The answer to the final question is the following (i think) : Solving the TIDSE does NOT give you the Wavefunction Ψ(χ,t) , but X(x). The Wavefunction will therefore be Ψ(χ,t)=Asin(ka)e^(-iE/hbar)t , which of course is a complex function.
"The infinite square well is called that because it's potential is infinite... and, well.. square" I see what you did there ;) Plus a good explanation, thanks!
I am watching this vedio in 2020 during Corona virus pandemic to improve my self in quantum physics and its realy very good with detailed explanation that help understand the subject thoroughly
Thank you. This was a super helpful review. My Stat-Mech professor brought up particle in a box today and I completely forgot everything except that it was a particle in an infinite square well.
So i was trying to find a video that explains infinite square well clearly, and this is the one i found! Extremely clear explanation, precise and beautifully presented, thank you so much for this video. I do want to point out, for those who get confused while watching the video, that at 17:15 there is a typo, the E should be replaced with E^2. Anyways, great video and i hope you keep doing videos like this to help others.
Brant - PLEASE write a Modern Physics text book on this stuff! Your presentation is perfectly clear at all times, the modern physics text by Krane on this is horrible in comparison.
+WellWellWell Hey there, if you notice he is going through Griffiths, Introduction to Quantum Mechanics, but of course he's showing extra steps but essentially is the same, however I still do prefer Brant hehe.
Thank you ! For the questions at the end: 1- I still did not get the relation between the infinite V(x) and the second spacial derivative of the wave function..why does the question assume that an infinite V(x) implies an infinite second spacial derivative of the wave function? 2- Complex numbers indeed include real numbers, so we are not breaking a QM postulate. More importantly, we can expand the exponential solution in terms of sin and cos, and continue solving using the boundary conditions and we will end up with our solution! Also, we assumed the solution of sin and cos due to the formula that resembles a simple harmonic motion that we got when we substituted with our conditions, and that can be expressed also exponentially, so our action is driven by a reason and does not contradict a postulate.
Ibrahim Hany for the first question the second spatial derivative of psi is related to V(x) by the schrodinger equation. But psi must be 0 outside the well (V infinite) is physically obligatory for all the wave functions
@@ahmedboubaker8514For the questions my take is: 1. All wavefunctions drawn at the end, end on the axes (regardless of whether odd or even). In the previous video it was stated that if V(x) > E which outside the bounds is cause infinity, it would move away from the axes, unless it was stuck on the axes. In this example, it is stuck on the axes so it cannot leave towards infinity and is stuck at 0. 2. Wavefunction is complex, but the solution gotten at the end of this video, is only the spatial part. The complete wavefunction adds the time evolution as well, which is a complex exponential, hence we have not broken any features of QM. Another take on 2 could be that this spatial part could be considered complex, but with an imaginary term 0i.
For the questions my take is: 1. All wavefunctions drawn at the end, end on the axes (regardless of whether odd or even). In the previous video it was stated that if V(x) > E which outside the bounds is cause infinity, it would move away from the axes, unless it was stuck on the axes. In this example, it is stuck on the axes so it cannot leave towards infinity and is stuck at 0. 2. Wavefunction is complex, but the solution gotten at the end of this video, is only the spatial part. The complete wavefunction adds the time evolution as well, which is a complex exponential, hence we have not broken any features of QM. Another take on 2 could be that this spatial part could be considered complex, but with an imaginary term 0i.
1) dψ/dx is continuous except at points where the potential is infinite. A particle outside the well would have infinite energy. Since is impossible to find the particle outside that boundary, we must have |ψ|²=0 in this region. ψ is always continuous. This brings us to the boundary conditions ψ(0)=ψ(a)=0. 2) Considering all possible Schrodinger equations, ψ is in general complex. But that doesn't mean that for any particular case we will encounter a complex solution. The general solution for a given potential V(x), is given by the linear combination: ψ(x,t) = ∑cnXnTn Our stationary states Xn remain the same!
