordinates is an orthogonal transformation.
I attempted the homework but I ended up with half a page of nonsense. I took second partial derivatives of both time and space under the Lorentz transform but ended up with half a page of jargon when I plugged the transformed derivatives into the Klein Gordon Eq. I couldn't get the derivatives to cancel each other out. I'm not sure if I made an error in the algebra (which is to be expected because I had to keep track of so many derivatives) or If my error came from incorrectly using the chain rule or miscalculating a transformation coefficient. I really want to solve the challenge but am stuck with a headache-inducing mess. Could you possibly help me? UPDATE: I am making progress. It looks much better than it did before, however, there is an annoying v^2/c^2 term I can't get rid of.
I think the we should get psi'(x',t')=psi(x,t)(1-v²/c²)^(¼) to make the probability in the same region to be equal in both frames. Because probability should be invariant. And this psi' satisfies Schroedinger equation in psi does. By the way I got that equation you wrote but d²/dx'dt' terms vanished. I think I did some calculation mistake.
Because you aren’t taking the partial derivative with respect to it. For example if taking the partial derivative with respect to x, y would be treated as a constant for the same reason
It's defined as d2/(cdt)2 - d2/dx2 - d2/dy2 - d2/dz2, so you can understand it as an extrapolation of the Laplacian operator to the Minkowski space-time (if you try to transform this operator you'll see that it remains invariant under Lorentz transformations (aka it is a scalar in the Minkowski spacetime). That's why it can appear in Relativistic equations like the wave equation (it respects Lorentz's Covariance).
A correct submission has been made!
Andrew Dotson How long did it take you to get to this point in your understanding of physics? My question comes from the fact that I am 27, but only recently decided to change directions in life to push towards my true passion of majoring in physics and seeing where it takes me. Anyways, just curious man. Thanks in advance.
@@K24_ej1 Romans, I can't speak for Andrew, obviously, but I can briefly tell you my experience. I decided to make a career change when I was 28. I was in a dead end job with no prospects, so I decided to start working on a Physics degree at my local University as a part time student (2 classes per semester). Bear in mind that I already had a degree (art related), so I did not need to take my General Education classes again, just the math and science courses. I kind of had to reshape my entire life. I needed to keep a full time job to pay the bills, so I applied for and finally found a full-time clerical position at the school I was attending. I was able to get a tuition discount from the school since I was an employee, use my lunch break for classes, etc. That was about 4 years ago. I'm almost 32 now. I graduate this coming May. I have a bunch of applications out right now to graduate schools to hopefully start my PhD work in the Fall. If you are dedicated, persistent, and perhaps a bit obsessive compulsive than anyone can learn the material. It's the whole "4 years of your life" thing that will stop most people. It's been a major undertaking, but worth every second for me, especially since I can finally see the light at the end of the tunnel. The material he is covering here is, truth be told, not overly complex. It's certainly not beginner material, but it's not the most complicated either. Good luck with whatever you decide.
Lyle Arnett, Jr Thank you for the reply Lyle. That definitely gives me a boost in confidence since we seem to be (were) living in the same boat. Dead end job. GE our of the way in college etc. I’ll take note of the things you said. I do appreciate it. 🙏🏻 thanks so much.
Lyle Arnett, Jr What kind of art did you study, and why did it lead you to a dead end job? Asking 5 a friend
@@K24_ej1 You'll do well, man. Go on!
Lvl 1 crook : Schrodinger
Lvl 100 Boss : Dirac
*THAT'S HOW QUANTUM MECHANICS WORKS*
Mike Warrecker that’s how mafia works
What does that make Feynman?
C'mon now, Schrödinger still did a very useful thing there!
Just as we don't need even Special Relativity to do a lot of physics, we can still do a lot of QM with Schrödinger's Equation.
It gives us quantized energy levels for confined particles, hydrogen wave functions, quantum tunneling, etc., etc.
Fred
@@ffggddss Yes I understand, it's just a joke going on around here on the internet with lvl 1 crooks and lvl 100 boss.
