Solving Schrodinger Equation

  • #1

Main Question or Discussion Point

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

Answers and Replies

  • #2
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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
I'm not really sure how to deal with the sums, when I plug in and evaluate [itex]H^0[/itex]I get
[itex]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> [/itex] and then if I act with a [itex]<E_n^0|[/itex] it will simply cancel out the [itex]|E_n^0>[/itex] (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 [itex]A[/itex] with eigenvalue [itex]a[/itex] at t=0. How do I find the probability that as t->infinity a measurement of [itex]A[/itex] will give a different eigenvalue?
 
  • #4
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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.
 
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  • #5
Right so because of orthogonality each term where [itex]n \neq m[/itex] will cancel leaving the left side as [itex]i \hbar \frac{d}{dt} c_n (t)[/itex] but wouldn't the same have to happen to the right side making the equation [itex]i \hbar \frac{d}{dt} c_n (t) = f(t)E_n^0 c_n (t)[/itex]. Is that right? Also what about my second question from post 3?
 
  • #6
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Right so because of orthogonality each term where [itex]n \neq m[/itex] will cancel leaving the left side as [itex]i \hbar \frac{d}{dt} c_n (t)[/itex] but wouldn't the same have to happen to the right side making the equation [itex]i \hbar \frac{d}{dt} c_n (t) = f(t)E_n^0 c_n (t)[/itex]. Is that right?
Yes, that is the idea.
Also what about my second question from post 3?
[tex] P = \lim_{t \to +\infty}\left(1-|\langle A(t)| A(0)\rangle|^2\right),[/tex]
where [itex]|A(t)\rangle [/itex] can be computed from Schrodinger's equation.
 
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  • #7
Yes, that is the idea.

[tex] P = \lim_{t \to +\infty}\left(1-|\langle A(t)| A(0)\rangle|^2\right),[/tex]
where [itex]|A(t)\rangle [/itex] can be computed from Schrodinger's equation.
Thanks, but how would I compute [itex]A(t)[/itex]. I'll give a little more detail. For this particular example, I am given that
[itex]A = \left( \begin{array}{cc}
1 & 0 \\
0 & -1 \end{array} \right)[/itex] with eigenvalue 1 at [itex]t= 0[/itex]. Since I've now got an expression for [itex]c_n(t)[/itex] I can calculate [itex]|\phi (t)>[/itex], but what about this [itex]A(t)[/itex].
 
  • #8
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Also if I am given a well-defined Hamiltonian in terms of time and a state [itex]A[/itex] with eigenvalue [itex]a[/itex] at t=0.
I thought that [itex]A[/itex] was a quantum state.

For this particular example, I am given that
[itex]A = \left( \begin{array}{cc}
1 & 0 \\
0 & -1 \end{array} \right)[/itex] with eigenvalue 1 at [itex]t= 0[/itex]. Since I've now got an expression for [itex]c_n(t)[/itex] I can calculate [itex]|\phi (t)>[/itex], but what about this [itex]A(t)[/itex].
Do you mean that system is in one of the eigen states of operator [itex]A[/itex]?
Also can you write down the Hamiltonian?
 
  • #9
Do you mean that system is in one of the eigen states of operator [itex]A[/itex]?
yes at t=0


Also can you write down the Hamiltonian?
[itex]H(t) = ae^{-t/b} \left( \begin{array}{cc}
0 & 1 \\
1 & 0 \end{array} \right)\[/itex]
 
  • #10
178
17
Ok.
So the time independent part of Hamiltonian is
[itex]H^0 = \left( \begin{array}{cc}
0 & 1 \\
1 & 0 \end{array} \right),[/itex]
with eigen-vectors and eigen-values as [itex] |e_1\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}
1 \\
1 \end{array} \right)[/itex], [itex] |e_2\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}
1 \\
-1 \end{array} \right)[/itex] and [itex]1,-1[/itex] respectively.

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

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

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