1. I only have the answer for number 1. I think it's because the probability of finding the wave outside of the well is equal to zero so the psi function would be equal to zero as well. 2. I'm guessing because we can re-arrange the equation in such a way that we can use the real number solution for simple harmonic motion and it still preserves equality.
For the first one the second derivative cannot be evaluated at x=0 or a because it is not smooth (it is continuous, but not smooth). The second one: The solution in general is complex, but it does not have to be. Another issue is that you could not use the coefficients A and B with it since it is one term, and it could not be set for the boundary conditions correctly. These are my guesses.
1. [h^2/(2m)ψ"(x)]/(V-E) = ψ(x) If general solution to the time-independent is still Asin(kx), it's clear the limit as v approaches infinity is 0. If the solution is not Asin(kx), then I don't know. 2. If ψ(x) = e^ikx, it will not be normalizable. I'm new to QM, and I'm still leaning. I don't think that I'm correct but I'd like to try.
Lol I remember writing those types of comments when I was 12. Keep up the good work and maybe one day you'll be able to say that without lying. Quantum Mechanics (let alone, QFT, QCM, and other more advanced fields) is a long journey and would usually require most people around 3 or 4 revisits to actually get the gist of it... anyone can follow along a basic math video but developing an intuition for this stuff is a totally different story. However, it's great that you're starting at 12, as long as you don't forget about some fundamentals that can aid your understanding even more! I know this comment is pretty random but this is something I needed to hear in my early days (I'm still 18 and have a long way to go as well). Good luck and keep learning!
Thank you for nice explanation. One thing I'm struck with, what really is or difference between single well and Double well potential. why do we need double well potential when we already have Single well potential well solution?? Any help.
Hmm. Technically, the solution to the TISE is not Asin(kx)+Bcos(kx), but is Aexp(ikx)+Bexp(-ikx). Remember the wave function can be a complex function.
You are absolutely correct, but you can go between sine-cosine and complex exponentials. In a recent video I used the exponential solution and then applying boundary conditions I showed how to turn it into a sine-cosine solution: ruclips.net/video/pbZN8Pd8kac/видео.html
but can't you only assume that B=0, if you have the potential well "starting" at x=0? i mean what's stopping you from putting the "borders", for lack of a better term, at per say, x=-2 and x=a where a is different to 0? or am i missing something crucial here?
Kendrick Lamar I realise I'm a bit late, but in my opinion, it doesn't matter where on the x-axis you place the "box". With reference to the other end of the "box", x is taken as zero. Thus, even if the points are x=2 and x=a, the width of the box is 2-a and B is taken as zero at x=2. Hope that made sense!
What you are proposing is a change of coordinate system, and the physics should not change. However, with your choice the potential better reflects the symmetry of the problem about the origin, and the mathematical solutions also reflect this symmetry. I recently discussed the solution to the well with the origin at the centre of the well: ruclips.net/video/pbZN8Pd8kac/видео.html
Countering the arguments at the end: First: V(x) and d/dx2 are not simply scalars, but operators on the wavefunction ψ. In general, what operators do to a function can depend on the function they operate on. V(x) can be infinite in a region, but d/dx2 can still be 0 in the case that ψ = 0 in that region (which is exactly what our solution says). Second: Complex numbers can be real numbers. e^ikx is a real number if kx = nπ, which again is exactly what our solutions are! Thus we can simplify our expression by dropping the imaginary part of the expression, which is always 0. Thoughts?
I agree with you and think you're right on the first. On the second, let's not forget the solution to the TISE is X(x) a function only of space, and as such is not the more general solution to the Schrodinger Eq. If it were time dependent, we surely would have gotten a complex number because of the T(t) which is a complex exponential. So we're not in fact missing out on some fundamental aspect of QM, we're just looking at a case where there is not time dependence and as such the function is real (we can also interpret this as having a complex wavefunction with no imaginary component; let's not forget that real numbers are also complex numbers). I think your answer for the second problem is right, I was just wondering if it would be tight to explain it/ phrase it this way? 🤔
To add to this, you are absolutely correct that in the case of time dependence it will become critical to include the complex nature of the wave function. I recently did a video explaning time dependence in the infinite square well, with animations and a Jupyter notebook that allows you to play with it: ruclips.net/video/CGbv9cVwJL0/видео.html
Your wave function is defined inside the well between the estabilished boundaries, in this case, 0 and a. This means that when you "visualize" or measure the system inside the well you will definitely find a particle in a given position, therefore, the probability of your wave function between said interval must equal to 1. In conclusion, the particle is, for sure, inside the well, so you are certain to find it, so probability=1
The only part I need help with is the simplification of sin((n*pi*x)/a)^2, A=srqt(2/a) really fell from the sky : what is the real mathematical proof ? I can't find it anywhere one the internet or in my lessons...
