# Matrix mechanics

If you have a Hamiltonian

$H =\begin{pmatrix} 2 & 5i\\ -5i & 2 \end{pmatrix}$

and a wave function

$|\psi> =\begin{pmatrix} i\\ 1 \end{pmatrix}$

describing a particle at t=0, will the state of the particle modify at a latter time t>0?

If you have an operator described at $t=0$ by the matrix

$\hat A =\begin{pmatrix} a & 0\\ 0 & 3a \end{pmatrix}$

and you make a measurement of the observable described by this operator at t=0 you obtain a value of 3a. Will this value of the measurement of this observable be the same or different at a latter time $t>0$?

a)

If the state of the particle is described at t=0 by the wave function

$|\psi>= |\psi(0)>= \begin{pmatrix} i\\ 1 \end{pmatrix}$

that satisfies the time independent Schrodinger equation

$H|\psi(0)> =E|\psi(0)>$   or on components $H_{n m}\psi_m =E_m\Psi_m$
then at a latter time t>0 the wave function of the same particle is defined as

$|\psi(t)>= e^{-i H t}*|\psi(0)>$ or on components $|\psi(t)> =e^{-i E_mt}*|\psi(0)>$

because the time dependent Schrodinger equation needs to be also satisfied:

$\frac {d|\psi(t)>}{d t} =-i H|\psi(t)>$

This means that the state of the particle at t>0 is not preserved but it is modified with a factor $e^{-i H t}$

b) If you have an operator

$\hat A(0) =(A_{n m}(0))$  at a latter time the definition of the operator is

$\hat A(t) =e^{i H t}\hat A(0)*e^{-i H t}$ or on components $A_{m n}(t) = e^{i E_mt}A_{m n}(0)*e^{-i E_n t}$

When measuring the observable A at a time t>0 we have

$\hat A(t)*|\psi(t)> =e^{i H t}\hat A(0)*e^{-i H t}*e{-i H t}*|\psi(0)> =\hat A(0)|\psi(t)>$

and because for any given wave function $|\psi>$  we have

$\hat A(0)|\psi> =3a*|\psi>$

we can say that the result of the measurement is the same $3a$ at a latter time t>0.