The wave for a free particle
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- Опубликовано: 30 июл 2017
- MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-04S16
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
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Thank you! greetings from a Mexican girl studying in Germany.
This video has been the best of the series so far. I'm looking forward to the rest (as soon as my brain cools down!)
So brilliantly dissected.
So clear! solved my question that lasted for so many years. Great teacher! Thank you so much.
+x is right -x is left
Very good lecture Sir. Thanks 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
thankyou from Italy
Great sir
using Eulers formula you could represent the e-functions as cosines and sines, but we could then represent these as we did in the first 2 examples, is it thus only because of the multiplication of the sine with i that this does work?
yup, the imaginary part is key, once again showing the neccesity of complex numbers
Got it.
why making left-moving wave is different with angle fuction when we suppose that wave function is exponential?
professor say that because of phase-changing , i dont understand what is it
In light of his previous statement that matter waves are not Galilean invariant, can I assume that if 1 and 2 are considered reasonable waves, then when we measure such waves, we must obtain phase information, and the phase information we obtain must be Galilean invariant. Therefore, 1 and 2 are unreasonable waves. On the other hand, 3 and 4, as waves in complex number form, we can only measure their norm. Clearly, during the process of taking the magnitude, the phase information is wiped out. Hence, this does not contradict the Galilean invariance of phase and is considered reasonable waves?
what is ment by (always an e⁻ⁱkt)
on 2:03
so that it moves only along +x axis
@
MIT OpenCourseWare
I have a question, @9:10 why cos(wt) is not accepted as an answer but cos(kx) is accepted? isn't cos(kx) will means the particle will vanish at some place which is unacceptale for a matter?
It can vanish for certain x, but not for all x at some t
Well it's not problematic that the wave isn't non-zero everywhere. If you think about it, a particle is typically in a very confined position and hence there are a whole number of places it (almost certainly) isn't. However, if the whole wave becomes zero, it means that at that time the particle isn't _anywhere_, which is problematic.
Are these wave functions are complex?
How do we decide a particle is moving left or right by seeing an equation?
Can anyone pls help me out.
Kx-wt in +x because to keep function value constant asvx increases t must increase and time factor wt must be subtracted.
I know only little bit about this mysterious concept. Hope you got it. I think you knew it but wanted someone to tell you.
@@TanveerAlam-oe7yt Thank you Tanveer.
Imagine shifting origin by amount
-omega*t but keeping do not move the wave. The overall effect is same as moving wave to right.
What if there are three plane waves superimposing
I do not get how this philosophy that by keeping the sign of the e^(omega*t) segment constant it is assumed the probability of wave propagating left and right is equal. my intuition tells me it should be the other way round.
I think it has to do with getting into the form of 2cos(kx)e^(-iwt). If we do change the phase we are unable to factor out a e^(-iwt). For example lets say instead of having e^(ikx-iwt) + e^(-ikx-iwt) we have e^(ikx-iwt) + e^(ikx + iwt). When we put this in simplest form we get 2cos(wt)e^(ikx). Once again we see that this x can be zero everywhere at a specific time and so we are back in the scenario of the sin(kx-wt) and cos(kx-wt) of an impossible matter wave. So its not a matter of we CANT change the phase but we dont want to because it doesnt get us the right answer where we have a matter wave that actually makes sense where at any time the particle exists SOMEWHERE. Thus we do not change the phase because we want to factor it out so the wt factor is always in the exponent where that pesky trigonometric function cannot get to it and make the particle not exist anywhere. I hope that clear it up.
Lol. The subtitle is quite right at 13:23
what is vague notion of probability on 3:56?
That the probability of the particle is moving to the right is the same as the probability of it moving to the left.
That there is a probability for the particle to be measured at some x at any time.
In cases (i) and (ii), there were times when the wave function was 0 for all x, implying that the probability of measuring the particle at any point x is 0 for all x. Since we believe this not to be the case, we rejected those two cases.
@Pradyumna S I had the same issue with this part. He's said previously "we don't yet know that Psi is a probability amplitude. We're going to work up to that." But then by this point he's already invoking that fact. It threw me off.
8:14 why can't we change that phase?
I think it has something to do with energy. Energy and frequency are related by de Broglie relation, so if a matter wave is moving with same speed in opposite direction, than energy should not change to opposite. He even says: "Always this energy." That's my guess, I might be wrong.
