time evolution

Heisenberg picture position commutator

August 14, 2015 phy1520 No comments , , , ,

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Question: Heisenberg picture position commutator ([1] pr. 2.5)

Evaluate

\begin{equation}\label{eqn:positionCommutator:20}
\antisymmetric{x(t)}{x(0)},
\end{equation}

for a Heisenberg picture operator \( x(t) \) for a free particle.

Answer

The free particle Hamiltonian is

\begin{equation}\label{eqn:positionCommutator:40}
H = \frac{p^2}{2m},
\end{equation}

so the time evolution operator is

\begin{equation}\label{eqn:positionCommutator:60}
U(t) = e^{-i p^2 t/(2 m \Hbar)}.
\end{equation}

The Heisenberg picture position operator is

\begin{equation}\label{eqn:positionCommutator:80}
\begin{aligned}
x^\textrm{H}
&= U^\dagger x U \\
&= e^{i p^2 t/(2 m \Hbar)} x e^{-i p^2 t/(2 m \Hbar)} \\
&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i p^2 t}{2 m \Hbar} }^k
x
e^{-i p^2 t/(2 m \Hbar)} \\
&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k p^{2k} x
e^{-i p^2 t/(2 m \Hbar)} \\
&=
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ \antisymmetric{p^{2k}}{x} + x p^{2k} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \antisymmetric{p^{2k}}{x}
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar \PD{p}{p^{2k}} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar 2 k p^{2 k -1} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
-2 i \Hbar p \frac{i t}{2 m \Hbar} \sum_{k = 1}^\infty \inv{(k-1)!} \lr{ \frac{i t}{2 m \Hbar} }^{k-1} p^{2(k – 1)}
e^{-i p^2 t/(2 m \Hbar)} \\
&= x + t \frac{p}{m}.
\end{aligned}
\end{equation}

This has the structure of a classical free particle \( x(t) = x + v t \), but in this case \( x,p \) are operators.

The evolved position commutator is
\begin{equation}\label{eqn:positionCommutator:100}
\begin{aligned}
\antisymmetric{x(t)}{x(0)}
&=
\antisymmetric{x + t p/m}{x} \\
&=
\frac{t}{m} \antisymmetric{p}{x} \\
&=
-i \Hbar \frac{t}{m}.
\end{aligned}
\end{equation}

Compare this to the classical Poisson bracket
\begin{equation}\label{eqn:positionCommutator:120}
\antisymmetric{x(t)}{x(0)}_{\textrm{classical}}
=
\PD{x}{}\lr{x + p t/m} \PD{p}{x} – \PD{p}{}\lr{x + p t/m} \PD{x}{x}
=
– \frac{t}{m}.
\end{equation}

This has the expected relation \( \antisymmetric{x(t)}{x(0)} = i \Hbar \antisymmetric{x(t)}{x(0)}_{\textrm{classical}} \).

References

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

Dynamics of non-Hermitian Hamiltonian

August 13, 2015 phy1520 No comments , , , , ,

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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.