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Vacuum state of the Klein-Gordon field

  1. Aug 2, 2008 #1
    Why are my formulas not displayed correctly?
  2. jcsd
  3. Aug 2, 2008 #2
    It's not [\itex] but it's [/itex] :smile:
  4. Aug 2, 2008 #3
    I have a two questions concering the vacuum state of the Klein-Gordon field. But let me first briefly introduce the topic.

    Suppose we are in special relativity, adopt the signature for the metric [itex]\eta_{\mu \nu} = (+,-,-,-)[/itex], and work in natural units, i.e. [itex]c=1[/itex] and [itex]\hbar=1[/itex]. Then the Lagrangian for the Klein-Gordon field is given by

    [tex] \mathcal{L} = \frac{1}{2} \dot{\Phi}^2- (\nabla \Phi)^2 - m^2 \Phi^2 = \frac{1}{2} g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- m^2 \Phi^2[/tex]

    and the equation of motion is the Klein-Gordon equation

    [tex]\left( \square \Phi + m^2 \right)\Phi = 0.[/tex]

    The canonical momentum [itex]\pi(x)[/itex] to [itex]\Phi(x)[/itex] is given by

    [tex]\pi(x) = \frac{\partial \mathcal{L}}{\partial \dot{\Phi}} = \dot{\Phi}.[/tex]

    Now we apply the canonical quantization to [itex]\Phi(x)[/itex] and [itex]\pi(x)[/itex]. This means [itex]\Phi \rightarrow \hat{\Phi}[/itex] and [itex]\pi \rightarrow \hat{\pi}[/itex] become operators and are subject to the following commutation rules:

    [tex]\left[ \hat{\Phi}(\vec{x},t),\hat{\pi}(\vec{x}',t) \right] = i \delta(\vec{x}-\vec{x}'),[/tex]
    [tex]\left[ \hat{\Phi}(\vec{x},t),\hat{\Phi}(\vec{x}',t) \right] = \left[ \hat{\pi}(\vec{x},t),\hat{\pi}(\vec{x}',t) \right] = 0.[/tex]

    This equivalent to

    [tex]\left[ \hat{a}(\vec{k}),\hat{a}^\dag(\vec{k}') \right] = (2 \pi)^3 \delta(\vec{k}-\vec{k}'),[/tex]
    [tex]\left[ \hat{a}(\vec{x}),\hat{a}(\vec{k}') \right] = \left[ \hat{a}^\dag(\vec{k}),\hat{a}^\dag(\vec{k}') \right] = 0,[/tex]

    whereas [itex]\hat{a}(\vec{k})[/itex] and [itex]\hat{a}^\dag(\vec{k})[/itex] are related to [itex]\hat{\Phi}(\vec{x},t)[/itex] and [itex]\hat{\pi}(\vec{x},t)[/itex] by

    [tex]\hat{\Phi}(\vec{x},t) = \frac{1}{(2 \pi)^3} \int \frac{1}{\sqrt{2 \omega_k}} \left( \hat{a}(\vec{k}) e^{i(\vec{k}\vec{x}-\omega_k t)} + \hat{a}^\dag(\vec{k}) e^{-i(\vec{k}\vec{x}-\omega_k t)} \right) dk^3[/tex]
    [tex]\hat{\pi}(\vec{x},t) = -\frac{i}{(2 \pi)^3} \int \sqrt{\frac{\omega_k}{2}} \left( \hat{a}(\vec{k}) e^{i(\vec{k}\vec{x}-\omega_k t)} - \hat{a}^\dag(\vec{k}) e^{-i(\vec{k}\vec{x}-\omega_k t)} \right) dk^3[/tex]


    [tex]\hat{a}(\vec{k}) = \sqrt{\frac{\omega_k}{2}} \int \left( \hat{\Phi}(\vec{x},t) + \frac{i}{\omega_k} \hat{\pi}(\vec{x},t) \right) e^{-i(\vec{k}\vec{x}-\omega_k t)} dx^3 [/tex]
    [tex]\hat{a}^\dag(\vec{k}) = \sqrt{\frac{\omega_k}{2}} \int \left( \hat{\Phi}(\vec{x},t) - \frac{i}{\omega_k} \hat{\pi}(\vec{x},t) \right) e^{+i(\vec{k}\vec{x}-\omega_k t)} dx^3. [/tex]

    (For a derivation look for instance at http://web.mit.edu/8.323/spring06/notes/ft1qsf-06.pdf.)