@@jibran8410 OK, thanks for the warning. I'll be on the lookout for that ;-)
Fred
Huh? Did I just get assigned homework?
You are going to be an incredible lecturer and physicist.
Hey epic math time
Why am I watching Wolverine's lost brother talk about quantum mechanics at 1 am?
We may never know.
And after this video, I watched...ruclips.net/video/5wCldWohDIg/видео.html
Do you hype yourself up before saying, “What’s going on smart people”? It’s always on point.
I'm just naturally hype af
The first time a youtube recommendation has actually been good. Subscribed!
Malcolm Hall awesome!
@@AndrewDotsonvideos 😭😂😂
Glad to have stumbled upon your channel. Physics is really all about asking yourself why something is wrong/ right and actually calculating the answer by yourself (and of course, coming up with a hand waving argument before ).
Schrodinger did an OPPSIE!
But still hoodwinked a generation of mathematicians, physicists, sundry academics , politicians, and generals and made off like a bandit.
Both the schrodinger equation and relativity work... so what point is anyone trying to prove here?
@@woofle4830 but not together.
Ohok
@@Nebukanezzer exactly and with this demostration we define one of the problems scientists want to solve
5:06 This is why we often express vectors of the holonomic basis as ∂i , since there is really an equivalence between vectors and directional derivatives as well as how vectors and partial derivatives transform.
In good old flat space this is just a curiosity, but this is crucial in Riemannian geometry where you want to define a tangent space without an embedding. The tangent space can be defined as some abstract vector space made out of directional derivative operators.
The only explanation my lecturers give us was SE doesn’t treat spacetime on equal footing, KG and Dirac does
Nice to see some more detailed explanations
Nice! Another pre-maths observation: The Schrodinger equation (for a free particle) has a fixed mass value, which is also a relativistic red flag. Another decent exercise: Showing the Schrodinger equation is invariant under Galilean transformations - this also lets you see that mass defines a superselection variable in nonrelativistic QM ('Bargmann mass superselection rule').
I’m proud of myself for being able to keep up with all the calculus going on here
1:34 You know this is fucked up when even the Unicode missing letter square is a parameter on your equations
I swear physics should also count as a second language.🐈🤯
Math is literally a language, and so is logic. Look up formal systems and formal languages for more information. English is a natural language (along with the rest of the spoken languages).
@@ClayonTutorials I was going to say something similar. Physics is a nation, they speak calculus there.
math, logic, geometrie are per definition languages. All Equations are sentences about quatitative or countable relations of individual elements of a defined universal set and so on. They all have the same structure as normal languages, they have a syntax, a semantics and a truth value and the only thing which can be true is a sentence (ok in normal language its utterances, which include context and sentence, because in normal languages there is also a context principle)
@@ClayonTutorials True, but I'd rather order my food in any other language than math. 😊
@@narata1541 But you should always "order" your internet using math: ruclips.net/video/Z3IPVWN-1ks/видео.html.
Instead of working out the coordinate transformation, one can glance briefly at the equation itself. It expresses the non-relativistic relation between total, kinetic and potential energy in the operator form required by Quantum Mechanics: E = p^2/2m + V. This is essentially different from the relation in Special Relativity. Done.
I love coming back to these videos and understanding them a little more each time
In anime world, mentioning Schrodinger will make you smart
Grandioso encontrar un canal así
y con tanta naturalidad
I like the initial explanation, and I love the fact that there's so much I'll get the chance to learn in some time to understand the following part. Greetings!
Extraordinarily clear explanation of the mathematics of both Lorentz transforms and how to apply it to an equation involving space and time. Absolutely fantastic. So many folks talking about these aspects of physics have never done the math. It is abundantly clear that you have, likely many times over.
For my class that I TA'd for, we actually did that as an extra credit assignment...
My god this looks and sounds so tough. I going through high school, studying mathematics and physics and I'm about to apply for a university and try to get my masters in medical physics. Feels like I have a long way to go when I compare my high school material to this.
Oh well.. I'm not afraid of hard work.
