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Integral kernel in vacuum wave functional

  1. Jun 5, 2015 #1

    ShayanJ

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    I'm trying to solve the exercise below in a book I'm reading.
    I inverted equation 1.3 to get ## \phi_{\mathbf k}(t)=\int \frac{e^{-i \mathbf k \cdot \mathbf x}}{(2\pi)^{\frac 3 2}} \phi(\mathbf x,t) d^3 \mathbf x ##. Then I put it in I to get:
    ## I=\int \int d^3 \mathbf x d^3 \mathbf y \phi(\mathbf x,t)\left[ \int \frac{d^3 \mathbf k}{(2\pi)^3} \sqrt{k^2+m^2} e^{i(\mathbf y-\mathbf x)\cdot \mathbf k} \right]\phi^*(\mathbf y,t)##.
    So it seems that I found the kernel but there are two problems:
    1) I'm sure that one of the fields should be a complex conjugate but in the form suggested by the exercise, no one of them is!
    2) The kernel is a divergent integral, both its real and imaginary parts diverge. Just try wolframalpha.com to see this.

    Can anybody help me on this?
    Thanks
     
  2. jcsd
  3. Jun 5, 2015 #2

    vanhees71

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    It's not clear to me, where this functional should be useful. Could you provide the context, where it is defined?
     
  4. Jun 5, 2015 #3

    ShayanJ

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    Its not going to be used in anything, at least not in the parts of the book that I've read.(I'm on chapter 2 now which is a reminder of classical and quantum mechanics.)
    The exercise is in the chapter 1 of the book "Introduction to Quantum Effects in Gravity" By Mukhanov and Winitzki after the discussion about ground state fluctuations.
     
  5. Jun 5, 2015 #4

    samalkhaiat

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    It seems to me the whole exercise is formulated up-side-down. Usually, you want to solve the Schrodinger functional equation, [tex]i \partial_{t} \Psi [\phi , t] = \int d^{3} x \mathcal{H} \left( \phi (\vec{x}) , \pi ( \vec{x}) ; t \right) \Psi [\phi , t]. \ \ \ \ (1) [/tex] If the Hamiltonian has no explicit time dependence, we look for solution to the energy eigen-functional equation [tex]\int d^{3} x \mathcal{H} ( \phi , \pi ) \Psi [\phi] = E \Psi[\phi] . \ \ \ \ \ (2)[/tex] Since we are interested in the vacuum wavefunctional, we may assume that [itex]\Psi_{0}[\phi][/itex] is positive everywhere and substitute [itex]\Psi_{0}[\phi] = N \exp ( - I[\phi] )[/itex] in Eq(2). Then, power counting suggests a general form for the functional [itex]I[\phi][/itex] which in your case (i.e., free real scalar field) must be quadratic in the field, [tex]I[\phi] = \int d^{3} x \ d^{3} y \ \phi ( \vec{x} ) K ( \vec{x} , \vec{y} ) \phi ( \vec{y} ) . \ \ \ \ (3)[/tex] Now, for massless real scalar, the unknown functional [itex]I[\phi][/itex] satisfies the following equation [tex]\frac{1}{2} \int d^{3} x \left[ \frac{\delta^{2} I[\phi]}{\delta \phi^{2} (\vec{x})} - \left( \frac{\delta I[\phi]}{\delta \phi (\vec{x})}\right)^{2} \right] = E_{0} - \frac{1}{2} \int d^{3} x \phi ( \vec{x} ) (- \nabla^{2}) \phi (\vec{x}) . \ \ \ (4)[/tex] If we insert (3) back into (4) and equate equal-power terms on both sides, we find [tex]E_{0} = \frac{1}{2} \int d^{3} x \frac{\delta^{2} I[\phi]}{\delta \phi^{2} (\vec{x})} = \int d^{3} x \ K( \vec{x} , \vec{x} ) , \ \ \ \ (5)[/tex] [tex]\int d^{3}x \ d^{3}y \ d^{3}z \ \phi(\vec{z}) K( \vec{z} , \vec{x}) K ( \vec{x} , \vec{y} ) \phi(\vec{y}) = \frac{1}{4} \int d^{3}x \ \phi(\vec{x}) (- \nabla^{2}) \phi(\vec{x}) .[/tex] Thus [tex]\int d^{3} x \ K( \vec{z} , \vec{x}) K ( \vec{x} , \vec{y} ) = - \frac{1}{4} \nabla^{2} \delta^{3} (\vec{z} - \vec{y}) . \ \ (6)[/tex]
    Since the right-hand depends only on [itex]\vec{z} - \vec{y}[/itex], we may assume that [itex]K[/itex] is translational invariant and Fourier transform it as [tex]K ( \vec{x} , \vec{y} ) = \int \frac{d^{3}k}{(2 \pi)^{3}} \bar{K} (\vec{k}) \ e^{i \vec{k} \cdot (\vec{x} - \vec{y})} .[/tex] Plugging this in (6), we find [itex]\bar{K} (\vec{k}) = \sqrt{\vec{k} \cdot \vec{k}} / 2 = \omega (\vec{k}) / 2[/itex]. Therefore, your kernel is [tex]K (\vec{x} , \vec{y}) = \frac{1}{2} \int \frac{d^{3}k}{(2 \pi)^{3}} \ \omega (\vec{k}) \ e^{i \vec{k} \cdot (\vec{x} - \vec{y})} .[/tex]
    I suppose they want you to keep track of the Fourier component which for real scalar field is [itex]\phi^{*}_{k} = \phi_{- k}[/itex]. So the vacuum wave functional is actually [tex]\Psi_{0}[\phi] = N \exp \left( - \frac{1}{2} \int \frac{d^{3} k}{(2 \pi)^{3}} \ \omega (\vec{k}) \phi (\vec{k}) \phi( - \vec{k}) \right) .[/tex]

