Mde decomposition of quantum field in a box

[DavidDC]

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)?

 

[nrqed]

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.