Do not worry this calculation Is very easy, just standard derivation and substitution. You will understand It fully in a few years
doing quantum physics at uni RN, hyped to ask this question in class to make everyone roll their eyes, and for the lecturer to get slightly annoyed
If you say that the Schrödinger equation is i d/dt [e, t> = H [e, t> it totally works for a special relativistic hamiltonian.
The familiar wave equation form of the Schrödinger equation is just it's representation in Position Basis with THE non-relativistic Hamiltonian:
H=p^2/2m + V(x)
In the case of a free relativistic Scalar particle:
H =1/2m^2 * (pI * pI + m^2)
where pI denotes the I-th element of Momentum, where I=2 or 3
This is so called lightcone quantization
Add a potential V(x) and you're good to go. (you then obviously have a non-free Scalar particle)
Thank you Andrew for showing the rigorous calculation . I like to point out that without going through the long functional derivative part for the 2nd order term you can just square it. (d/dx)^2 = (rhs)^2. That will give the same result and a big time saver.
I don’t know what the hell is happening, and I’m not a physics major, but I continue to watch anyways.
Aha I love these videos. Tell me who doesn't want to watch that after waking up. Great video!!!
Schroedinger started with relativistic equation, the calculations following this approach were inconsistent with experimental results of that time, so he did a non-relativistic approximation, which is exactly the Schroedinger equation we know. When the experiments became more precise, the need for relativistic approach became obvious.
You could've made this way easier by approximating the whole thing to be 3.1
I was always confused when I heard that the Schrödinger equation was non-relativistic, when I knew you could just sub in a relativistic Hamiltonian and it just works (being identical to the Dirac equation); I didn't realise that 'the Schrödinger equation' and 'the general Schrödinger equation' were considered distinct, because I'd always only been taught the latter i.e. id/dt=H. I guess this makes sense now.
1:36 why physicists always love to forcefully shorten their equation by using alien symbols
Operators are used elsewhere too
It gives you a better understanding of what happening as you solve the physical problems.
With that particular notation it has two nice uses: it tells you in a millisecond that the equation is Lorentz invariant and makes finding the Green's function extremely simple.
It's a lot easier to remember the following:
1) No free Lorentz indices means Lorentz invariance
2) box +m^2 -> 1/(w^2-p^2-m^2) in momentum space. This is just a Fourier transform.
Why write more down then you have to? General relativity can get tedious enough to work with using index notation/ differential forms. It would be an absolute nightmare without it.
I am proud of myself that I understood almost 96% of what was presented. I am a Physics major with minor in Mathematics. I still have more work to do.
Kudos for the brilliant explanation.
Me too...🎉. I am really proud of myself. I did have to rewind on some parts. But I got it. Thank you Andrew.
One thing I did not realize until now was how we could apply all the mathematical physics stuff into every derivation and understand that these are correlated. I am so inspired.🎉🎉🎉
me to, I'm going 5th grade and I am proud of myself for finally getting what is going on after a lot of work.
There was a mistake in your proof. You assumed that the wavefunction is a scalar -- it is not. Let's take psy to be a Gaussian which, at it's peak, has the value p. In a primed frame the value at the peak cannot be the same as p because if you Lorentz contract a function that preserves probability, the peak will be larger. Taking this into account does not save the Schrodinger's equation from being non-relativistic, but people should be aware of this argument.
But the wave function is scalar. Complex, but scalar.
@@danielplacido8746 Nope. It changes under a Lorentz transformation.
@MetraMan09 For a constant velocity, gamma is a constant; it only varies with time when you have acceleration.
@@dXoverdteqprogress Wish I had seen this comment before I spent time writing my own.
It's worth noting that not only is the Schrodinger restricted to the non-relativistic domain, but it is also restricted to the description of electrons in a spin eigenstate! The Pauli equation is a non-relativistic approximation of the Dirac equation that reduces to the Schrodinger equation when in an eigenstate of spin.
