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Quantum Particle in a Box / Infinite Potential Well / Infinite Square Well (One-Dimensional Case) 

Elucyda
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I use the time-independent Schrodinger equation to find the energy eigenfunctions and eigenvalues for a single quantum particle in a box. Then, I construct the general solution and describe the time evolution given initial conditions.
Link to Quantum Playlist:
• Elucidating Quantum Ph...
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2 окт 2024

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Комментарии : 10   
@Elucyda
@Elucyda 3 года назад
When determining the normalization constant A, the positive real root was considered. This is because the phase of A has no physical significance.
@HermioneCosslett
@HermioneCosslett 5 месяцев назад
I got confused when I tried to verify the normalisation constant, and I see that someone else in the comments was also confused by the same thing, so I'll post the steps that I eventually worked out in case this helps to clear the confusion for anyone else ! So, we want to normalise the wavefunction, psi(a) = Asin(ka) Take the square of the magnitude of the wavefunction to get the integrand, A^(2)sin^(2)(ka), and integrate it with respect to " a " with bounds of " a " to 0 The result of this integral is (A^(2)ka - 0.5*A^(2)sin(2ka)) / 2k + C C is absorbed into the constant k So, we get (A^(2)ka - 0.5*A^(2)sin(2ka)) / 2k We have found the value of k, defined as " k_n=((n)(pi)) / a " for which k satisfies the continuity boundary of the wavefunction. Substitute k_n into every k term in the result of the integral This gives ((A^(2)(n)(pi) - A^(2)sin(2(n)(pi)))a) / 2(n)(pi) The second term of the nominator is equal to 0 because n is any natural number so sin(2(n)(pi))=0 Now we have (A^(2)(n)(pi)) / 2(n)(pi), which we want to be equal to 1 to satisfy the normalisation condition Rearrange (A^(2)(n)(pi)) / 2(n)(pi) = 1 and solve for A This gives sqrt(2/a) And, as Elucyda said, "When determining the normalization constant A, the positive real root was considered. This is because the phase of A has no physical significance."
@BeckCaesar-r8l
@BeckCaesar-r8l 20 дней назад
Miller Helen Jackson John Gonzalez Ronald
@mubeenparvaiz-0788
@mubeenparvaiz-0788 6 месяцев назад
Amazing ❤
@dalegriffiths3628
@dalegriffiths3628 Год назад
Great video - thanks, you explain things really clearly ... in my text book it talks about a particle in a box with standing waves as an analogy to an electron in an atom what I don't get is that here we get E proportional to n^2 whereas energy levels in an atom converge because E is proportional to 1/n^2 so this suggests particle in a box is not a good way to think of schrodinger's atomic energy levels. Is there any way that this approach could be modified so it works?
@Szszsecr
@Szszsecr 3 года назад
Great video series really helping with my Quantum module! Thankyou so much!
@bayanalshehri7818
@bayanalshehri7818 Год назад
Thank you for your great explanation
@sophiagershaft1986
@sophiagershaft1986 3 года назад
This is really awesome!! Thank you for explaining each step thoroughly, it makes it so much easier to understand! I still don't understand the normalization constant though, sqrt(2/a)
@Elucyda
@Elucyda 3 года назад
Glad it was helpful! There's another video on normalization, which you can find here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE--y2GF5K6mHE.html
@hershyfishman2929
@hershyfishman2929 3 года назад
The integral from -inf to inf of |ψ|^2 dx = 1 = the integral from 0 to a of A^2 sin^2(kx) dx = A^2 a/2 (as the integral of sin^2(x) dx over integer number of cycles = x/2), --> A = sqrt(2/a) (Griffiths 2nd ed p. 32,ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-nFHhLJGDNHA.html)
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