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Quantum Mechanics and Group Theory questions

  1. Jun 4, 2015 #1
    Hello all. I am new here. I am in the last quarter of a 3 quarter sequence of undergrad quantum mechanics and I just had some conceptual questions (nothing pertaining to homework). We just recently covered Berry's Phase and the Dynamical Phase. Now I wanted to start with a more basic quantum mechanical question before moving to Berry's Phase:

    When determining a wavefunction $$\psi$$ we can pretty much derive a wavefunction "up to" a phase, i.e $$\psi = A \left(functions\right) e^{i \chi}$$ Now, if I am understanding elementary group theory correctly, a U(1) transformation would be multiplying our initial function by a complex exponential with a phase. So, is the wavefunction I wrote "up to a phase" basically transformed by U(1)?

    Okay, assuming what I first asked is legitimate, I wanted to ask about Berry and Dynamical Phases. Through the adiabatic approximation we can write $$\Psi_n\left(t\right) = \psi_n \left(t \right) e^{i \theta_n \left(t \right)} e^{i \gamma_n \left(t \right)} $$. My question here is if any of this may be similar or is a U(1) transformation, or is it not so because of the possible time dependence?

    Relevant equations:
    $$\theta_n \left(t \right) =-\frac{1}{\hbar} \int_{0}^{t} E_n\left(t'\right) dt'$$

    $$\gamma_{n} \left(t \right) = i \int_{0}^{t} \langle \psi_m\left(t'\right) | \frac{\partial}{\partial t'} \psi_{m}\left(t' \right) \rangle dt'$$

    Thank you all for your patience and help!
     
  2. jcsd
  3. Jun 4, 2015 #2
    I'm not sure I'm understanding your two questions but I will try to do the best with my (not that extensive) knowledge of the matter

    - Yes, multiplying by a phase is equivalent to doing a U(1) transformation.

    - if the adiabatic approximation is valid and If the path is closed, meaning H(T)=H(t=0), then yes the acquired geometrical phase is a very specific U(1) transformation linking the psi(t=0) original vector to the psi(T) final vector. It is specific in the sense that it depend on the path traveled by the hamiltonian (or more generally the path traveled in the projective hilbert space).
     
  4. Jun 4, 2015 #3
    Thank you! That was basically what I was looking for. We are using Griffiths, which when it comes to showing me how to do quantum problems, is great. Conceptually, I am not too happy with it all the time.
     
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