QM: Groundstates and changing Hamiltonians

  • Thread starter Niles
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  • #1
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Homework Statement


At time t<0 I have a Hamiltonian given by:

[tex]
H = \left( {\begin{array}{*{20}c}
{\varepsilon _1 } & { - V} \\
{ - V} & {\varepsilon _2 } \\
\end{array}} \right)
[/tex]

with the ground state [tex]\left| {v_ - } \right\rangle = A\left( {\sqrt 2 - 1} \right)\left| 1 \right\rangle + \left| 2 \right\rangle [/tex], where A is some constant.

Now at t=0 the Hamiltonain changes, since [tex]\varepsilon_1=\varepsilon_2=0[/tex] for t>0.

I have to find the probability of finding the particle in the groundstate of the new Hamiltonian at the time t=0.


The Attempt at a Solution



Ok, first I have found the groundstate [tex]\left| {m_0 } \right\rangle[/tex] of the new Hamiltonian. Since I have to find the probability of the particle being in the state [tex]\left| {m_0 } \right\rangle[/tex] at t=0, I still need the Hamiltonian for t<0, i.e. I need:


[tex]
H = \left( {\begin{array}{*{20}c}
{\varepsilon _1 } & { - V} \\
{ - V} & {\varepsilon _2 } \\
\end{array}} \right)
[/tex]

I am a little uncertain of what should be done now. Can you push me in the right direction?
 

Answers and Replies

  • #2
207
1
I'm also uncertain, but my guess is that you should first normalize your ground states and then take the square modulus of their inner product to get the probability:

[tex]P = |\langle v_- | m_0\rangle|^2[/tex]
 
  • #3
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I think we're on the right track, Irid.

I'm quite sure I have to use [tex]P = |\langle v_- | \Psi\rangle|^2 [/tex], but how can I extract \Psi from the Hamiltonian at t=0?
 
  • #4
207
1
You said that you have Psi, only named it m_0. Anyway, I think one needs to set up an eigenvalue equation and you'll obtain eigenvectors. The one which corresponds to the lowest energy is termed "the ground state".
 
  • #5
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I have already found (and normalized) the groundstate [tex]\left| {m_0 } \right\rangle[/tex]. But the groundstate is not the same as the wavefunction itself, i.e. the solution of the time-dependant Schrödinger-equation?

Please correct me if I am wrong.
 
  • #6
1,868
0
Ok, I worked it out. The particle is in the groundstate, so [tex]\Psi(x,0) = \left| {v__- } \right\rangle[/tex] and the new stationary state is [tex]\psi(x) = \left| {m_0 } \right\rangle[/tex]. That is, we can find the probability as:

[tex]
P = |\langle m_0 | v_-\rangle|^2.
[/tex]

Thanks.
 

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