Finite Quantum Well Explained - Part 1

/ edmundsj
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Here I go over how to set up the solution to the finite quantum well, including what the solutions look like inside and outside the well using Schrodinger's equation. I go over what boundary conditions to use to solve Schrodinger's equation. Next time I will show you how to 'stitch' together these solutions using these boundary conditions, and how to solve Schrodinger's equation graphically. MIT's OpenCourseWare also has a truly excellent lecture on this subject which can be found here: ocw.mit.edu/courses/physics/8...
This is part of my series on semiconductor physics (often called Electronics 1 at university). This is based on the book Semiconductor Physics and Devices by Donald Neamen, as well as the EECS 170A/174 courses taught at UC Irvine.
Hope you found this video helpful, please post in the comments below anything I can do to improve future videos, or suggestions you have for future videos.

[@rohitnimje9]

Thank you, sir, for the brief explanation.

[@ruoyangchang6835]

Thanks. Useful for UCSB Physics 5 and ECE132

[@tae-babytigerbts4890]

Really helpful for we engineers. ✌🏻

[@anasalsa4858]

nice explanation!! thanks :) keep going (Y)

[@vijgenboom2843]

Thank you! ❤️

[@farzanasnigdha7048]

Really nice 👍

[@FD-rt3rv]

thanks so much!

[@mdjasimmiah7713]

It's really helpful man!

[@JordanEdmundsEECS]

Thanks!

[@Lanz8088]

bless you man

[@emillytabara9410]

Hi, Jordan. Such a great explanation. I have one question. When you said that energy of a infinity square well is inaccurate for the real barriers of potential in MQ, why is that? Because n goes to infinity? I hope I'm being clarifying

[@robertwagner7139]

Hey there, first of all thank you so much for making these intuitive explanations, you present all concepts in such an intruiging manner! Secondly, can you please elaborate, why we can assume C1 continuity on the wave function, but not any higher than that?

[@aneesrehman6004]

Pl send part2 it is nice video

[@manaoharsam4211]

Nice.

[@johnkubala2772]

Do the energy levels of the states inside the well represent potential energy or kinetic energy?

[@JordanEdmundsEECS]

The kinetic energy in quantum mechanics is defined in terms of an operator (the p^2/2m part), so if you take the potential to be zero inside the well, the energy is all kinetic.

[@yuchuanwei]

Jordan, do you have videos about the boundary continuous condition? if not, guide me to somewhere else.

[@yuchuanwei]

i means in case the effective mass are different inside and outside the well.

[@brucelu4782]

V(x)-E is always negative in QW? if E>V(x) always true inside QW would the electron escape the QW?

[@brucelu4782]

I know the results of the derivation is correct, but this assumption puzzles me.

[@TheCoachjoel]

Inside the quantum well, the potential is considered to be V(x)=0. Then E>V(x) and the term V(x)-E is negative. Outside the well, them term V(x)-E is positive. That is V(x) > E, then the electrons cannot escape the walls, i.e. they are 'bound'.

[@LL-mq7gj]

How does one learn how to know what the general solutions are?

[@jacobharris5894]

A standard differential equations book will explain how to find a general solution to a ordinary differential equation. If you haven't read any, I recommend A First Course in Differential Equation with Modeling Applications by Denis G. Zill. For linear differential equations it's analogous to linear algebra if you've studied it before. For a vector space like R^n, for example, the vector space is n dimensional and you need n basis vectors in order to describe it. For the three dimensional vector space R^3 one set of basis vectors you can have is i hat, j hat and k hat or x hat, y hat and z hat if you prefer. But for any vector space there are an infinite number of sets of basis vectors you can choose. However once you pick one, any vector in that vector space can be written as a linear combination of those vectors. So for an arbitrary position vector, r, in say R^3, r = xxhat+yyhat+zzhat. A necessary property for a set of basis vectors is that they are linearly independent which means you can not write one of the basis vectors as a sum of the other basis vectors. So, for example, if you have some set of vectors, {e1,e2,e3} and e1=ae2+be2 where a and b are nonzero, that set is not linearly independent, it's linearly dependent. To generalize, the set is linearly independent if ae1+be2+ce3=0 if and only if a=b=c=0. Linear ordinary differential equations work in a similar way but instead of having basis vectors you have what you can call basis functions. The highest order derivative in the differential equation is called the order of the differential equations and this tells you how many basis functions you will need for your general solution. It's analogous to the dimension of a vector space. If you have an nth order differential equation, you have found n functions that satisfy the differential, and you have confirmed they are linearly independent then you can simply write your general solution as a linear combination of all these functions. (For a easy way to check if functions are linearly independent it might be worth looking up the Wronskian. It comes from linear algebra but all you really need to know is how to compute them which isn't too difficult.) It doesn't matter what set of basis functions you end up with. As long as they are linearly independent and each satisfy the differential equation then it's linear combination can describe any possible function that satisfies it. How you actually find these functions will depend on the differential equation. Some differential equations will have a method or multiple methods to systematically solve for them. If all else fails or it's just easier, you can guess a function and check if it salsifies the differential equation. The method of undetermined coefficients is a method to make better guesses and is usually faster than the more systematic method, variation of parameters. For nonlinear differential equations finding the general solution analytically depends on the equation and sometimes it simply isn't possible. The Schrodinger equation is fortunately linear however. Any differential equation that takes the form a0(x)y+a1(x)y'+a2(x)y''+...+an(x)y^(n) = b(x) is linear. Linear partial differential equations take on a similar form. The most common way to solve pde's for physicists and engineers is separation of variables which assumes that a function, f(x1,x2,x3,...,xn), that salsifies the pde can be written as a product of functions for each variable. f(x1,x2,x3,...xn)=f1(x1)*f2(x2)*f3(x3)*...*fn(xn). Substituting this back into the pde will allow you to break up the pde into a system of ode's which is exactly what's done for the Schrodinger equation. Once you solve for each of the functions you just multiply them together. The general solution, like before, is the sum of all basis function. In this case there are an infinite number of basis functions or products of functions that satisfy the pde, so the general solution becomes an infinite sum. This is a very powerful technique and can usually be done if the pde is linear and homogeneous.

[@clashultimate1339]

India

[@arnavagarwal7871]

feeble audio

[@JordanEdmundsEECS]

Thanks! I have since fixed this.

[@arnavagarwal7871]

@@JordanEdmundsEECS 👍