    According to the commutation rules, [itex]\hat{a}(\vec{k})[/itex] and [itex]\hat{a}^\dag(\vec{k})[/itex] can be interpreted as annihilation and creation operators respectively for particles with momentum [itex]\vec{k}[/itex]. We can now define the vacuum state [itex]\left|0\right\rangle[/itex] by

    [tex] \hat{a}\left|0\right\rangle = 0[/tex].

    Now, I can ask my questions. From what I have written here, how can I prove that for a given [itex]\vec{k}[/itex] in the ground state [itex]\left|0\right\rangle[/itex] both the real and imaginary part of [itex]\Phi(\vec{k},t) = \int \Phi(\vec{x},t) e^{-i\vec{k}\vec{x}} dx^3[/itex] are independent Gaussian distributed with zero mean, i.e. by means of repeted measurements I would find the values of the real and imaginary part to be independent Gaussian random variables?

    And a second question: How can I prove formally that [itex]\left|0\right\rangle[/itex] is translational and rotational invariant?

    Thanks a lot for any suggestion!
    Last edited by a moderator: Apr 23, 2017
  5. Aug 2, 2008 #4
    How about reviewing how the exact same thing is done with the harmonic oscillator ?
  6. Aug 2, 2008 #5
    I did this of course. If we have the simple quantum mechanical harmonic oscillator, the annihilation operator reads as

    [tex]\hat{a} = \sqrt{\frac{\omega}{2}} \left(\hat{x} + \frac{i}{\omega} \hat{p} \right) = \sqrt{\frac{\omega}{2}} \left(x + \frac{1}{\omega} \frac{d}{dx} \right), [/tex]

    where in the last step we have inserted the spatial representation [itex]\hat{x} \rightarrow x[/itex] and [itex]\hat{p} \rightarrow \frac{1}{i}\frac{d}{dx} [/itex]. So the differential equation

    [tex] \hat{a} \Psi_0(x) = 0[/tex]

    is so solved by

    [tex]\Psi_0(x) \propto \exp\left(-\frac{1}{2 \omega} x^2\right). [/tex]

    So the probability in the groundstate to find the particle at the point x is

    [tex]\left| \Psi_0(x) \right|^2 \propto \exp\left(-\frac{1}{\omega} x^2\right), [/tex]

    thus the probability density function (pdf) it is Gaussian distributed with zero mean.

    But in the case of the Klein-Gordon field, we don't have a representation analogue to [itex]\hat{x} \rightarrow x[/itex] and [itex]\hat{p} \rightarrow \frac{1}{i}\frac{d}{dx} [/itex], don't we? Furthermore what is [itex]x[/itex] in the case of the simple harmonic oscillator is now a complex number [itex]\Phi(\vec{k},t) = \int \Phi(\vec{x},t) e^{-i\vec{k}\vec{x}} dx^3[/itex] for a given [itex]\vec{k}[/itex]. (Note that in my introduction to the Klein-Gordon field, I have worked in the Heisenberg picture while for the simple harmonic oscillator I have worked in the Schrödinger Picture. But since the time dependence is trivial anyway, time can just be neglected.)

    So let me repete my question: How can I show in case of the Klein-Gordon field, that both, real part and imaginary part of [itex]\Phi(\vec{k},t)[/itex] are independent Gaussian distributed? I don't expect it to be very difficult, but I don't see yet the formal calculation.
    Last edited: Aug 2, 2008
  7. Aug 2, 2008 #6


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    This is problem 8.8 in Srednicki's book:

    A harmonic oscillator (in units with [itex]m=\hbar=1[/itex]) has a ground-state wave function [itex]\langle q|0\rangle \propto e^{-\omega q^2/2}[/itex]. Now consider a real scalar field [itex]\varphi(x)[/itex], and define a field eigenstate [itex]|A\rangle[/itex] that obeys

    \varphi({\bf x},0)|A\rangle = A({\bf x})|A\rangle \;,

    where the function [itex]A({\bf x})[/itex] is everywhere real. For a free-field theory, show that the ground-state wave functional is

