Mde decomposition of quantum field in a box

In summary, the conversation discusses the Klein Gordon Equation in a box with Dirichlet conditions and the use of 1-D functions to obtain a mode decomposition of the equation. The attempt at a solution involves writing the equation in terms of sine functions and reducing it to a Klein Gordon equation form. The solution is obtained by showing that the equation must be valid for any choice of integers n, n1, n2, and n3, proving the final equation to be true for all possible values of n.
  • #1
DavidDC
1
0

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.
 

Related to Mde decomposition of quantum field in a box

1. What is the Mde decomposition of quantum field in a box?

The Mde decomposition of quantum field in a box is a mathematical technique used to study the behavior of a quantum field confined within a box. It involves breaking down the field into a sum of simple modes or oscillations, which can then be analyzed separately.

2. Why is the Mde decomposition important in quantum field theory?

The Mde decomposition is important because it allows us to understand the complex behavior of a quantum field in a more manageable way. By breaking down the field into simpler modes, we can study each mode individually and then combine them to understand the overall behavior of the field.

3. How does the Mde decomposition affect the quantization of the field?

The Mde decomposition does not affect the quantization of the field itself, but it provides a useful tool for quantizing the field in a confined system such as a box. By analyzing each mode separately, we can determine the quantized energy levels of the field within the box.

4. Can the Mde decomposition be applied to any type of quantum field?

Yes, the Mde decomposition can be applied to any type of quantum field, whether it is a scalar field, spinor field, or vector field. It is a general technique that can be used to study the behavior of any quantum field in a confined system.

5. What are the limitations of the Mde decomposition of quantum field in a box?

The Mde decomposition is most useful for studying the behavior of a quantum field in a confined system, such as a box. It may not be as applicable in more complex or open systems. Additionally, the Mde decomposition assumes that the field is homogeneous and isotropic, which may not always be the case in real-world systems.

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