I'm so dumb I literally don't understand any of this but I'm still here to support.
i just saw partial derivatives at university,im feeling really smart for understanding ate least 10% of this video
I wish I was able to understand this video so I could do the challenge at the end
What there anything specific that was unclear that I can work on?
@@AndrewDotsonvideos oh no, it's not your bad, just a bit of a lack of understanding on my part.
@@AndrewDotsonvideos formulations in applications is algorithms.
I am not smart enough to follow formulations, always.
Takes me long time to get to the applications, then guess the algorithms and then look at the formulations and definitions of the universes and multiverses.
The attempted long term goals of the algorithms is to change the definitions.
Water is wet. Change my mind.
you cant spell water without wet + ar because water ar wet
*Galaxy brain*
Water can't be wet. Wetness is a property of something with water on it. If water has water on it there is just more water, not wet water. You inherently need a second object in which to frame the state of wetness. So a bowl with water in it could be described as wet. But a globe of water floating in the vacuum of space is just water.
@@waveexistence5742 C-C-C-COMBO killed it lol
Amazing!😂
I like your videos. You definitely know your stuff. Also i like your shortcut videos an they way you do integrals & differentiating. Thank you.
I'm so happy I found this channel!!!
Love the medium level of rigour. I subscribed. Keep it up.
Hi Andrew! Regarding the Schrodinger work and the process he used to find his equation it may be useful to read his article on Physical Review n.6 Vol.28 December, 1926. In pragraph n.10 he explains why he didn’t include a relativistic analisis.
Which equations should we use for large systems of molecules?
Newtons. If wayy to big, then Einsteins GR.
Can I ask for a book to study tensors intuitively
I think you can directly square the partial x differentiation operator? It'll be easier this way, since we don't need to apply the definiation of it again.
On the same line of thought you will see that gallelian transformation doesn't keep SE invariant. But actually it does. While proving such invariance we can't simply take psi going to psi prime.
We have to keep a phase factor then this problem gets super duper complicated
awsome video, try to add some mic to your place so that sound can be captured when you are not faceing the camera.
Awesome man! I'm new here! Didn't know about your channel! it's amazing! And of course, already subscribed!! Keep the work going!!
subbed. (decided by the 4min mark) Please keep doing this Andrew!
Hey Andrew. Here's my main confusion: it looks like the Schrodinger Equation isn't even invariant under Galilean transformations! In particular, looking at the final equation you wrote, taking the limits as v/c approaches zero and gamma approaches 1, there still remains the rightmost term of the right side of the equation. Any ideas on why this lingering term appears?
Ohhh I've figured it out. That "lingering" term actually needs to be there! (tldr: the Schrodinger equation is in fact invariant under Galilean transformations.) In taking the partial with respect to t', we're taking x' to be constant, not x. But in verifying the validity of the transformation, we have to start with the non-transformed Schrodinger equation, which originally uses psi(x,t) not psi(x',t'). In taking the partial of psi(x,t) with respect to t' (keeping x' constant, not x!), there is an extra term v(delPsi/delx) which is exactly what is cancelled out by the "lingering" term.
Thank you so much .
Love from Pakistan ❤
Great work. Keep it up. Enjoy your video very much.
Watching this I'm so proud of you bro, you understand this shit like it's second nature. And I know that took crazy effort. Will meet you at the top someday
I've got a real question. Love your vids, bringing some real physics on youtube, that's juste pure kindness and it is quiet satisfying to watch :) But in your video you said multiple time that "Schrödinger's eq. leads to wave equation" I'm not quite sure, this is what De Broglie said "stationary wave" but in fact all the Schrödinger's eq. solution's aren't wave, the solution are just "stationary state" without the proper form of a wave eq. Anyway that's what I've been told. I would to hear you on that one :) !
3:25 It was at this point that i gave the video a like, and settled in for the long haul
Looks like ur daily upload schedule is always on the dot(son)
Andrew Son cosθ
On 8:04 shouldn't it be d^2/dx'dt'?
I forgot how to do the chain rule, this was a great explanation/refresher. P.s. I never knew that a mu could be written as a little m.
