What Do I Calculate the Probability Of?

[Saptarshi Sarkar]

Homework Statement

Given the Hamiltonian H = Ho + iT where Ho and T are Hermitian operators, find whether H is a Hermitian operator. Also find whether the total probability is conserved or not.

Relevant Equations

Η† = (Ho+iT)† = H† - iT† = H - iT != H + iT

I found out that the operator H is not a Hermitian operator but I didn't understand the second part of the question. What do I calculate the probability of?

 

[PeroK]

Homework Statement:: Given the Hamiltonian H = Ho + iT where Ho and T are Hermitian operators, find whether H is a Hermitian operator. Also find whether the total probability is conserved or not.
Homework Equations:: Η† = (Ho+iT)† = H† - iT† = H - iT != H + iT

I found out that the operator H is not a Hermitian operator but I didn't understand the second part of the question. What do I calculate the probability of?

Conservation of probability is equivalent to the wave function remaining normalised under time evolution.

 

[julian]

I know how to prove conservation of total probability when the $H$ is Hermitian. So conservation of total probability reads

$$
\int d^3x |\psi (x,t)|^2 = \int d^3x |\psi (x,t=0)|^2 .
$$

In terms of bra-ket notation the above equation reads:

\begin{align*}
<\psi (t) | \psi (t)> &\equiv \int d^3x <\psi (t) |x> <x| \psi (t)>\\
&\equiv \int d^3x |\psi (x,t)|^2 \\
&= \int d^3x |\psi (x,t=0)|^2\\
&= \int d^3x <\psi (0) |x> <x | \psi (0)>\\
&= <\psi (0) | \psi (0)> .
\end{align*}

Let us first take $H$ to be Hermitian. Assuming a dynamical equation:

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

where $H$ is time-independent, we can solve this with:

$$
|\psi (t) > = \exp (-i t H) |\psi (0)> .
$$

Now if $H$ is Hermitian then conservation of total probability is obvious:

\begin{align*}
<\psi (t) | \psi (t)> & <\psi (0) | \exp (i t H^\dagger) \exp (-itH) |\psi (0)>\\
& <\psi (0) | \exp (i t H) \exp (-itH) |\psi (0)>\\
&= <\psi (0) | \psi (0)> .
\end{align*}

This can also be proven in terms of eigenstates of $H$ ($H |E_n> = E_n |E_n>$ where $E_n$ is real as $H$ is Hermitian)...Then

\begin{align*}
|\psi (t)> &= \exp (-itH) |\psi (0)>\\
&= \exp (-itH) \sum_n|E_n><E_n|\psi (0)>\\
&= \sum_n e^{-it E_n} |E_n><E_n| \psi(0)>
\end{align*}

and

\begin{align*}
<\psi (t) | \psi (t)> &= \sum_{m,n} (e^{i E_m t} <\psi (0) | E_m> <E_m|) ( e^{-iE_nt} |E_n> <E_n |\psi (0)>)\\
&= \sum_n <\psi (0) | E_n> <E_n |\psi (0)>\\
&= <\psi (0) | \psi (0)>
\end{align*}

where we used $<E_m|E_n> = \delta_{mn}$.

Now, what if $H$ is not Hermitian? Take $H = H_0 + iT$, and assuming again that $H$ is time-independent, then:

\begin{align*}
<\psi (t) | \psi (t)> & <\psi (0) | \exp (i t H^\dagger) \exp (-itH) |\psi (0)>\\
& <\psi (0) | \exp ( t T + i t H_0) \exp (-itH_0 + t T) |\psi (0)> .
\end{align*}