    Of course it is divergent. What is the first divergent quantity you encounter in QFT? Just calculate the ground state energy from (5). It is the sum of zero point energies of all of the field oscillators [tex]E_{0} = \frac{1}{2} \int d^{3}k \ \omega (\vec{k}) \delta^{3} (0) .[/tex]

    Sam
     
  6. Jun 5, 2015 #5

    ShayanJ

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    Nice reply, Thank you very much Sam!

    But I have three problems:

    1) I suppose you got equation 4 from putting ## \Psi_0[\phi]=N \exp(-I[\phi]) ## in equation 2. But I can't do it. Could you give me only a plan of the way for its calculation? Thanks again.

    2) After inserting ## K(\vec x, \vec x) ## in 5 to calculate ## E_0##, How did you get the delta? It seems strange to me because to get the delta, we need to integrate w.r.t. k but you got the delta before that!

    Thanks again.
     
    Last edited: Jun 6, 2015
  7. Jun 6, 2015 #6

    samalkhaiat

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    For massless real scalar field, the Hamiltonian operator is given by [tex]H = \frac{1}{2} \int d^{3} x \left( \left( \hat{\pi} ( t , x ) \right)^{2} + \left( \nabla \hat{\phi}( t , x ) \right)^{2} \right) .[/tex] In the Schrodinger picture, the operators [itex]( \hat{\phi} , \hat{\pi} )[/itex] are time independent. So, in the coordinate (Schroginger) representation we have [tex]\hat{\phi} (0 , \vec{x}) | \phi \rangle = \phi ( \vec{x} ) | \phi \rangle , \ \ \hat{\pi} (0 , \vec{x}) = - i \frac{\delta}{\delta \phi (\vec{x})} ,[/tex] and [tex]\Psi [\phi , t] = \langle \phi | \Psi ( t ) \rangle ,[/tex]where [itex]|\phi \rangle[/itex] is the coordinate basis. In the coordinate representation, the Hamiltonian is diagonal [tex]\langle \phi | H | \bar{\phi} \rangle = \delta (\phi - \bar{\phi}) \ H (\phi (x) , \delta / \delta \phi (x)) .[/tex] So the Schrodinger equation for the state vector [itex]|\Psi (t)\rangle[/itex] becomes functional differential equation[tex]i \partial_{t} \Psi[\phi , t] = \int (D\bar{\phi})\ \langle \phi | H | \bar{\phi}\rangle \langle \bar{\phi}| \Psi (t) \rangle = H( \phi , \delta / \delta \phi ) \Psi[\phi , t].[/tex] Writing [tex]\Psi [ \phi , t] = \Psi[\phi] e^{- i E t} ,[/tex] we obtain the time-independent Schrodinger equation [tex]\frac{1}{2} \int d^{3}x \left( - \frac{\delta^{2}}{\delta \phi^{2}(\vec{x})} + \vec{\nabla} \phi(\vec{x}) \cdot \vec{\nabla} \phi(\vec{x}) \right) \Psi[\phi] = E\Psi[\phi] .[/tex] Integrating the second term by parts, we find [tex]- \frac{1}{2} \int d^{3}x \frac{\delta^{2} \Psi[\phi]}{\delta \phi^{2}(\vec{x})} = \left( E + \frac{1}{2} \int d^3 x \ \phi(\vec{x}) \nabla^{2} \phi(\vec{x}) \right) \Psi[\phi] .[/tex] Okay, now put [itex]\Psi_{0}[\phi] = N \exp( - I[\phi])[/itex] and use the chain rule to obtain [tex]\frac{\delta^{2}\Psi_{0}}{\delta \phi^{2}} = \frac{\partial^{2} \Psi_{0}}{\partial I^{2}} \left( \frac{\delta I}{\delta \phi} \right)^{2} + \frac{\partial \Psi_{0}}{\partial I} \frac{\delta^{2}I}{\delta \phi^{2}} .[/tex]
    *****************
    [tex](2 \pi)^{3} \delta^{3} (q) = \int d^{3}x \ e^{ i \vec{q} \cdot \vec{x}} .[/tex] So [tex](2 \pi)^{3} \delta^{3} (0) = \int d^{3}x .[/tex]
     
    Last edited: Jun 6, 2015
  8. Jun 6, 2015 #7

    ShayanJ

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    Its always a great pleasure(and of course unbelievably helpful) to receive replies from you Sam. Thanks a lot.
     
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