- #1

uxioq99

- 11

- 4

- Homework Statement:
- Determine a wave function as a column vector of two states that solves the Schrodinger equation given by the Hamiltonian ##H=-\omega S_y##.

- Relevant Equations:
- ##H=-\omega S_y##

By the statement of the question, a solution must take the form

##\begin{pmatrix} \Psi_1 \\ \Psi_2 \end{pmatrix}##

and the energy operator will be as per usual ##\hat{E} = i\hbar \frac{\partial}{\partial t}##. I am confused by the fact that

##S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}##

because the Schrodinger equation implies

##\begin{pmatrix}\frac{\partial \Psi_1}{\partial t}\\\frac{\partial \Psi_2}{\partial t}\end{pmatrix}=\frac{\omega}{2}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}

\begin{pmatrix}

\Psi_1\\

\Psi_2

\end{pmatrix}

##

So that exponential solutions as opposed to oscillatory solutions are observed.

##

\begin{pmatrix}

\Psi_1(x) \\

\Psi_2(x)

\end{pmatrix}

=

\frac{1}{\sqrt{e^{\omega/2 t} + e^{-\omega/2 t}}}

\begin{pmatrix}

e^{-\omega/2 t} \\

e^{\omega/2 t}

\end{pmatrix}

##

Is this really a physically possible solution? Is it akin to when the energy of a bound state in a finite square well is below the depth of the well leading to an exponential decay in the solution? Now the decay appears to be in time, suggesting that the final state of the system will simply be ##\Psi_1 = 0## and ##\Psi_2 = 1##.

In general, if ##A \in \text{GL}_2(\mathbb R)## and

##

i\hbar\begin{pmatrix}

\frac{\partial \Psi_1}{\partial t}\\

\frac{\partial \Psi_2}{\partial t}

\end{pmatrix}=

i\hbar

\begin{pmatrix}a & b \\ c & d\end{pmatrix}

\begin{pmatrix}

\Psi_1\\

\Psi_2

\end{pmatrix}

##

Could one perform an eigendecomposition to exponentiate the spectrum of the matrix? The corresponding characteristic polynomial will be

##\lambda^2 - \lambda(a + d) + |A| = 0##

yielding

##\lambda_\pm = \frac{a+d}{2} \pm \sqrt{\frac{a+d}{2}^2 - |A|}##

The corresponding eigenvector matrix ##S## is too tedious to include here, but can be readily solved from the expression for the eigenvalues. Does that imply then, that the general solution is

##\begin{pmatrix}\Psi_1 \\ \Psi_2 \end{pmatrix} = S e^{\Lambda t} S^{-1}##

where ##\Lambda## is the diagonal matrix formed from ##\lambda_+## and ##\lambda_-##?

For either of these evolution equations, how would somebody specify the initial conditions so that the spin is taken along some vector in ##\mathbb{R}^3##?

##\begin{pmatrix} \Psi_1 \\ \Psi_2 \end{pmatrix}##

and the energy operator will be as per usual ##\hat{E} = i\hbar \frac{\partial}{\partial t}##. I am confused by the fact that

##S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}##

because the Schrodinger equation implies

##\begin{pmatrix}\frac{\partial \Psi_1}{\partial t}\\\frac{\partial \Psi_2}{\partial t}\end{pmatrix}=\frac{\omega}{2}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}

\begin{pmatrix}

\Psi_1\\

\Psi_2

\end{pmatrix}

##

So that exponential solutions as opposed to oscillatory solutions are observed.

##

\begin{pmatrix}

\Psi_1(x) \\

\Psi_2(x)

\end{pmatrix}

=

\frac{1}{\sqrt{e^{\omega/2 t} + e^{-\omega/2 t}}}

\begin{pmatrix}

e^{-\omega/2 t} \\

e^{\omega/2 t}

\end{pmatrix}

##

Is this really a physically possible solution? Is it akin to when the energy of a bound state in a finite square well is below the depth of the well leading to an exponential decay in the solution? Now the decay appears to be in time, suggesting that the final state of the system will simply be ##\Psi_1 = 0## and ##\Psi_2 = 1##.

In general, if ##A \in \text{GL}_2(\mathbb R)## and

##

i\hbar\begin{pmatrix}

\frac{\partial \Psi_1}{\partial t}\\

\frac{\partial \Psi_2}{\partial t}

\end{pmatrix}=

i\hbar

\begin{pmatrix}a & b \\ c & d\end{pmatrix}

\begin{pmatrix}

\Psi_1\\

\Psi_2

\end{pmatrix}

##

Could one perform an eigendecomposition to exponentiate the spectrum of the matrix? The corresponding characteristic polynomial will be

##\lambda^2 - \lambda(a + d) + |A| = 0##

yielding

##\lambda_\pm = \frac{a+d}{2} \pm \sqrt{\frac{a+d}{2}^2 - |A|}##

The corresponding eigenvector matrix ##S## is too tedious to include here, but can be readily solved from the expression for the eigenvalues. Does that imply then, that the general solution is

##\begin{pmatrix}\Psi_1 \\ \Psi_2 \end{pmatrix} = S e^{\Lambda t} S^{-1}##

where ##\Lambda## is the diagonal matrix formed from ##\lambda_+## and ##\lambda_-##?

For either of these evolution equations, how would somebody specify the initial conditions so that the spin is taken along some vector in ##\mathbb{R}^3##?

Last edited: