# Finding unitary operator associated with a given Hamiltonian

• ubergewehr273
In summary, the equation for the unitary operator in the Hamiltonian is easy to compute provided the Hamiltonian is diagonalized.

#### ubergewehr273

Homework Statement
Given a Hamiltonian ##H = \hbar \omega \sigma_1##, you are supposed to find the associated time-evolution unitary operator ##U(t)##.
Relevant Equations
Time-independent Schrodinger's equation

$$-i \hbar \frac{\partial | \psi (t) \rangle}{\partial t} = \hat{H} | \psi (t) \rangle$$

The associated unitary operator is

$$U(t) = \exp (\frac{-i \hat{H} t}{\hbar})$$
Now from the relevant equations,
$$U(t) = \exp(-i \omega \sigma_1 t)$$

which is easy to compute provided the Hamiltonian is diagonalized. Writing ##\sigma_1## in its eigenbasis, we get

$$\sigma_1 = \begin{pmatrix} 1 & 0\\ 0 & -1\\ \end{pmatrix}$$

and hence the unitary ##U(t)## becomes

$$U(t) = \begin{pmatrix} e^{-i \omega t} & 0\\ 0 & e^{i \omega t}\\ \end{pmatrix}$$

Mind you that the above representation of $U(t)$ is in the basis ##\{ |+\rangle, |-\rangle\}## where

$$|+\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1\\ \end{pmatrix}$$

$$|-\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ -1\\ \end{pmatrix}$$

Now, I need to write ##U(t)## back in the original basis ##\{|0\rangle, |1\rangle\}## (which is where I'm facing an issue). Finding the components of the above expression for ##U(t)## in the original basis,

$$\langle 0 | U(t) | 0 \rangle = e^{-i \omega t} \qquad \langle 1 | U(t) | 1 \rangle = e^{i \omega t}$$

with ##\langle 0 | U(t) | 1 \rangle = 0## and ##\langle 1 | U(t) | 0 \rangle = 0##. This gives me the exact same matrix representation in the original basis. Obviously this is not true and I'm doing something wrong.

Last edited:
For inline Latex you need to use double hashes rather than single dollars.

ubergewehr273 said:
Finding the components of the above expression for ##U(t)## in the original basis,

$$\langle 0 | U(t) | 0 \rangle = e^{-i \omega t} \qquad \langle 1 | U(t) | 1 \rangle = e^{i \omega t}$$

with ##\langle 0 | U(t) | 1 \rangle = 0## and ##\langle 1 | U(t) | 0 \rangle = 0##. This gives me the exact same matrix representation in the original basis. Obviously this is not true and I'm doing something wrong.
You'll have to show your work, for instance ##\langle 0 | U(t) | 0 \rangle##, because the result you get is incorrect.

You can write your solution in terms of
$$\hat{U}=\exp(-\mathrm{i} \omega t) |+ \rangle \langle +| + \exp(\mathrm{i} \omega t) |- \rangle \langle -|.$$
Now simply express ##|+ \rangle## and ##|- \rangle## as linear combinations of ##|1 \rangle## and ##|0 \rangle## and multiply out the dyadic products. Then you can read out the matrix elements easily.

A much easier way is to directly use the properties of the Pauli matrix. All you need is ##\hat{\sigma}_1^2=1## and then write down the power series for the exponential function.