JVPierre142 try cos(a+b) and substitute 'b' by a , then substitute the cos^2 that you 'll get by 1-sin^2 and there it is , when you'll integrate it you will have sin2x/n goes to zero for great values of n
i understand that you're dealing with a time-independent case, but it seems very odd to me that an equation of a standing wave (a vibrating wave) can be described by an equation that does NOT depend on time. as a vibrating wave, it is surely EVOLVING over time, i.e. changing in shape. i think i've seen it's something to do with phase, i.e. that it's a wave of fixed shape which is effectively turning. but it is turning OVER time, surely! i'd be really grateful for any help please - please bear in mind i have no background in physics, but have done maths at uni
Hi Dylan. The most general form of the Schrodinger equation is the "time dependent Schrodinger equation" (TDSE), which does have explicit time dependence. Typically, we employ separation of variables to derive the "time independent Schrodinger equation" (TISE), which is used here. If a wavefunction is an eigenfunction of the Hamiltonian operator, then it is a standing wave, fixed in shape and only varying in amplitude. The time part is found by multiplying the spatial part by exp(i E t / hbar), but this has no effect on any physical properties (expectation values), as this standing wave isn't moving over time.
You're welcome. No worries about not understanding everything the first time through. QM gets very abstract as you go down the rabbit hole until you see enough examples to see how all the pieces fit together. It's mostly a matter of effort and time.
Such a wonderful video. Thank you for making it so easy to re-familiarize myself with what I learned ages ago!
The answer to the final question is the following (i think) : Solving the TIDSE does NOT give you the Wavefunction Ψ(χ,t) , but X(x). The Wavefunction will therefore be Ψ(χ,t)=Asin(ka)e^(-iE/hbar)t , which of course is a complex function.
Thanks for the video. I have struggled with Schrodinger's equation for years but you have explained it very well.
"The infinite square well is called that because it's potential is infinite... and, well.. square" I see what you did there ;) Plus a good explanation, thanks!
I am watching this vedio in 2020 during Corona virus pandemic to improve my self in quantum physics and its realy very good with detailed explanation that help understand the subject thoroughly
Thank you. This was a super helpful review. My Stat-Mech professor brought up particle in a box today and I completely forgot everything except that it was a particle in an infinite square well.
So i was trying to find a video that explains infinite square well clearly, and this is the one i found! Extremely clear explanation, precise and beautifully presented, thank you so much for this video. I do want to point out, for those who get confused while watching the video, that at 17:15 there is a typo, the E should be replaced with E^2. Anyways, great video and i hope you keep doing videos like this to help others.
Brant - PLEASE write a Modern Physics text book on this stuff! Your presentation is perfectly clear at all times, the modern physics text by Krane on this is horrible in comparison.
+WellWellWell Hey there, if you notice he is going through Griffiths, Introduction to Quantum Mechanics, but of course he's showing extra steps but essentially is the same, however I still do prefer Brant hehe.
Thank you !
For the questions at the end:
1- I still did not get the relation between the infinite V(x) and the second spacial derivative of the wave function..why does the question assume that an infinite V(x) implies an infinite second spacial derivative of the wave function?
2- Complex numbers indeed include real numbers, so we are not breaking a QM postulate. More importantly, we can expand the exponential solution in terms of sin and cos, and continue solving using the boundary conditions and we will end up with our solution! Also, we assumed the solution of sin and cos due to the formula that resembles a simple harmonic motion that we got when we substituted with our conditions, and that can be expressed also exponentially, so our action is driven by a reason and does not contradict a postulate.
