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An interpretation of hamiltonians, states and fields.

  1. Jun 22, 2011 #1
    Hi all. Here is an n-particle hamltonian.

    [tex]H=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}+U(x)\Bigr)a(x)+\int d^{3}xd^{3}yV(x-y)a^{\dagger}(x)a^{\dagger}(y)a(x)a(y)[/tex]

    Here is an n-particle state.

    [itex]\Bigr|\psi,t\Bigl\rangle=\int d^{3}x_{1}...d^{3}x_{n}\psi(x_{1},....,x_{n};t)a^{\dagger}(x_{1}).....a^{\dagger}(x_{n})\Bigr|0\Bigl\rangle[/itex]

    I am wondering how this hamiltonian acts on this state. Let us first look at the kinetic term. Is it true to say that first the [itex]a(x)[/itex] in the hamiltonian searches through the state for it's counterpart [itex]a^{\dagger}(x)[/itex] and annihalates the particle at [itex]x[/itex]. Then the [itex]\nabla^{2}[/itex] acts on an [itex](n-1)[/itex]-particle state. Finally the [itex]a^{\dagger}(x)[/itex] restores the particle at [itex]x[/itex]? In the same way, the interaction hamiltonian acts on the state by removing two particles, operating with the interaction operator [itex]V[/itex] and finally restoring particles at the two points in question? If this is correct why do we think of it this way?

    I am also worried about the dimension of the fields [itex]a(x)[/itex] and [itex]a^{\dagger}(x)[/itex]. I compared dimensions on the left and right hand sides of the hamiltonian and got something strange. What should I be expecting for the dimension of such a field?

    I would also like a better understanding of the state above. I am used to thinking of states as a linear combination of chosen base states, the coefficients being the amplitudes for the state to be in one of the base states. For example,

    [itex]\Bigl|\psi\Bigl\rangle=\underset{i}{\sum}\Bigl|i\Bigl\rangle\Bigr\langle i\Bigl|\psi\Bigl\rangle[/itex]


    So how am I to compare the two states? What is similar what is different? I can see how the sum and integral are related. Is the [itex]\psi(x_{1},....,x_{2};t)[/itex] just like the coefficients and the [itex]a^{\dagger}[/itex]'s acting on the vacuum state like the base states?

    Cheers.
     
    Last edited: Jun 22, 2011
  2. jcsd
  3. Jun 22, 2011 #2
    Erm, how do I make that code into something legible?
     
  4. Jun 22, 2011 #3

    kith

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    Put it into [*tex] [*/tex] or [*itex] [*/itex] (without the *s)
     
  5. Jun 23, 2011 #4

    strangerep

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    First, do you understand how H acts on a very simple single-particle state?
    What are your commutation relations? To check that you understand this much,
    what is the result of the following?
    [tex]
    H_0 a^\dagger(x) |0\rangle
    [/tex]
    where H_0 is your H with U = V = 0.

    And what is
    [tex]
    H_U \; a^\dagger(x) |0\rangle ~~~~~~~ ?
    [/tex]
    (where H_U is just the part of H containing U).

    That's the general idea.
     
  6. Jun 23, 2011 #5
    Thanks for the reply strangerep.

    Acting the hamiltonian on the single particle state, we get [tex]a(x)a^{\dagger}(x)[/tex] hitting the vacuum state. Using the bosonic commutation relation [tex]\left[a(x),a^{\dagger}(x^{\prime})\right]=\delta(x-x^{\prime})[/tex] we can write [tex]a^{\dagger}(x)a(x)+1[/tex] instead. [tex]a(x)[/tex] kills the vacuum state so we are left with

    [tex]H_{0}a^{\dagger}(x)\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\nabla^{2}\Bigr)\Bigl|0\Bigl\rangle[/tex]

    So from here I suppose we need the explicit form of the vacuum state and the field to do the differentiation and integration. Any ideas what they may look like? For a particular case even?

    I cannot see how [tex]H_{U}[/tex] would be any different

    [tex]H_{U}a^{\dagger}(x)\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)U(x)\Bigl|0\Bigl\rangle[/tex]

    Able.
     
    Last edited: Jun 23, 2011
  7. Jun 23, 2011 #6

    strangerep

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    No. The x inside the Hamiltonian is a dummy variable of integration.
    Choose a different symbol for one of the variables.

    That doesn't follow -- but fix up the first mistake and it might become clearer.

    You have a free variable x on the LHS but a dummy integration variable x on
    the RHS. So something has gone wrong. Try doing it again, but taking more
    care to distinguish between free and dummy variables.
     
  8. Jun 25, 2011 #7
    Ok, what about this

    [tex]H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)a(x)a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle[/tex]

    [tex]H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)\Bigl(a^{\dagger}(x_{1})a(x)+\delta^{3}(x-x_{1})\Bigr)\Bigl|0\Bigl\rangle[/tex]

    [tex]H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)a^{\dagger}(x_{1})a(x)\Bigl|0\Bigl\rangle+\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)\delta^{3}(x-x_{1})\Bigl|0\Bigl\rangle[/tex]

    The first term in this is zero because the a(x) hits the vacuum state. Integrating the delta function gives

    [itex]H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=a^{\dagger}(x_{1})\Bigl(-\nabla^{2}\Bigr)\Bigl|0\Bigl\rangle[/itex]

    Sorry about the inconsistency of factors.
     
    Last edited: Jun 25, 2011
  9. Jun 25, 2011 #8

    strangerep

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    No. The derivative of the delta was wrt x, which is a parameter inside the delta,
    so this must be handled correctly.

    As a simpler warm-up exercise, do you know how the derivative of a delta function
    [tex]
    \partial_x \; \delta(x - y)
    [/tex]
    is defined (and handled in practical situations)? E.g., can you evaluate the following?
    [tex]
    \int dx\; f(x) \; \partial_x \delta(x - y)
    [/tex]
     
    Last edited: Jun 25, 2011
  10. Jun 25, 2011 #9
    Well I didn't off hand but I did some research. It seems to only be defined in conjunction with another function inside an integral. Then you use integration by parts to put derivatives on the other function.

    [tex]\int g(x)\delta^{\prime}(x-y)dx=-g^{\prime}(y)[/tex]

    [tex]\int g(x)\delta^{\prime\prime}(x-y)dx=g^{\prime\prime}(y)[/tex]

    So what I want is this

    [tex]\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)\delta^{3}(x-x_{1})\Bigl|0\Bigl\rangle=-\frac{\hbar^{2}}{2m}\nabla^{2}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle[/tex]
     
    Last edited: Jun 25, 2011
  11. Jun 25, 2011 #10

    strangerep

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    Good. BTW, it would be less error-prone if you include a subscript on the derivatives
    to indicate which variable they're for. E.g.,
    [tex]
    \nabla^{2}_x ~~~\mbox{or}~~~ \nabla^{2}_{x_1} ~~~~ \mbox{(etc)}
    [/tex]
    to make your next equation's rhs more explicit.

    Now, what equation does your free field satisfy in the absence of interactions?
     
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