CJ Isham, Lectures on Quantum Theory, Gap Filled: p. 124

Mark Vuletic

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Source:

CJ Isham. 1995. Lectures on Quantum Theory: Mathematical and Structural Foundations. London: Imperial College. p. 124

Problem:

In the "alternative method" section on p. 124, Isham gets us to \( |\psi_t\rangle = \frac{1}{2}e^{-it(1+\alpha)/\hbar}\begin{pmatrix}1\\i\end{pmatrix} + \frac{1}{2}e^{-it(1-\alpha)/\hbar}\begin{pmatrix}1\\-i\end{pmatrix} \) and then says that we "thus" have \( |\psi_t\rangle = e^{-it/\hbar} \begin{pmatrix}\cos(\frac{\alpha t}{\hbar}) \\ \sin(\frac{\alpha t}{\hbar})\end{pmatrix} \). I am unpacking the word "thus" into steps:

Solution:

$$ |\psi_t\rangle = \frac{1}{2}e^{-it(1+\alpha)/\hbar}\begin{pmatrix}1\\i\end{pmatrix} + \frac{1}{2}e^{-it(1-\alpha)/\hbar}\begin{pmatrix}1\\-i\end{pmatrix} $$

.

$$ = \frac{1}{2} \left[ e^{-it(1+\alpha)/\hbar}\begin{pmatrix}1\\i\end{pmatrix} + e^{-it(1-\alpha)/\hbar}\begin{pmatrix}1\\-i\end{pmatrix} \right] $$

$$ = \frac{1}{2} \left[ \begin{pmatrix}e^{-it(1+\alpha)/\hbar}\\ie^{-it(1+\alpha)/\hbar}\end{pmatrix} + \begin{pmatrix}e^{-it(1-\alpha)/\hbar}\\-ie^{-it(1-\alpha)/\hbar}\end{pmatrix} \right] $$

$$ = \frac{1}{2} \begin{pmatrix}e^{-it(1+\alpha)/\hbar}+e^{-it(1-\alpha)/\hbar}\\ie^{-it(1+\alpha)/\hbar} -ie^{-it(1-\alpha)/\hbar}\end{pmatrix} $$

Using \( e^{a+b} = e^ae^b \) :

$$ = \frac{1}{2} \begin{pmatrix}e^{-it/\hbar}e^{-it\alpha/\hbar}+e^{-it/\hbar}e^{it \alpha/\hbar}\\ie^{-it/\hbar}e^{-it\alpha/\hbar} -ie^{-it/\hbar}e^{it\alpha/\hbar}\end{pmatrix}$$

$$ = \frac{1}{2} e^{-it/\hbar} \begin{pmatrix}e^{-it\alpha/\hbar}+e^{it \alpha/\hbar}\\ie^{-it\alpha/\hbar} -ie^{it\alpha/\hbar}\end{pmatrix}$$

At this point, you can use Euler's relations \( \frac{e^{ix} + e^{-ix}}{2} = \cos{x} \) and \( \frac{e^{ix} - e^{-ix}}{2i} = \sin{x} \) to get Isham's result almost directly (you have to multiply the second row of the matrix by \( i/i \) first). If you do not know the relations, you can follow the next steps:

Using \( e^{i\theta} = \cos\theta + i\sin\theta \) and \( e^{-i\theta} = \cos\theta - i\sin\theta \) :

$$ = \frac{1}{2}e^{-it/\hbar} \begin{pmatrix}\cos(t\alpha/\hbar)-i\sin(t\alpha/\hbar) + \cos(t\alpha/\hbar) + i\sin(t\alpha/\hbar) \\ i[\cos(t\alpha/\hbar)-i\sin(t\alpha/\hbar) - \cos(t\alpha/\hbar) - i\sin(t\alpha/\hbar))]\end{pmatrix}$$

$$ = \frac{1}{2}e^{-it/\hbar} \begin{pmatrix}2\cos(t\alpha/\hbar) \\ i(-2i\sin(t\alpha/\hbar))\end{pmatrix}$$

$$ = \frac{1}{2}e^{-it/\hbar} \begin{pmatrix}2\cos(t\alpha/\hbar) \\ 2\sin(t\alpha/\hbar)\end{pmatrix}$$

$$ = e^{-it/\hbar} \begin{pmatrix}\cos(t\alpha/\hbar) \\ \sin(t\alpha/\hbar)\end{pmatrix}$$

The last line is Isham's result.

Last updated: 9 Nov 2014

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