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Raising and lowering operators of the Hamiltonian

  1. May 8, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    a) The operators ##a## and ##a^{\dagger}## satisfy the commutation relation ##[a,a^{\dagger}] = 1##. Find the normalization of the state ##|\psi \rangle = C (a^{\dagger} )^2 |0\rangle##, where the vacuum state ##|0\rangle## is such that ##a|0\rangle = 0##

    b)A one dimensional simple harmonic oscillator has the Hamiltonian $$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}mw^2 \hat{x}^2.$$ The operator ##\hat{a}## is defined as $$\hat{a} = \left(\frac{mw}{2\hbar}\right)^{1/2}\hat{x} + \frac{i}{\sqrt{2mw\hbar}} \hat{p}$$

    Suppose that the oscillator frequency is changed slightly to ##w \rightarrow \sqrt{1+\epsilon} w## What are the exact new energy eigenvalues? Now calculate the first order perturbation in the energy and compare with the exact answer.
    2. Relevant equations
    In a), perhaps that ##a|n\rangle = \sqrt{n}|n-1\rangle## and that ##a^{\dagger} |n\rangle = \sqrt{n+1} |n+1\rangle##

    3. The attempt at a solution
    In a), write out the commutator explicitly ##aa^{\dagger} - a^{\dagger}a = 1##. Now I tried various manipulations of this equation to get something that resembled the equation given. Multiply by ##(a^{\dagger})^2## on the left and ##|0\rangle## on the right would give ##(a^{\dagger})^2 a a^{\dagger} |0\rangle - (a^{\dagger})^3 a|0\rangle = (a^{\dagger})^2 |0\rangle.## But this simplifies to LHS=RHS, so it is of no use.

    b)New energy eigenvalues would be ##E_n = \hbar (\sqrt{1+\epsilon} w) (n+1/2)## and the shift is ##E' = \langle n | \hat{H'} | n\rangle## where ##\hat{H'}## is the Hamiltonian above but with ##w## replaced with ##\sqrt{1+\epsilon}w## Do we treat ##\epsilon## as a constant here?

    Many thanks.
     
  2. jcsd
  3. May 8, 2014 #2

    dextercioby

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    For a), just compute the vector itself, a†|0> = √0+1 |1> = |1>, etc.
     
  4. May 9, 2014 #3

    CAF123

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    Applying a† again would give √2 |2>, not some constant premultiplying the state |0>. So I think somehow we have to incorporate the commutation relation given, which I tried to do in the OP.
     
  5. May 9, 2014 #4

    vela

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    Try using ##\|C(a^\dagger)^2\vert 0 \rangle\|^2=1##.
     
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