- #1
dark_matter_is_neat
- 34
- 1
- Homework Statement
- Consider an infinite square well in one dimension extending from -a/2 to a/2. A particle of mass m is sitting in the ground state. At time t = 0 the size of the well doubles so it now goes from -a to a. If one then measures the energy, what is the most probable result and what is its probability.
- Relevant Equations
- E##\Psi(x)## = ##-\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi(x)}{\partial x^{2}}##
For wave function of the initial well:
##\Psi(\frac{a}{2}) = \Psi(-\frac{a}{2}) = 0##
For the wave function of the final square well:
##\Psi(a) = \Psi(-a) = 0##
Going through the schrodinger wave equation, ##-\frac{2mE}{\hbar^{2}}\Psi(x) = \frac{\partial^{2} \Psi(x)}{\partial x^{2}}##, so ##Psi(x) = C_{1}sin(\frac{\sqrt{2mE}}{\hbar}x) + C_{2}cos(\frac{\sqrt{2mE}}{\hbar}x)##.
Enforcing the boundary conditions:
##cos(\frac{\sqrt{2mE}}{\hbar} \frac{a}{2}) = 0##, so ##\frac{\sqrt(2mE)}{\hbar} \frac{a}{2} = n \frac{\pi}{2}## for odd n.
##sin(\frac{\sqrt{2mE}}{\hbar} \frac{a}{2}) = 0##, so ##\frac{\sqrt(2mE)}{\hbar} \frac{a}{2} = n \pi ##.
So this leads to:
##\Psi(x) = C_{2}cos(\frac{n \pi x}{a})## for odd n
##\Psi(x) = C_{1}sin(\frac{n \pi x}{a})## for even n
##E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2 m a^{2}}##
Enforcing normalization, ##C_{1} = C_{2} = \sqrt{\frac{2}{a}}##.
The wave function for the expanded well is identical except you replace a with 2a.
My first thought for what would be the most probable energy, was the energy of the ground state of the initial square well, ##\frac{\pi^{2} \hbar^{2}}{2 m a^{2}}## which corresponds to the n = 2 state for the widened well , however when I calculate this probability I get:
P = ##|\int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sqrt{2}}{a} cos(\frac{\pi x}{a}) sin(\frac{\pi x}{a}) dx|^{2}## which is obviously 0.
I believe the actual most probable energy is the ground state energy of the new well, which has probability:
P = ##|\int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sqrt{2}}{a} cos(\frac{\pi x }{a}) cos(\frac{\pi x}{2a}) dx|^{2}## = ##\frac{64}{9 \pi^{2}}##
Which I've checked against the probabilities for n>1 up to n = 3 and it had the highest probability.
What confuses me about this answer is if I were to look at a physically identical well that instead starts at 0 (so the first well would go from 0 to a and the widened well would go from 0 to 2a), I would get that the probability of E = ##\frac{\pi^{2} \hbar^{2}}{2ma^{2}}## to be non-zero and in fact be the most probable energy.
These to wells are identical, I just changed my coordinate system, but I end up getting two different results for the most probable energy. I find this really strange, shouldn't the system be translationally invariant? What is going on here? Did I just make a mistake in my calculations or is there something else I'm not seeing?
Enforcing the boundary conditions:
##cos(\frac{\sqrt{2mE}}{\hbar} \frac{a}{2}) = 0##, so ##\frac{\sqrt(2mE)}{\hbar} \frac{a}{2} = n \frac{\pi}{2}## for odd n.
##sin(\frac{\sqrt{2mE}}{\hbar} \frac{a}{2}) = 0##, so ##\frac{\sqrt(2mE)}{\hbar} \frac{a}{2} = n \pi ##.
So this leads to:
##\Psi(x) = C_{2}cos(\frac{n \pi x}{a})## for odd n
##\Psi(x) = C_{1}sin(\frac{n \pi x}{a})## for even n
##E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2 m a^{2}}##
Enforcing normalization, ##C_{1} = C_{2} = \sqrt{\frac{2}{a}}##.
The wave function for the expanded well is identical except you replace a with 2a.
My first thought for what would be the most probable energy, was the energy of the ground state of the initial square well, ##\frac{\pi^{2} \hbar^{2}}{2 m a^{2}}## which corresponds to the n = 2 state for the widened well , however when I calculate this probability I get:
P = ##|\int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sqrt{2}}{a} cos(\frac{\pi x}{a}) sin(\frac{\pi x}{a}) dx|^{2}## which is obviously 0.
I believe the actual most probable energy is the ground state energy of the new well, which has probability:
P = ##|\int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sqrt{2}}{a} cos(\frac{\pi x }{a}) cos(\frac{\pi x}{2a}) dx|^{2}## = ##\frac{64}{9 \pi^{2}}##
Which I've checked against the probabilities for n>1 up to n = 3 and it had the highest probability.
What confuses me about this answer is if I were to look at a physically identical well that instead starts at 0 (so the first well would go from 0 to a and the widened well would go from 0 to 2a), I would get that the probability of E = ##\frac{\pi^{2} \hbar^{2}}{2ma^{2}}## to be non-zero and in fact be the most probable energy.
These to wells are identical, I just changed my coordinate system, but I end up getting two different results for the most probable energy. I find this really strange, shouldn't the system be translationally invariant? What is going on here? Did I just make a mistake in my calculations or is there something else I'm not seeing?
