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modphysnoob
 3 years ago
1. Consider an infinite square well of width 2L centered
at x = 0; i.e.,
V(x) = for x L
V(x) = 0 for –L < x < L
V(x) = for x L
a) Using the general solution,y(x) = Asin kx + Bcos kx ,
for an infinite square well, find the normalized ground
state wave function 1(x) and its energy E1.
b) Repeat part (a) for the first excited state 2(x).
2.
modphysnoob
 3 years ago
1. Consider an infinite square well of width 2L centered at x = 0; i.e., V(x) = for x L V(x) = 0 for –L < x < L V(x) = for x L a) Using the general solution,y(x) = Asin kx + Bcos kx , for an infinite square well, find the normalized ground state wave function 1(x) and its energy E1. b) Repeat part (a) for the first excited state 2(x). 2.

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Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0The Schrodinger equation for the infinite square well is \[ \frac{\hbar^2}{2m} y''(x) = E y(x)\] or \[y''(x) = \frac{2mE}{\hbar^2}y(x) = k^2y(x)\] where \[k = \sqrt{\frac{2mE}{\hbar^2}} \] As mentioned in the problem, the most general solution to this problem is \[y(x) = A \sin(kx) + B\cos(kx) \] where A and B are constants that we haven't found yet. From here, we must apply our boundary conditions. From the fact that the potential is infinite outside of the well, it must be that y(L) = y(L) = 0: \[y(L) = A \sin(k\cdot L) + B\cos(k\cdot L) =A\sin(k\cdot L) + B\cos(k \cdot L) \] since sin(x) = sin(x) and cos(x) = cos(x). Also, at x= L, \[y(L) = A\sin(k\cdot L) + B\cos(k \cdot L) \] Notice for this to be true, A must be zero because there's no other way for these two expressions to be the same. Therefore, you can revise your wave function to be \[y(x) = B\cos(kx) \] now, there are two ways for y(L) = y(L) = 0. First, B could be zero ... but that would be silly, because then our function would just be zero everywhere and that's not a good choice. The other possibility is for the cosine function to be zero. Now, you know that cos(x) = 0 at pi/2, 3pi/2, 5pi/2, etc... So this means that \[ kL = (2n1)\frac{\pi}{2} \] where n = 1, 2, 3....... etc But this tells us what k may be. It appears that \[ k = \frac{(2n1)\pi}{2L} \] so our wave function becomes \[y(x) = B\cos\left( \frac{(2n1)\pi x}{2L}\right) \] And our allowed energies are \[E_n = \frac{\hbar^2 k^2}{2m} = \frac{(2n1)^2 \pi^2 \hbar^2}{4mL^2}\] Our last step is to normalize our wavefunction  that is, make sure that \[ \int_{L}^L y_n(x)^2 dx = 1\] so we have \[\int_{L}^L B^2 \cos^2\left(\frac{(2n1)\pi x}{2L}\right) dx = B^2L = 1\] so \[B = \frac{1}{\sqrt{L}} \] and at long last, we have our completely normalized wavefunctions: \[ y_n(x) = \frac{1}{\sqrt{L}} \cos\left(\frac{(2n1)\pi x}{2L} \right)\] and allowed energies: \[E_n = \frac{(2n1)^2\pi^2 \hbar^2}{4mL^2} \] The ground state ( lowest energy) is n=1. n=2 is the first excited state, etc etc etc.
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