How Does the Delta Emerge in the Time Evolution Operator's Exponential Form?

[Gregg]

If you have some Hamiltonian represented by a 2x2 matrix

$H = \left( \begin{array}{cc} 0 & \Delta \\ \Delta & 0 \end{array} \right)$

And you want to use the time evolution operator

$U = \exp ( - \frac{i}{\hbar} H t )$

it says that

$U = \exp (- \frac{i \Delta}{\hbar} t)$

Why is this?
How did the $\Delta$ get out?

 

[Gregg]

The was an error in the latex.

Is it just true that

$e^A | \psi_\alpha > e^{\alpha} | \psi_{\alpha} >$

Where

$A | \psi_\alpha>= \alpha |\psi_{\alpha}>$?

 

[Steely Dan]

The was an error in the latex.

Is it just true that

$e^A | \psi_\alpha > e^{\alpha} | \psi_{\alpha} >$

Where

$A | \psi_\alpha>= \alpha |\psi_{\alpha}>$?

The exponential of a matrix is really just short hand for the power series representing the exponential function, $$e^A = \sum_{n=0}^\infty \frac{A^n}{n!}.$$
So if you apply this power series operator to the wave function, you should see that in fact yes, what you've said is true.

 

[dextercioby]

Just a few LaTex hints: use \left( and \right) before brackets containing fractions and \langle and \rangle for bra-ket notation. (No bold in the LaTex code, of course)

As for
U=exp(−iΔt/ℏ)

in your initial post, it's incomplete, it misses the unit matrix 2x2 in the RHS.

 

[Gregg]

The exponential of a matrix is really just short hand for the power series representing the exponential function, $$e^A = \sum_{n=0}^\infty \frac{A^n}{n!}.$$
So if you apply this power series operator to the wave function, you should see that in fact yes, what you've said is true.

Thanks that's so obvious now