    \langle A|0\rangle \propto \exp\!\left[-\,{1\over2}\int{d^3k\over(2\pi)^3}\,
    \omega({\bf k})\tilde A({\bf k})\tilde A(-{\bf k})\right],

    where [itex]\tilde A({\bf k}) \equiv \int d^3x\,e^{-i{\bf k}\cdot{\bf x}}A({\bf x})[/itex] and [itex]\omega({\bf k})\equiv({\bf k}^2+m^2)^{1/2}[/itex].
  8. Aug 3, 2008 #7
    I'm not sure how to help you here, and nobody else seems to really do so either. First I noted that
    refers to a real KG (neutral), and this is also the case in your notes. Complex KG (charged) would have had
    (or whatever you use to note hermitian conjugate, but the operator will not be equal to [itex]\dot{\Phi}[/itex] unless you have a real field)

    Honestly I think I don't understand your question :frown:
    There is no "[itex]\vec{k}[/itex] in the ground state", is there ? Your HO-like construction creates plane waves with [itex]\vec{k}[/itex] on the vacuum [itex]\left|0\right\rangle[/itex].

    Not sure of being any help here...
  9. Aug 3, 2008 #8
    Let me clarify a few things. First of all, [itex]\Phi[/itex] is a real scalar field (we don't consider complex scalar field hier), thus the step [itex]\Phi \rightarrow \hat{\Phi}[/itex] yields a hermitian operator [itex]\hat{\Phi}[/itex]. This means [itex]\Phi[/itex] is an observable whose eigenvalues are measurable, i.e.

    [tex]\hat{\Phi} \left| \Phi\right\rangle = \Phi \left| \Phi\right\rangle.[/tex]

    But here, I am NOT interested in [itex]\Phi[/itex], but in its Fourier transformed [itex]\Phi_k[/itex]. (Note: For clarification I flag now the Fourier transformed explicitly by a [itex]k[/itex]. Maybe this was confusing in my earlier posts.) However, since the Fourier transformed [itex]\Phi_k[/itex] is a complex function in general, it can't correspond to a hermitian operator and so it can't correspond to an observable. However the real part and the imaginary part of [itex]\Phi_k[/itex], i.e. [itex]\Phi_k = \rm{Re}(\Phi_k) + i \rm{Im}(\Phi_k)[/itex], are both real and thus correspond to hermitian operators [itex]\widehat{\rm{Re}(\Phi_k)}[/itex] and [itex]\widehat{\rm{Im}(\Phi_k)}[/itex] respectively. Thus the two quantities, real and imaginary parts, are both observables and can, in principle, be measured. Furthermore they can be measured independently if

    [tex]\left[\widehat{\rm{Re}(\Phi_k)},\widehat{\rm{Im}(\Phi_{k'})} \right] = 0.[/tex]

    So this is a relation we have to show. Now we introduce the eigenstates for these two operators

    [tex]\widehat{\rm{Re}(\Phi_k)} \left| \rm{Re}(\Phi_k)\right\rangle = \rm{Re}(\Phi_k) \left| \rm{Re}(\Phi_k)\right\rangle[/tex]
    [tex]\widehat{\rm{Im}(\Phi_k)} \left| \rm{Im}(\Phi_k)\right\rangle = \rm{Im}(\Phi_k) \left| \rm{Im}(\Phi_k)\right\rangle.[/tex]

    The quantity I am interested in is basically the probability of measuring [itex]\rm{Re}(\Phi_k)[/itex] and [itex]\rm{Im}(\Phi_k)[/itex] while we are in the groundstate of the system, i.e.

    [tex]\left\langle \rm{Re}(\Phi_k) | 0 \right\rangle = ?[/tex]
    [tex]\left\langle \rm{Im}(\Phi_k) | 0 \right\rangle = ?[/tex]

    The operators [itex]\hat{\rm{Re}(\Phi_k)}[/itex] and [itex]\hat{\rm{Im}(\Phi_k)}[/itex] should be easily expressible in terms of [itex]\hat{a_k}[/itex] and [itex]\hat{a_k^\dag}[/itex].

    This is actually a simple basic question of quantum field theory and I hoped that somebody could quickly show me how the calculation works since I am not an expert in quantum field theory. Furthermore, if anything I have written here is not correct, please let me know. I hope the problem is clearer now.
  10. Aug 4, 2008 #9


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    The thing to do is to write the classical hamiltonian in terms of the real and imaginary parts of Phi_k and their conjugate momenta. You should find that each is an independent harmonic oscillator. The ground state wave function(al) then follows immediately, by applying QM to each of these independent oscillators.
  11. Aug 5, 2008 #10
    I have now found the formal proof (hopefully). I will give the thread of the solution soon.
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