Ibrahim Hany for the first question the second spatial derivative of psi is related to V(x) by the schrodinger equation.
But psi must be 0 outside the well (V infinite) is physically obligatory for all the wave functions
@@ahmedboubaker8514For the questions my take is:
1. All wavefunctions drawn at the end, end on the axes (regardless of whether odd or even). In the previous video it was stated that if V(x) > E which outside the bounds is cause infinity, it would move away from the axes, unless it was stuck on the axes. In this example, it is stuck on the axes so it cannot leave towards infinity and is stuck at 0.
2. Wavefunction is complex, but the solution gotten at the end of this video, is only the spatial part. The complete wavefunction adds the time evolution as well, which is a complex exponential, hence we have not broken any features of QM. Another take on 2 could be that this spatial part could be considered complex, but with an imaginary term 0i.
Very clear and definitive. Much easier to understand than my textbook. Thank you.
If you mean griffiths, yes it is hahaha
For the questions my take is:
1. All wavefunctions drawn at the end, end on the axes (regardless of whether odd or even). In the previous video it was stated that if V(x) > E which outside the bounds is cause infinity, it would move away from the axes, unless it was stuck on the axes. In this example, it is stuck on the axes so it cannot leave towards infinity and is stuck at 0.
2. Wavefunction is complex, but the solution gotten at the end of this video, is only the spatial part. The complete wavefunction adds the time evolution as well, which is a complex exponential, hence we have not broken any features of QM. Another take on 2 could be that this spatial part could be considered complex, but with an imaginary term 0i.
you sound like Eric from That 70's Show
never going to unhear this now thx
1) dψ/dx is continuous except at points where the potential is infinite. A particle outside the well would have infinite energy. Since is impossible to find the particle outside that boundary, we must have |ψ|²=0 in this region.
ψ is always continuous. This brings us to the boundary conditions ψ(0)=ψ(a)=0.
2) Considering all possible Schrodinger equations, ψ is in general complex. But that doesn't mean that for any particular case we will encounter a complex solution.
The general solution for a given potential V(x), is given by the linear combination:
ψ(x,t) = ∑cnXnTn
Our stationary states Xn remain the same!
you are a virtuoso at melding mathematical models with real like examples and diagrams, ty sir
1. I only have the answer for number 1. I think it's because the probability of finding the wave outside of the well is equal to zero so the psi function would be equal to zero as well.
2. I'm guessing because we can re-arrange the equation in such a way that we can use the real number solution for simple harmonic motion and it still preserves equality.
Thank you very much. ALL your presentations are very clear and helpful. Do you have presentation on electro dynamic Jackson as well?
Thank youuu!!!
For the first one the second derivative cannot be evaluated at x=0 or a because it is not smooth (it is continuous, but not smooth).
The second one: The solution in general is complex, but it does not have to be. Another issue is that you could not use the coefficients A and B with it since it is one term, and it could not be set for the boundary conditions correctly.
These are my guesses.
Answer according to me
Q1 d2Ψ/dx2 will tend towards zero even if V tends to infinity as Ψ tends to 0.
Q2 Ψ(x,t) = Asin(kx)exp(-iEt/h)
Mr. Carlson is awesome thanks so much for these videos
Such a good video! you explained the concepts very well. Thank you so much!
you did a great job explanation the equation
You are an amazing person sir
1. [h^2/(2m)ψ"(x)]/(V-E) = ψ(x)
If general solution to the time-independent is still Asin(kx), it's clear the limit as v approaches infinity is 0. If the solution is not Asin(kx), then I don't know.
2. If ψ(x) = e^ikx, it will not be normalizable.
I'm new to QM, and I'm still leaning. I don't think that I'm correct but I'd like to try.
+Samera Pornpan Would be better to use this solution to the dif ec. ψ(x) = Ae^ikx + Be^-ikx,, then when ψ(0) , B=-A, so ψ(x) = 2AiSin(kx).