Now if we assume that $[H_0 , T]= 0$, then first of all we can say that

$$
\exp (-itH_0 + t T) = \exp (itH_0) \exp (t T)
$$

and we can also say that $H_0$ and $T$ have simultaneous eigenstates:

\begin{align*}
H_0 |E_n> &= E_n |E_n>\\
T |E_n> &= T_n |E_n>
\end{align*}

where $E_n$ and $T_n$ are real as $H$ and $T$ are Hermitian. It follows that:

\begin{align*}
|\psi (t)> &= \exp (-itH) |\psi (0)>\\
&= \exp (-itH) \sum_n|E_n><E_n|\psi (0)>\\
&= \exp (t T) \exp (-itH_0) \sum_n|E_n><E_n|\psi (0)>\\
&= \sum_n e^{-it E_n + t T_n} |E_n><E_n| \psi(0)> .
\end{align*}

We could then calculate:

\begin{align*}
<\psi (t) | \psi (t)> &= \sum_{m,n} (e^{i E_m t + t T_m} <\psi (0) | E_m> <E_m|) ( e^{-iE_nt + t T_n} |E_n> <E_n |\psi (0)>)\\
&= \sum_n e^{2 t T_n} <\psi (0) | E_n> <E_n |\psi (0)> .
\end{align*}

which will, for general values of $t$, not be equal to $<\psi (0) | \psi (0)>$.I'm not entirely sure how to proceed if $[H_0 , T] \not= 0$.

 

[kdv]

It is easier to differentiate with respect to time $\int dx |\psi(x,t)^2$ and then use Schrodinger's equation for the derivatives of $\psi$ and $\psi^*$.

 

[Bo-Chien Huang]

From $i \hbar | \dot \psi \rangle = H | \psi \rangle = (H_0 + i T)| \psi \rangle$, we know $| \dot \psi \rangle = \frac{1}{i \hbar} (H_0 + i T)| \psi \rangle$ as well as its dual $\langle \dot \psi | = \frac{1}{-i \hbar} \langle \psi |(H_0 - i T)$.
Directly consider $\frac{d}{dt} \langle \psi | \psi \rangle = \langle \dot \psi | \psi \rangle + \langle \psi | \dot \psi \rangle$.
$$ \begin{align*}\langle \dot \psi | \psi \rangle + \langle \psi | \dot \psi \rangle &= \frac{1}{-i \hbar} \langle \psi |(H_0 - i T) | \psi \rangle + | \dot \psi \rangle + \frac{1}{i \hbar} \langle \psi | (H_0 + i T)| \psi \rangle \\ &= \frac{1}{i \hbar} \langle \psi | 2 i T | \psi \rangle = \frac{2}{\hbar} \langle T \rangle \end{align*}$$
As long as the expectation value $\langle T \rangle$ is non-zero, the probability conservation fails.

 

[nrqed]

From $i \hbar | \dot \psi \rangle = H | \psi \rangle = (H_0 + i T)| \psi \rangle$, we know $| \dot \psi \rangle = \frac{1}{i \hbar} (H_0 + i T)| \psi \rangle$ as well as its dual $\langle \dot \psi | = \frac{1}{-i \hbar} \langle \psi |(H_0 - i T)$.
Directly consider $\frac{d}{dt} \langle \psi | \psi \rangle = \langle \dot \psi | \psi \rangle + \langle \psi | \dot \psi \rangle$.
$$ \begin{align*}\langle \dot \psi | \psi \rangle + \langle \psi | \dot \psi \rangle &= \frac{1}{-i \hbar} \langle \psi |(H_0 - i T) | \psi \rangle + | \dot \psi \rangle + \frac{1}{i \hbar} \langle \psi | (H_0 + i T)| \psi \rangle \\ &= \frac{1}{i \hbar} \langle \psi | 2 i T | \psi \rangle = \frac{2}{\hbar} \langle T \rangle \end{align*}$$
As long as the expectation value $\langle T \rangle$ is non-zero, the probability conservation fails.

We usually prefer to not do all the work for the original poster. It is much better to give some guiding help and make the person work through the calculation themselves.

 

[Bo-Chien Huang]

We usually prefer to not do all the work for the original poster. It is much better to give some guiding help and make the person work through the calculation themselves.

Ok, I get it.