Mde decomposition of quantum field in a box

  • #1
DavidDC
1
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Homework Statement


I am considering the Klein Gordon Equation in a box with Dirichlet conditions (i.e., ##\hat{\phi}(x,t)|_{boundary} = 0 ##). 1-D functions that obey the Dirichlet condition on interval ##[0,L]## are of the form below (using the discrete Fourier sine transform)
$$f(x) = \sum_{n=1}^{\infty}f(n)\sqrt{\frac{2}{L}}\sin(\frac{n \pi}{L}x) \tag 1$$I need to find a mode decomposition of ##\hat{\phi}(\overrightarrow{x},t)## in terms of ##\hat{\phi}(\overrightarrow{n},t)## and obtain the Klein Gordon equation
$$(\frac{\partial^2 }{\partial t^2} - \Delta + m^2)\hat{\phi}(\overrightarrow{x},t) =0$$
in terms of ##\hat{\phi}(\overrightarrow{n},t)##.

The Attempt at a Solution


I decided (and this is a guess) to write ##\hat{\phi}(\overrightarrow{x},t)## as

$$\hat{\phi}(\overrightarrow{x},t) = (\frac{2}{L})^{\frac{3}{2}}\sum_{n_1,n_2,n_3=1}^{\infty}\hat{\phi}(\overrightarrow{n},t)\sin(\frac{n_1 \pi}{L}x)\sin(\frac{n_2 \pi}{L}y)\sin(\frac{n_3 \pi}{L}z)\tag2$$
where ##\overrightarrow{n} = (n_1,n_2,n_3)##
and ##\overrightarrow{x} = (x,y,z)##

This obeys the Dirichlet boundary condition for a box of dimensions ##[0,L], [0,L], [0,L]##. However, I'm not sure if it encompasses all possible ##\hat{\phi}(\overrightarrow{x},t)## that obey the Dirichlet boundary condition. Is this correct?

I stuck this into the Klein Gordon Equation:
$$(\frac{\partial^2 }{\partial t^2} - \Delta + m^2)\hat{\phi}(\overrightarrow{x},t) =0$$
where ##\Delta## is the Laplacian ##\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2}##

and obtained
$$\sum_{n_1,n_2,n_3=1}^{\infty}\sin(\frac{n_1 \pi}{L}x)\sin(\frac{n_2 \pi}{L}y)\sin(\frac{n_3 \pi}{L}z)(\frac{\partial^2 }{\partial t^2} - \frac{n^2\pi^2}{L^2} + m^2)\hat{\phi}(\overrightarrow{n},t)=0\tag3$$
where ##n^2 = n_1^2 + n_2^2 + n_3^2##

It looks like this is the form for uncoupled harmonic oscillators of frequency ##\sqrt(m^2-\frac{n^2\pi^2}{L^2})##, uncoupled because the Laplacian ##\Delta## isn't present. But how do I go about reducing this to a Klein Gordon equation form without the ##\sum## and all the ##\sin## functions. If I can prove that every term in the summation in eqn. ##(3)## is equal to zero, I can argue that since the ##\sin## terms cannot be zero for all ##\overrightarrow{x}##, the Klein Gordon equation for uncoupled harmonic oscillators would follow. i.e.,
$$(\frac{\partial^2 }{\partial t^2} - \frac{n^2\pi^2}{L^2} + m^2)\hat{\phi}(\overrightarrow{n},t)=0\tag4$$
But how do I show that each of the terms in the summation must be equal to zero (if at all it is true)?
 
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  • #2
DavidDC said:

Homework Statement


I am considering the Klein Gordon Equation in a box with Dirichlet conditions (i.e., ##\hat{\phi}(x,t)|_{boundary} = 0 ##). 1-D functions that obey the Dirichlet condition on interval ##[0,L]## are of the form below (using the discrete Fourier sine transform)
$$f(x) = \sum_{n=1}^{\infty}f(n)\sqrt{\frac{2}{L}}\sin(\frac{n \pi}{L}x) \tag 1$$I need to find a mode decomposition of ##\hat{\phi}(\overrightarrow{x},t)## in terms of ##\hat{\phi}(\overrightarrow{n},t)## and obtain the Klein Gordon equation
$$(\frac{\partial^2 }{\partial t^2} - \Delta + m^2)\hat{\phi}(\overrightarrow{x},t) =0$$
in terms of ##\hat{\phi}(\overrightarrow{n},t)##.

The Attempt at a Solution


I decided (and this is a guess) to write ##\hat{\phi}(\overrightarrow{x},t)## as

$$\hat{\phi}(\overrightarrow{x},t) = (\frac{2}{L})^{\frac{3}{2}}\sum_{n_1,n_2,n_3=1}^{\infty}\hat{\phi}(\overrightarrow{n},t)\sin(\frac{n_1 \pi}{L}x)\sin(\frac{n_2 \pi}{L}y)\sin(\frac{n_3 \pi}{L}z)\tag2$$
where ##\overrightarrow{n} = (n_1,n_2,n_3)##
and ##\overrightarrow{x} = (x,y,z)##

This obeys the Dirichlet boundary condition for a box of dimensions ##[0,L], [0,L], [0,L]##. However, I'm not sure if it encompasses all possible ##\hat{\phi}(\overrightarrow{x},t)## that obey the Dirichlet boundary condition. Is this correct?

I stuck this into the Klein Gordon Equation:
$$(\frac{\partial^2 }{\partial t^2} - \Delta + m^2)\hat{\phi}(\overrightarrow{x},t) =0$$
where ##\Delta## is the Laplacian ##\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2}##

and obtained
$$\sum_{n_1,n_2,n_3=1}^{\infty}\sin(\frac{n_1 \pi}{L}x)\sin(\frac{n_2 \pi}{L}y)\sin(\frac{n_3 \pi}{L}z)(\frac{\partial^2 }{\partial t^2} - \frac{n^2\pi^2}{L^2} + m^2)\hat{\phi}(\overrightarrow{n},t)=0\tag3$$
where ##n^2 = n_1^2 + n_2^2 + n_3^2##

It looks like this is the form for uncoupled harmonic oscillators of frequency ##\sqrt(m^2-\frac{n^2\pi^2}{L^2})##, uncoupled because the Laplacian ##\Delta## isn't present. But how do I go about reducing this to a Klein Gordon equation form without the ##\sum## and all the ##\sin## functions. If I can prove that every term in the summation in eqn. ##(3)## is equal to zero, I can argue that since the ##\sin## terms cannot be zero for all ##\overrightarrow{x}##, the Klein Gordon equation for uncoupled harmonic oscillators would follow. i.e.,
$$(\frac{\partial^2 }{\partial t^2} - \frac{n^2\pi^2}{L^2} + m^2)\hat{\phi}(\overrightarrow{n},t)=0\tag4$$
But how do I show that each of the terms in the summation must be equal to zero (if at all it is true)?
Your equation (3) must be valid for any choice of ##n_1,n_2## and ##n_3##, right? (with only condition that they are integers and that ##n^2=n_1^2 + n_2^2 +n_3^2##). Therefore pick ##n_1=1,n_2=n_3=0##, for example. This will prove that your equation 4 must be satisfied with ##n=1##. Then you can repeat the argument for any choice of ##n_1,n_2,n_3##, which proves that your equation 4 must be valid for any choice of ##n^2## given as the sum of the squares of three integers.
 
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