Finding unitary operator associated with a given Hamiltonian

[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.

 

[PeroK]

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

 

[DrClaude]

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.

 

[vanhees71]

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.