Question:
What is the probability that the particle in the ground state of a 1 dimensional delta potential get to an excited state when the potential suddenly double its strength?
Solution:
I attempted the exam before I watch week 8 lecture 16, and the approach to this problem is discussed there. I am really thrilled because I used exactly the same procedure as the lecture told us to without watching it first!
The key idea is that the wave function is unchanged when the potential suddenly change, but then it suddenly becomes a linear combination of eigen functions in the doubled potential. To find out if it is excited, we compute the probability that it is still in the ground state using the inner product.
The probability that it is still in the ground state is $ (\langle \psi_{0, \alpha} | \psi_{0, 2\alpha} \rangle )^2 $.
We are given that
$ \psi_{0,\alpha}(x) = \frac{\sqrt{m\alpha}}{\hbar}\exp\left(-\frac{m\alpha|x|}{\hbar^2}\right) $
So the required integral can be computed as:
$ \begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{\frac{\sqrt{m\alpha}}{\hbar}\exp\left(-\frac{m\alpha|x|}{\hbar^2}\right)\frac{\sqrt{2m\alpha}}{\hbar}\exp\left(-\frac{2m\alpha|x|} {\hbar^2}\right)dx} \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\int\limits_{-\infty}^{\infty}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx} \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\int\limits_{-\infty}^{0}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx} + \int\limits_{0}^{\infty}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\int\limits_{-\infty}^{0}{\exp\left(\frac{3m\alpha x}{\hbar^2}\right)dx} + \int\limits_{0}^{\infty}{\exp\left(-\frac{3m\alpha x}{\hbar^2}\right)dx}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\frac{\hbar^2}{3m\alpha}\exp\left(\frac{3m\alpha x}{\hbar^2}\right)|_{-\infty}^{0} + \frac{-\hbar^2}{3m\alpha}\exp\left(-\frac{3m\alpha x}{\hbar^2}\right)|_{0}^{\infty}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\frac{2\hbar^2}{3m\alpha} \\ &=& \frac{2\sqrt{2}}{3} \\ \end{eqnarray*} $
So the probability that the particle stay in ground state is $ \left(\frac{2\sqrt{2}}{3}\right)^2 = \frac{8}{9} $, and the probability that it get excited is $ \frac{1}{9} $.
What is the probability that the particle in the ground state of a 1 dimensional delta potential get to an excited state when the potential suddenly double its strength?
Solution:
I attempted the exam before I watch week 8 lecture 16, and the approach to this problem is discussed there. I am really thrilled because I used exactly the same procedure as the lecture told us to without watching it first!
The key idea is that the wave function is unchanged when the potential suddenly change, but then it suddenly becomes a linear combination of eigen functions in the doubled potential. To find out if it is excited, we compute the probability that it is still in the ground state using the inner product.
The probability that it is still in the ground state is $ (\langle \psi_{0, \alpha} | \psi_{0, 2\alpha} \rangle )^2 $.
We are given that
$ \psi_{0,\alpha}(x) = \frac{\sqrt{m\alpha}}{\hbar}\exp\left(-\frac{m\alpha|x|}{\hbar^2}\right) $
So the required integral can be computed as:
$ \begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{\frac{\sqrt{m\alpha}}{\hbar}\exp\left(-\frac{m\alpha|x|}{\hbar^2}\right)\frac{\sqrt{2m\alpha}}{\hbar}\exp\left(-\frac{2m\alpha|x|} {\hbar^2}\right)dx} \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\int\limits_{-\infty}^{\infty}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx} \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\int\limits_{-\infty}^{0}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx} + \int\limits_{0}^{\infty}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\int\limits_{-\infty}^{0}{\exp\left(\frac{3m\alpha x}{\hbar^2}\right)dx} + \int\limits_{0}^{\infty}{\exp\left(-\frac{3m\alpha x}{\hbar^2}\right)dx}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\frac{\hbar^2}{3m\alpha}\exp\left(\frac{3m\alpha x}{\hbar^2}\right)|_{-\infty}^{0} + \frac{-\hbar^2}{3m\alpha}\exp\left(-\frac{3m\alpha x}{\hbar^2}\right)|_{0}^{\infty}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\frac{2\hbar^2}{3m\alpha} \\ &=& \frac{2\sqrt{2}}{3} \\ \end{eqnarray*} $
So the probability that the particle stay in ground state is $ \left(\frac{2\sqrt{2}}{3}\right)^2 = \frac{8}{9} $, and the probability that it get excited is $ \frac{1}{9} $.
No comments:
Post a Comment