Also it's not pronounced d
Nice and concise. I could think of a few lecturers who would do well to emulate this performance. :)
but lorentz transform is only for the special relativity right? should we work in functional form for the coordinate transforms, it would be more general i suppose.
Which brings up a question. Where is the line between relativistic and non relativistic. For example up to what limit does Schrodinger equation work? ie, what is the fastest speed an electron can go and Schrodinger still works?
Go Mezant The answer is always “it depends on the error you’re willing to accept”
What role would velocity "v" play if your final result? I know that v isn't well defined in QM, and that the Schrodinger equation doesn't use velocity. So how can you combine the v from the Lorentz transform with the Schrodinger equation?
The 'v' is the velocity of the moving frame of the observer, with respect to the frame of the previous observer who was previously viewing the system and describing it via the Schrodinger equation. Particle velocity is, indeed, ill-defined in QM, but the velocity of the frame is well-defined. If Schrodinger equation were a relativistic equation, it should have been of the same form, as viewed from any non-accelerating frame, which is the first principle of relativity. Now, from the second principle of relativity (speed of light is same as viewed from any non-accelerating frame), we obtain that the correct transformation law between the space-time coordinates is the Lorentz transformation. So, Schrodinger equation should have been invariant under these transformations if it wanted to pass the test of being relativistic. The whole point of the exercise was that it fails to do so.
P.S. - I apologise if I am not able to explain it properly. In that I hope someone can do it better than me and to your satisfaction. All the best.
@@beetehotraroy3468 Yes, this answer makes sense. Thank you.
From the view of someone that is also studying physics that was very well explained ^
I have a feeling there are three types of people who watch these videos:
1. The people who actually understand
2. The people who don’t fully understand but want to do physics in the future
3. The people who don’t understand and don’t know why they are here
Very interesting video! Although I think it will be simpler to derive it if considering the relation of ∂²/∂x²=(∂/∂x)² =(γ∂/∂x’-γv∂/c²∂t‘)²
which is to regard the partial derivatives as operators.
I've subscribed to your channel at the first minute of the video.
I don't understand why the chain rule is used. My understanding is that it's used to solve composite functions, but i only did Calc one so this level of math is a bit beyond me...
I love quantum physics after I saw ur lecture.
Lovely, thanks.
Can someone help me with one question? At the end of the video, Andrew replaces V(x) with V(x') and also Psi(x,t) with Psi(x' , t' ). Why did that happen?
Given the Lorentz transformation relating x, x', t and t' we can show x = r (x' + vt' ) . Where r= gamma but I can't write greek symbols.
Then, it would seem, V(x) is replaced by V( r(x' + vt') ) and not V(x' ).
Similarly Psi (x, t) = Psi (r(x' + vt') , r(t' + (v/c^2)x' ) and not just Psi (x' , t' ).
Is it essential that the wave function and potential function are (separately) invariant functions if we were only interested in checking that the Schrodinger wave equation as a whole is preserved under a Lorentz transformation?
Completed this entire major in physics, but still not much info is taught about equations being general relativity or special relativity besides the obvious such as Maxwell's equations and the E=mc2 equations. But I think Schrodinger's equation describes only the wave and simply gives a probability for where the particles is but does not tell the speed of the particle, just its energy and mass are involved. There is particle wave duality and Schrodinger equation focuses on the wave part.
Great video Andrew!
One question though:
Since v(x,t) = \frac{ \partial x}{ \partial t }, should't you also differentiate the v^2 terms when you Lorentz transform?
Cheers,
Henrik
No, he doesn't differentiate them, because v is constant, so that makes the whole γ factor a constant. Because v is the velocity of the moving inertial reference frame S'. If v was a function of t that would mean, that S' is not inertial, because it would have acceleration.
Really liked this video! Would like to see more like this.
Andrew, is there a way to do the same with Dirac's equation without using the relativistic notation?
Why haven't you supposed that Psi will also (after Lorentz transf) change to Psi (prime)?
you are a very good teacher you know that?