As per my knowledge ..
A wave can have only one phase.. if we change the sign it would have two phases and that will be like having two waves superimposed to each other.
@@agrisbuzs3892 It is pretty idea. Thank you!
I think it has to do with getting into the form of 2cos(kx)e^(-iwt). If we do change the phase we are unable to factor out a e^(-iwt). For example lets say instead of having e^(ikx-iwt) + e^(-ikx-iwt) we have e^(ikx-iwt) + e^(ikx + iwt). When we put this in simplest form we get 2cos(wt)e^(ikx). Once again we see that this x can be zero everywhere at a specific time and so we are back in the scenario of the sin(kx-wt) and cos(kx-wt) of an impossible matter wave. So its not a matter of we CANT change the phase but we dont want to because it doesnt get us the right answer where we have a matter wave that actually makes sense where at any time the particle exists SOMEWHERE. Thus we do not change the phase because we want to factor it out so the wt factor is always in the exponent where that pesky trigonometric function cannot get to it and make the particle not exist anywhere. I hope that clear it up.
How can a particle move to the right and to the left at the same time ?
I don't understand this probability trick he is using...
How can we create it experimentally? Beam Splitter ?
Same way that when you throw a stone into a pond, the wave propagates in multiple directions.
The particle isn't moving to the right and left simultaneously. We don't use this concept in QM. Instead, we say that it is possible for a free particle to be in a really weird state represented by superposed states of the same particle moving to the left and right. Any classical point of view cannot represent this state, but this is how the world really works. This idea was verified in several experiments that Professor Barton explained in previous lectures.
esto ya lo entiendo
(ikx-iwt) describes a "continuous wave" (cw), not a finite dimension proton (neutron, electron...). Further, protons (neutrons, electrons...) are measurable objects, there should be no need to resort to statistical probabilistic functions to describe them? Does anyone doubt a PDF function such as Psi, could be used to solve Maxwell's wave equation? Remember, any solution that validly solves a differential equation, is a valid solution? So Maxwell (who also uses the (ikx-iwt) construct) could've equally utilized Psi, instead of E&H? And how much less would we know about E/M and photons by using a PDF? (more...)
Both E&H vs Psi are valid solutions of Maxwell's wave equation, but Maxwell's solutions are considered "solved," or "settled-science," while QM is nowhere near settled? Why? Why is one "settled" and the other not? I'll tell you - Because one solved with measured quantities and the other got lazy and used statistical (PDF) function, Psi.
QM makes predictions the other cannot. It predicts the amount of energy required to free an electron from the potential well around an atom of silicon. In other words, QM makes predictions that allowed engineers to assemble the device you wrote this comment on. If protons, neutrons, and particularly electrons *were* "measurable objects", QM would not work, and would not make valid predictions.
You're talking shit.
@@FaxingtheTaxes Your last sentence wasn't necessary. The message was already clear enough.
@@jacobvandijk6525 I think it's ok to slap down people who aggressively talk bollocks with certainty
@@FaxingtheTaxes you pretend to understand QM very well
2:57 to 3:17 Explain Why only this method is available to decide form of wave? And why are these strange and surprising
13:59? How we trust that it gonna to be true for matter wave?
4:24 to 4:26 But why? What is the logic behind it?
12:32 why do I get in trouble?
Same reason for (ii) i guess
Look carefully it's same as eqn ii , that's why
Quantum Mechanics is linear correct? So if we have 2 solutions, the superposition of the 2 solutions MUST be true. The equations of 3 and 4 when superimposed are not, so how can we just choose one of them to be correct? Doesnt that violate linearity of quantum mechanics?
well that is the point! that's why we rejected (4) as our solution. if it is omitted as a solution, then there is nothing to worry about.
8:47 Why? Sinnce we take real part at the end of the day, So take real part annd you get cos(wt) which may be zero. Explain.
1) we do not just take the real part at the end of the day. We learn that all wavefunction MUST be complex. Refer back to the earlier video when he talked about the neccesity of complex numbers. So we cannot just take the real part of a given wave function but must take it in its entirety. You cannot just disregard the complex part.
2) its not cos(wt), it is cos(kx)e^(-iwt) and so for any given t there is at least one x that is nonzero so the particle may exist
13:12 Why?
3:39 to 4:00 Why only this?
4:05 to 4:17 But why?
9:07 What? Explain
2:02 Explain