You're so good I can understand and I'm 12
Lol I remember writing those types of comments when I was 12. Keep up the good work and maybe one day you'll be able to say that without lying. Quantum Mechanics (let alone, QFT, QCM, and other more advanced fields) is a long journey and would usually require most people around 3 or 4 revisits to actually get the gist of it... anyone can follow along a basic math video but developing an intuition for this stuff is a totally different story. However, it's great that you're starting at 12, as long as you don't forget about some fundamentals that can aid your understanding even more! I know this comment is pretty random but this is something I needed to hear in my early days (I'm still 18 and have a long way to go as well). Good luck and keep learning!
Thank you for nice explanation. One thing I'm struck with, what really is or difference between single well and Double well potential. why do we need double well potential when we already have Single well potential well solution?? Any help.
Thank you! Keep doing what you're doing!
Excellent exposition
Nicely explained
Very nice Tutorial man...
thank you so much for that awesome video .. i undersood the exercice very well
Wow, well done and explained.
Hmm. Technically, the solution to the TISE is not Asin(kx)+Bcos(kx), but is Aexp(ikx)+Bexp(-ikx). Remember the wave function can be a complex function.
You are absolutely correct, but you can go between sine-cosine and complex exponentials. In a recent video I used the exponential solution and then applying boundary conditions I showed how to turn it into a sine-cosine solution: ruclips.net/video/pbZN8Pd8kac/видео.html
Wonderful walkthrough. Thanks! :-D
U r the man!!!!
Thank you! You are a life saver.
You are awesome man 👍👍👍👍
Great video! Helps a lot!
thx alot mr brant , your vid was very helpful
What will happen if you have potential V(x) = 0, -a
Dennis W Hello, you would simply enforce the boundary conditions at -a for wavefunction to be 0 and at 2a for wavefunction to be 0
Is there such kind of lucid teaching videos on other topics( Like- relativity, Quantum field theory,)?
but can't you only assume that B=0, if you have the potential well "starting" at x=0? i mean what's stopping you from putting the "borders", for lack of a better term, at per say, x=-2 and x=a where a is different to 0? or am i missing something crucial here?
Kendrick Lamar I realise I'm a bit late, but in my opinion, it doesn't matter where on the x-axis you place the "box". With reference to the other end of the "box", x is taken as zero. Thus, even if the points are x=2 and x=a, the width of the box is 2-a and B is taken as zero at x=2. Hope that made sense!
Can you able to find electron between the two limits within the well?Even if the limited are not given??
I believe the boundary conditions are a necessity for the differential equation to be solved
why is the solution to the infinite square well not (A+B)e^(-xk^2)?
how to calculate probability of for given energy
Thank you! Very helpful video!!
what happens when x=-L and x=L find average of position, square of position and momentum in ground state n=1
What you are proposing is a change of coordinate system, and the physics should not change. However, with your choice the potential better reflects the symmetry of the problem about the origin, and the mathematical solutions also reflect this symmetry. I recently discussed the solution to the well with the origin at the centre of the well: ruclips.net/video/pbZN8Pd8kac/видео.html
SO CLEAR> THANKS SO MUCH!!!!
be careful: d2 psi/dx2 = -k2 psi would be equation of a harmonic oscillator if dx2 would be dt2 ....
Thank you you helped me alot
at 13:45 how do you get Psi(x)=A Sin(npi/a x)? I don't understand how you reached the brackets of sin???
the general soln: psi(x) = A sin(kx) + B cos(kx). cosine part gone. so put k = n pi/a here
+Andriy Boubriak in the equation he states that ka=npi so he just rearranged it for k, where k=npi/a
thank you so much for this
This is great!
Countering the arguments at the end:
First: V(x) and d/dx2 are not simply scalars, but operators on the wavefunction ψ. In general, what operators do to a function can depend on the function they operate on. V(x) can be infinite in a region, but d/dx2 can still be 0 in the case that ψ = 0 in that region (which is exactly what our solution says).
