The Schrodinger Equation: How to Solve for Time Evolution and Eigenstates

[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.

 

[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 <E_n⁰|, use the orthonormality of those states, and solve the ODE.

 

[wileecoyote]

I'm not really sure how to deal with the sums, when I plug in and evaluate H^0I get
i \hbar \frac{d}{dt} \sum\limits_{n=1}^N c_n (t)|E_n^0&gt; = f(t) \sum\limits_{n=1}^N c_n (t) E_n^0 |E_n^0&gt; and then if I act with a &lt;E_n^0| it will simply cancel out the |E_n^0&gt; (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?

 

[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.

 

[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?

 

[Ravi Mohan]

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?

Yes, that is the idea.

Also what about my second question from post 3?

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.

 

[wileecoyote]

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.

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}<br /> 1 &amp; 0 \\<br /> 0 &amp; -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)&gt;, but what about this A(t).

 

[Ravi Mohan]

Also if I am given a well-defined Hamiltonian in terms of time and a state A with eigenvalue a at t=0.

I thought that A was a quantum state.

For this particular example, I am given that
A = \left( \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 0 &amp; -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)&gt;, but what about this A(t).

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

 

[wileecoyote]

Do you mean that system is in one of the eigen states of operator A?

yes at t=0

Also can you write down the Hamiltonian?

H(t) = ae^{-t/b} \left( \begin{array}{cc}<br /> 0 &amp; 1 \\<br /> 1 &amp; 0 \end{array} \right)\

 

[Ravi Mohan]

Ok.
So the time independent part of Hamiltonian is
H^0 = \left( \begin{array}{cc}<br /> 0 &amp; 1 \\<br /> 1 &amp; 0 \end{array} \right),
with eigen-vectors and eigen-values as |e_1\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}<br /> 1 \\<br /> 1 \end{array} \right), |e_2\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}<br /> 1 \\<br /> -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}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1 \end{array} \right),
with eigen-vectors as |a_1\rangle = \left( \begin{array}{c}<br /> 1 \\<br /> 0 \end{array} \right), |a_2\rangle = \left( \begin{array}{c}<br /> 0 \\<br /> 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).