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## Question: Dynamics of non-Hermitian Hamiltonian ([1] pr. 2.2)

Revisiting an earlier Hamiltonian, but assuming it was entered incorrectly as

\begin{equation}\label{eqn:dynamicsNonHermitian:20}

H = H_{11} \ket{1}\bra{1}

+ H_{22} \ket{2}\bra{2}

+ H_{12} \ket{1}\bra{2}.

\end{equation}

What principle is now violated? Illustrate your point explicitly by attempting to solve the most generaqtl time-dependent problem using an illegal Hamiltonian of this kind. You may assume that \( H_{11} = H_{22} \) for simplicity.

## Answer

In matrix form this Hamiltonian is

\begin{equation}\label{eqn:dynamicsNonHermitian:40}

\begin{aligned}

H

&=

\begin{bmatrix}

\bra{1} H \ket{1} & \bra{1} H \ket{2} \\

\bra{2} H \ket{1} & \bra{2} H \ket{2} \\

\end{bmatrix} \\

&=

\begin{bmatrix}

H_{11} & H_{12} \\

0 & H_{22} \\

\end{bmatrix}.

\end{aligned}

\end{equation}

This is not a Hermitian operator. What is the physical implication of this non-Hermicity? Consider the simpler case where \( H_{11} = H_{22} \). Such a Hamiltonian has the form

\begin{equation}\label{eqn:dynamicsNonHermitian:60}

H =

\begin{bmatrix}

a & b \\

0 & a

\end{bmatrix}.

\end{equation}

This has only one unique eigenvector ( \( (1,0) \), but we can still solve the time evolution equation

\begin{equation}\label{eqn:dynamicsNonHermitian:80}

i \Hbar \PD{t}{U} = H U,

\end{equation}

since for constant \( H \), we have

\begin{equation}\label{eqn:dynamicsNonHermitian:100}

U = e^{-i H t/\Hbar}.

\end{equation}

To exponentiate, note that we have

\begin{equation}\label{eqn:dynamicsNonHermitian:120}

{\begin{bmatrix}

a & b \\

0 & a

\end{bmatrix}}^n

=

\begin{bmatrix}

a^n & n a^{n-1} b \\

0 & a^n

\end{bmatrix}.

\end{equation}

To prove the induction, the \( n = 2 \) case follows easily

\begin{equation}\label{eqn:dynamicsNonHermitian:140}

\begin{bmatrix}

a & b \\

0 & a

\end{bmatrix}

\begin{bmatrix}

a & b \\

0 & a

\end{bmatrix}

=

\begin{bmatrix}

a^2 & 2 a b \\

0 & a^2

\end{bmatrix},

\end{equation}

as does the general case

\begin{equation}\label{eqn:dynamicsNonHermitian:160}

\begin{bmatrix}

a^n & n a^{n-1} b \\

0 & a^n

\end{bmatrix}

\begin{bmatrix}

a & b \\

0 & a

\end{bmatrix}

=

\begin{bmatrix}

a^{n+1} & (n +1 ) a^{n} b \\

0 & a^{n+1}

\end{bmatrix}.

\end{equation}

The exponential sum is thus

\begin{equation}\label{eqn:dynamicsNonHermitian:180}

e^{H \tau}

=

\begin{bmatrix}

e^{a \tau} & 0 + \frac{b \tau}{1!} + \frac{2 a b \tau^2}{2!} + \frac{3 a^2 b \tau^3}{3!} + \cdots \\

0 & e^{a \tau}

\end{bmatrix}.

\end{equation}

That sum simplifies to

\begin{equation}\label{eqn:dynamicsNonHermitian:200}

\frac{b \tau}{0!} + \frac{a b \tau^2}{1!} + \frac{a^2 b \tau^3}{2!} + \cdots \\

=

b \tau \lr{ 1 + \frac{a \tau}{1!} + \frac{(a \tau)^2}{2!} + \cdots }

=

b \tau e^{a \tau}.

\end{equation}

The exponential is thus

\begin{equation}\label{eqn:dynamicsNonHermitian:220}

e^{H \tau}

=

\begin{bmatrix}

e^{a\tau} & b \tau e^{a\tau} \\

0 & e^{a\tau}

\end{bmatrix}

=

\begin{bmatrix}

1 & b \tau \\

0 & 1

\end{bmatrix}

e^{a\tau}.

\end{equation}

In particular

\begin{equation}\label{eqn:dynamicsNonHermitian:240}

U = e^{-i H t/\Hbar} =

\begin{bmatrix}

1 & -i b t/\Hbar \\

0 & 1

\end{bmatrix}

e^{-i a t /\Hbar }.

\end{equation}

We can verify that this is a solution to \ref{eqn:dynamicsNonHermitian:80}. The left hand side is

\begin{equation}\label{eqn:dynamicsNonHermitian:260}

\begin{aligned}

i \Hbar \PD{t}{U}

&=

i \Hbar

\begin{bmatrix}

-i a/\Hbar & -i b /\Hbar + (-i b t/\Hbar)(-i a/\Hbar) \\

0 & -i a /\Hbar

\end{bmatrix}

e^{-i a t /\Hbar } \\

&=

\begin{bmatrix}

a & b – i a b t/\Hbar \\

0 & a

\end{bmatrix}

e^{-i a t /\Hbar },

\end{aligned}

\end{equation}

and for the right hand side

\begin{equation}\label{eqn:dynamicsNonHermitian:280}

\begin{aligned}

H U

&=

\begin{bmatrix}

a & b \\

0 & a

\end{bmatrix}

\begin{bmatrix}

1 & -i b t/\Hbar \\

0 & 1

\end{bmatrix}

e^{-i a t /\Hbar } \\

&=

\begin{bmatrix}

a & b – i a b t/\Hbar \\

0 & a

\end{bmatrix}

e^{-i a t /\Hbar } \\

&=

i \Hbar \PD{t}{U}.

\end{aligned}

\end{equation}

While the Schr\”{o}dinger is satisfied, we don’t have the unitary invertion physical property that is desired for the time evolution operator \( U \). Namely

\begin{equation}\label{eqn:dynamicsNonHermitian:300}

\begin{aligned}

U^\dagger U

&=

\begin{bmatrix}

1 & 0 \\

i b t/\Hbar & 1

\end{bmatrix}

e^{i a t /\Hbar }

\begin{bmatrix}

1 & -i b t/\Hbar \\

0 & 1

\end{bmatrix}

e^{-i a t /\Hbar } \\

&=

\begin{bmatrix}

1 & -i b t/\Hbar \\

i b t/\Hbar & (b t)^2/\Hbar^2

\end{bmatrix} \\

&\ne I.

\end{aligned}

\end{equation}

We required \( U^\dagger U = I \) for the time evolution operator, but don’t have that property for this non-Hermitian Hamiltonian.

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.