Another thing i stumbled across is, that when you take the stationary SE and lets say you have an elektron which 1. is in no force field and therefore has no potential energy
And 2. has no kinetic energy.
Then the resulting energy should be m*c^2, but the stationary SE says its 0.
Is this correct??
I'm wondering if we should interpret this as the Schrodinger Hamiltonian is invalid at high energies, or that the energy is observer-frame dependent...Watching your Klein Gordon derivation next, thanks for the great videos.
Question:If entanglement is a defining aspect of QM,why would you expect relativistic terms in equations describing it. Isn't that essentially why folks like the whole " The universe is a hologram " idea? Doesn't entanglement mean there is no spacetime at the am level????
I am glad to see that some American youth are not destroying their brains with music and drugs but are doing something constructive, productive and illuminating.
How about gauge transformation, locality and the principle of least(stationary) action?
goddamn it you sound like an awesome teacher to have
tysm
Construction Tensors from non-Tensors... what does this refer to in physics?
Can u derive free particle solutions of the dirac equation
9:09 how the hell did you add it!
The only way I see the equations working is if v=0. What would that mean?
Schrödinger was looking after a relativistic equation, like Klein-Gordon, but couldn't find one. Then he let down his work and went on vacation. On return, he deemed that a non relativistic equation was "good enough." And even today, we use the Schrödinger equation because it encompasses all of the features of quantum mechanics, the Dirac equation never really worked correctly, and it is still good enough.
Shouldn't you consider a multiplicative phase onto the wavefunction? When showing the Schrödinger equation is invariant under guage transformations for an electromagnetic potential we pick up such a factor in the wavefunction!
What's that greek letter nu/upsilon with a slash?
Glad to have just stumbled upon this!
How the transformation coefficients can be figured out just like as if lorrentz transformation is an orthogonal transformation? I only know that these coefficients can be represented in derivative form if the transformation between two coordinates is an orthogonal transformation.
Muhammad Shatla just take the derivative of the Lorentz transformation and yes it is orthogonal
Is it even possible to solve PDE's with mixed second partial derivatives?
WHY? Here is the answer: 3:18
"Things are already going painfully wrong, and that's a good thing." Welcome to physics
I attempted the homework but I ended up with half a page of nonsense. I took second partial derivatives of both time and space under the Lorentz transform but ended up with half a page of jargon when I plugged the transformed derivatives into the Klein Gordon Eq. I couldn't get the derivatives to cancel each other out. I'm not sure if I made an error in the algebra (which is to be expected because I had to keep track of so many derivatives) or If my error came from incorrectly using the chain rule or miscalculating a transformation coefficient. I really want to solve the challenge but am stuck with a headache-inducing mess. Could you possibly help me?
UPDATE: I am making progress. It looks much better than it did before, however, there is an annoying v^2/c^2 term I can't get rid of.
I think the we should get psi'(x',t')=psi(x,t)(1-v²/c²)^(¼) to make the probability in the same region to be equal in both frames. Because probability should be invariant. And this psi' satisfies Schroedinger equation in psi does.
By the way I got that equation you wrote but d²/dx'dt' terms vanished. I think I did some calculation mistake.
When taking the derivatives, why is v (and therefore gamma) considered constant?
Because you aren’t taking the partial derivative with respect to it. For example if taking the partial derivative with respect to x, y would be treated as a constant for the same reason
What the heck is that square in 1:33
The d'Alembertian.
It's defined as d2/(cdt)2 - d2/dx2 - d2/dy2 - d2/dz2, so you can understand it as an extrapolation of the Laplacian operator to the Minkowski space-time (if you try to transform this operator you'll see that it remains invariant under Lorentz transformations (aka it is a scalar in the Minkowski spacetime). That's why it can appear in Relativistic equations like the wave equation (it respects Lorentz's Covariance).
Oh so is like a relativistic version of that upside down delta
@@rafaelaassuncao9729 Exactly.
If there was a lorentzian invariant hamiltonian would it work? Like for example H=cP for photons?
Fantastic video!