# Defining the Schrodinger Equation for Pauli Spin Matrix along the y-axis

uxioq99
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##?

Last edited:

Homework Helper
2022 Award
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}##

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}
##

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}\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}##
What happened to the ##\Psi## column vector on the Right Hand Side?

• topsquark and uxioq99
uxioq99
Thanks, I added it back into the equation, right at the end of the editing window.

Homework Helper
2022 Award
The rest is just incorrect. The solution you write down is not a solution. The exponent is not dimensionless and then it won't work anyway . At that point I stopped.
Get a copy of any good quantum text and work it through. Give it another attempt.
It is not nonsense......just full of mistakes Might be good to number your equations.

• topsquark
uxioq99
@hutchphd Sorry, I realized when I typed "\exp(-\omega/2t)" it produced ##\exp(-\omega/2t)## instead of ##\exp(-\frac{\omega}{2} t)## Wouldn't the units of ##\omega## cancel those of time given that it is an angular frequency? Would

##
\begin{pmatrix}
\Psi_1(x,t) \\
\Psi_2(x,t)
\end{pmatrix}
=
\begin{pmatrix}
\exp(\frac{\omega}{2} t) \\
\exp(-\frac{\omega}{2} t)
\end{pmatrix}
##

Be a valid solution of ##E\vec\Psi = H \vec \Psi## where ##H = -\omega S_y##?

Wouldn't

##
-i\hbar
\begin{pmatrix}
\frac{\partial \Psi_1}{\partial t} \\
\frac{\partial \Psi_2}{\partial t}
\end{pmatrix} =
-\frac{\hbar \omega}{2} \begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}
\begin{pmatrix}
\Psi_1 (x,t) \\
\Psi_2 (x,t)
\end{pmatrix}
##

because

##
-i\hbar \frac{\partial}{\partial t} \left(e^{\pm\frac{\omega}{2} t}\right) =
\mp i \frac{\hbar \omega}{2} e^{\pm\frac{\omega}{2} t}
##

and

##
-\frac{\hbar \omega}{2}
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}
\begin{pmatrix}
e^{-\frac{\omega}{2} t} \\
e^{\frac{\omega}{2} t}
\end{pmatrix} =
i\frac{\hbar \omega}{2}
\begin{pmatrix}
-e^{\frac{\omega}{2} t} \\
e^{-\frac{\omega}{2} t}
\end{pmatrix}
##

I believe that a solution to the Schrodinger equation cannot always be real, so my selected form for ##\Psi(x,t)## does not appear to be correct. Additionally, the normalization factor I mentioned earlier

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

is time dependent. I know that I neglected the appropriate derivatives in the energy operator, but I was unsure of what else to do.

When I consider the similar problem for ##H = -\omega S_z##, I had that the Hamiltonian became

##
H = -\frac{\hbar \omega}{2} \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
##

##
\begin{pmatrix}
\Psi_1(x,t) \\
\Psi_2(x,t)
\end{pmatrix} =
\frac{1}{\sqrt{2}}\begin{pmatrix}
e^{i\frac{\omega}{2} t} \\
e^{-i\frac{\omega}{2} t}
\end{pmatrix}
##

The normalization ##\frac{1}{\sqrt{2}}## remained constant in time because of the presence of these oscillators. I am confused by the presence of the imaginary unit ##i## in the ##S_y##. It must be present by definition, but it cancels the ##i## in the energy operator ##E = i \hbar \frac{\partial}{\partial t}##. Then, the solutions appear to be exponential instead of oscillatory. This fact conflicted my intuition because -- as said previously -- the wavefunction is real and does not appear to be easily normalizable.

Also regarding my generalization to any 2 by 2 matrix. I am aware that the Pauli matrices span ##\mathbb{C}^{2\times 2}## and so the general solution could be written by decomposing the Hamiltonian into these constituent matrices. I had hoped that an alternate characterization in purely the language of introductory linear algebra could be fashioned by examining the equation as a system of ODEs and considering the spectrum. If
##
\frac{d\vec x}{dt} = A \vec x(t)
##
then
##
\vec x (t) = S\exp(\Lambda t) S^{-1} \vec x_0
##
by considering ##\exp## as a matrix power series.

Last edited:
Homework Helper
Gold Member
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}
##
On the right side, check to see if you missed an overall minus sign.

If you multiply out the right hand side, what column vector do you get?

• uxioq99
uxioq99
@TSny Thank you, dividing by ##i## introduces the negative sign. I make more mistakes when I type things in Latex. Is the solution still supposed to be exponential, or should I have instead looked for a oscillatory behavior? I don't see how that is possible, given that the matrix defining the system becomes real. (Unless, I forgot an ##i## in addition to a negative sign.) My solution appears wrong because it no longer carries a phase.

Homework Helper
Gold Member
You shouldn't get exponentially growing or damping solutions.

What do you get after performing the matrix multiplication on the right-hand side of the equation below?

##\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}##

• hutchphd
uxioq99
I'm quite embarrassed. The values are on the off-diagonal. I apologize for asking this in the first place. I've been studying too long and now am just making stupid mistakes. Even by my standards, this was an incredible mistake -- sorry.

• hutchphd
Homework Helper
Gold Member
No problem. I’m glad it’s resolved.

• hutchphd