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Relativistic Quantum Mechanics

  1. Jun 16, 2013 #1
    the problem is on page 26 of "relativistic quantum mechanics and field theory" by Franz Gross.

    consider the lagrangian density:

    L=(1/2)[(∂ψ/∂t)^2 -(∂ψ/∂z)^2 -m^2ψ^2]

    a) find the momentum conjugate.
    b) find the equation of motion for the fields and the solution. use periodic boundary conditions.

    Attempt at solution:

    taking partial derivative with respect to (∂ψ/∂t) and (∂ψ/∂z)

    (a) the momentum conjugate is: π(z,t)= (∂ψ/∂t)-(∂ψ/∂z)

    (b) Using Euler-Lagrange for function with more than one variable we get:

    -m^2ψ=(d^2ψ/dt^2)- (d^2ψ/dz^2) using separation of variable

    ψ=Z(z)T(t)

    The part which is only a function of "t" equate it to square of constant "ω" then we get:

    T= Ae^(iωt) + Be^(-iωt)

    and equation becomes:

    -m^2=ω^2 - (1/Z)(d^2Z/dz^2) → (m^2 + ω^2)Z= (d^2Z/dz^2)

    call m^2 + ω^2 = κ^2 the

    Z=Ce^(ikz) + De^(-ikz)

    at this step I don't know how to apply the periodic boundary conditions. the periodic conditions are given as following given in the page 4 :


    ψ0=ψN
    dψ0/dt=dψN/dt

    where 0 and N indicate 0th and Nth oscillators.


    I would be really grateful if anyone could help me out with this.

    the rest of the question is related to this part and i can't do it without getting this part right.
     
  2. jcsd
  3. Jun 16, 2013 #2

    Fredrik

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    This would be much easier to read if you used LaTeX. Here's the FAQ post about it: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3.

    What definition of "momentum" are you using? I'm getting a different result. Shouldn't it be a vector by the way? Or are you just looking for the spatial component?

    I think you have a sign error in the Euler-Lagrange equation.

    I don't understand your comment before you separate the variables. What you're supposed to do is to insert ##\psi(t,z)=T(t)Z(z)## into the equation, and divide everything by T(t)Z(z). Then you can make arguments like "this term is independent of the t, so if I solve for it, what ends up on the other side of the equality sign must be independent of t too." (Same thing with z). This is what allows you to introduce that constant.
     
  4. Jun 16, 2013 #3
    The boundary conditions that you write look - use latex - like Dirichlet's and Neumann's

    Periodic boundary conditions in time mean

    [itex]\phi(x,t)=\phi(x,t+T)[/itex]

    and this implies a quantization of the energy

    [itex]\phi(x,t)=\exp[-i E_n T / \hbar][/itex]

    with

    [itex]E_n = n h / T[/itex]
     
  5. Jun 18, 2013 #4
    @Fredrik

    what I mean is that we have two partial derivatives (with respect to t,z) present in the Lagrangian so we have to apply the following formula:


    [tex]
    \frac{\partial L}{\partial ψ} = \sum_{i=1}^\infty \frac{\partial (\frac{\partial L}{\partial ψ_xi})}{\partial x_i} [/tex]


    where [tex] x_i [/tex] represent different vairables (ex: t,z)
     
  6. Jun 18, 2013 #5

    Fredrik

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    Your notation is a bit odd. I would write it as $$\frac{\partial\mathcal L}{\partial\psi} =\sum_{\mu=0}^1\partial_\mu\left( \frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\right),$$ where
    $$\partial_\mu=\frac{\partial}{\partial x^\mu}
    =\begin{cases}
    \frac{\partial}{\partial t} & \text{if }\mu=0\\
    \frac{\partial}{\partial z} & \text{if }\mu=1.
    \end{cases}$$ But it looks like the sign error was on my end.

    Your statement about the momentum still doesn't make sense. Maybe you have the right answer, but you're not writing it in a way that makes sense.
     
  7. Jun 18, 2013 #6
    @Fredrik

    well Im sorry I'm not really that good with -Latex- since i have just started using it.
    for the momentum Im just applying the following formula:
    [tex]π_μ = \frac{∂L}{∂(∂_μψ)} [/tex]
    and applying this to lagrangian
    [tex]L=(1/2)[(∂ψ/∂t)^2 -(∂ψ/∂z)^2 -m^2ψ^2][/tex]
    we get the follwing :
    [tex]π_0 = \frac{∂ψ}{∂t} [/tex]
    [tex]π_1 =- \frac{∂ψ}{∂z} [/tex]
    I think it is right now, isn't it?
     
    Last edited: Jun 18, 2013
  8. Jun 18, 2013 #7
    @naturale

    and by the way naturale can you please show me the steps how you get that?

    we put the solution into

    [tex] ψ=Z(z)T(t) [/tex]

    get the follwoing:
    [tex]ψ(z,t)=(Ae^{iωt} + Be^{-iωt})(Ce^{ikz} + De^{-ikz})[/tex]
    how do we apply the periodic conditions to this??
     
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