Second: Complex numbers can be real numbers. e^ikx is a real number if kx = nπ, which again is exactly what our solutions are! Thus we can simplify our expression by dropping the imaginary part of the expression, which is always 0.
Thoughts?
I agree with you and think you're right on the first. On the second, let's not forget the solution to the TISE is X(x) a function only of space, and as such is not the more general solution to the Schrodinger Eq. If it were time dependent, we surely would have gotten a complex number because of the T(t) which is a complex exponential. So we're not in fact missing out on some fundamental aspect of QM, we're just looking at a case where there is not time dependence and as such the function is real (we can also interpret this as having a complex wavefunction with no imaginary component; let's not forget that real numbers are also complex numbers). I think your answer for the second problem is right, I was just wondering if it would be tight to explain it/ phrase it this way? 🤔
To add to this, you are absolutely correct that in the case of time dependence it will become critical to include the complex nature of the wave function. I recently did a video explaning time dependence in the infinite square well, with animations and a Jupyter notebook that allows you to play with it: ruclips.net/video/CGbv9cVwJL0/видео.html
How does he write like that, some kind of touchscreen? I'd love to have something like that.
digitiser pen and tablet, it looks like. Things like this: www.wacom.com/en-us/products/pen-tablets
21:16 ...answers please?
Did you get them?
what is an amazing way thank you alot
How probability wave function is equal to one?? What is the Physical meaning of it😐🤔
Your wave function is defined inside the well between the estabilished boundaries, in this case, 0 and a. This means that when you "visualize" or measure the system inside the well you will definitely find a particle in a given position, therefore, the probability of your wave function between said interval must equal to 1. In conclusion, the particle is, for sure, inside the well, so you are certain to find it, so probability=1
thank you so much
The only part I need help with is the simplification of sin((n*pi*x)/a)^2, A=srqt(2/a) really fell from the sky : what is the real mathematical proof ? I can't find it anywhere one the internet or in my lessons...
JVPierre142 try cos(a+b) and substitute 'b' by a , then substitute the cos^2 that you 'll get by 1-sin^2 and there it is , when you'll integrate it you will have sin2x/n goes to zero for great values of n
That part was just to normalize the eigenfunctions. Search for 'inner product' to understand it.
Good one
ruclips.net/video/iWYAERgqnrQ/видео.html
thank you
good
I love you.
What is m?
m is the mass of the particle.
i understand that you're dealing with a time-independent case, but it seems very odd to me that an equation of a standing wave (a vibrating wave) can be described by an equation that does NOT depend on time. as a vibrating wave, it is surely EVOLVING over time, i.e. changing in shape.
i think i've seen it's something to do with phase, i.e. that it's a wave of fixed shape which is effectively turning. but it is turning OVER time, surely!
i'd be really grateful for any help please - please bear in mind i have no background in physics, but have done maths at uni
Hi Dylan. The most general form of the Schrodinger equation is the "time dependent Schrodinger equation" (TDSE), which does have explicit time dependence. Typically, we employ separation of variables to derive the "time independent Schrodinger equation" (TISE), which is used here. If a wavefunction is an eigenfunction of the Hamiltonian operator, then it is a standing wave, fixed in shape and only varying in amplitude. The time part is found by multiplying the spatial part by exp(i E t / hbar), but this has no effect on any physical properties (expectation values), as this standing wave isn't moving over time.
thank you for explaining - i'm still not sure i fully understand but that's my fault, not yours. thanks again
You're welcome. No worries about not understanding everything the first time through. QM gets very abstract as you go down the rabbit hole until you see enough examples to see how all the pieces fit together. It's mostly a matter of effort and time.
Ooooooooh man
Thanks a lot your video helped me a lot
thx aloooot, you are great XD !!
Thanks a lot for saving my ass
man wanna learn literally whole quantum mechanics! it is interesting AF, but can't :(
bcz of my syllabus
ruclips.net/video/iWYAERgqnrQ/видео.html
Please make a font of your handwriting xD
=)))
ruclips.net/video/iWYAERgqnrQ/видео.html
thank you