# Solving Schrodinger Equation

1. Nov 9, 2013

### wileecoyote

I'm working my way through some QM problems for self-study and this one has stumped me. Given the Hamiltonian as $H(t) = f(t)H^0$ where $f(t)$ is a real function and $H^0$ is Hermitian with a complete set of eigenstates $H^0|E_n^0> = E_n^0|E_n^0>$. Time evolution is given by the Schrodinger equation $i \hbar \frac{d}{dt}|\phi (t)> = H(t)|\phi (t)>$. I am supposed to write a solution to the Schrodinger equation as a linear combination of the eigenstates of $H^0$. Now clearly
$|\phi (t)> = \sum\limits_{n=1}^N c_n (t)|E_n^0>$. But where do I go from there. The second part is to convert the Schrodinger equation into a first order diff eq and solve for the $c_n (t)$. Any help is appreciated. Thanks.

2. Nov 9, 2013

### dauto

Seems pretty straight forward to me. Plug in the expressions for |phi(t)> and H(t) into the Schroedinger Equation, act on it with a bra of one of the eigenstates <En0|, use the orthonormality of those states, and solve the ODE.

3. Nov 9, 2013

### wileecoyote

I'm not really sure how to deal with the sums, when I plug in and evaluate $H^0$I get
$i \hbar \frac{d}{dt} \sum\limits_{n=1}^N c_n (t)|E_n^0> = f(t) \sum\limits_{n=1}^N c_n (t) E_n^0 |E_n^0>$ and then if I act with a $<E_n^0|$ it will simply cancel out the $|E_n^0>$ (completeness relation), but then how do I solve it from there? Also if I am given a well-defined Hamiltonian in terms of time and a state $A$ with eigenvalue $a$ at t=0. How do I find the probability that as t->infinity a measurement of $A$ will give a different eigenvalue?

4. Nov 9, 2013

### dauto

You should apply the bra on the left side and use the orthonormality relation - not the completeness relation. There will be only one single non-vanishing term left in the summation. That term will give you an ODE.

5. Nov 9, 2013

### wileecoyote

Right so because of orthogonality each term where $n \neq m$ will cancel leaving the left side as $i \hbar \frac{d}{dt} c_n (t)$ but wouldn't the same have to happen to the right side making the equation $i \hbar \frac{d}{dt} c_n (t) = f(t)E_n^0 c_n (t)$. Is that right? Also what about my second question from post 3?

6. Nov 9, 2013

### Ravi Mohan

Yes, that is the idea.
$$P = \lim_{t \to +\infty}\left(1-|\langle A(t)| A(0)\rangle|^2\right),$$
where $|A(t)\rangle$ can be computed from Schrodinger's equation.

Last edited: Nov 9, 2013
7. Nov 9, 2013

### wileecoyote

Thanks, but how would I compute $A(t)$. I'll give a little more detail. For this particular example, I am given that
$A = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)$ with eigenvalue 1 at $t= 0$. Since I've now got an expression for $c_n(t)$ I can calculate $|\phi (t)>$, but what about this $A(t)$.

8. Nov 10, 2013

### Ravi Mohan

I thought that $A$ was a quantum state.

Do you mean that system is in one of the eigen states of operator $A$?
Also can you write down the Hamiltonian?

9. Nov 10, 2013

### wileecoyote

yes at t=0

$H(t) = ae^{-t/b} \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)\$

10. Nov 10, 2013

### Ravi Mohan

Ok.
So the time independent part of Hamiltonian is
$H^0 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right),$
with eigen-vectors and eigen-values as $|e_1\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 1 \end{array} \right)$, $|e_2\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ -1 \end{array} \right)$ and $1,-1$ respectively.

Now after calculating $C_n(t)$, you have $|\phi(t)\rangle = \sum\limits_{i=1}^{2}C_i(t) |e_i \rangle$.

The operator corresponding to observable $A$ is
$A = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right),$
with eigen-vectors as $|a_1\rangle = \left( \begin{array}{c} 1 \\ 0 \end{array} \right)$, $|a_2\rangle = \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$.

When the time approaches infinity the probability that the measurement will not result in eigen-value = 1 is given by evaluating
$$P = \lim_{t \to +\infty}\left(1-|\langle a_1| \phi(t)\rangle|^2\right).$$