# What is a plane wave?

## Main Question or Discussion Point

Where does the plane wave solution (cos (ax-bt) and e 2pi(ax-bt)) come from?

I know that any solution of the form f(ax-bt) is a solution to the general wave equation but why are the two particular ones above used for describing photons/ particles in quantum mechanics?

Anyone here know?

Ive got an exam in quantum mechanics in about a month and dont know anything yet!

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Mentz114
Gold Member
Eto to you too.

the plane wave equation

$$\phi = cos(ax - bt)$$ is a solution of

$$a^2\frac{d^2\phi}{dt^2} - b^2\frac{d^2\phi}{dx^2} = 0$$

OK, I might have got a sign wrong but it's close.

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I think you mean either Acos(ax-bt)+Bsin(ax-bt) or e^i(ax-bt). I'd go to wikipedia and look up fourier transforms or fourier sums.

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nicksauce
Homework Helper
For a general wave equation with homogeneous boundary conditions, the only solutions will be sines and cosines. The actual solution is then found by applying fourier series to get the appropriate initial condition.

Trig functions and imaginary exponentials are used because they are simple, well-understood, and easy to work with.

What else would you use? One could try polynomials: ax-bt, (ax-bt)^2, etc. They blow up for large x or t, so they're not very useful for describing particle states. And anything else you might try would be more complicated than a trig function.

jtbell
Mentor
I know that any solution of the form f(ax-bt) is a solution to the general wave equation but why are the two particular ones above used for describing photons/ particles in quantum mechanics?
Plane-wave states have definite momentum $p = \hbar a$ (using your notation). They're actually physically impossible because you can't normalize them in the usual way, but they're useful as idealized approximations, and for constructing physically realistic solutions by superposition (wave packets).

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The wave equation can be written as
$$\left(\frac{\partial}{\partial t} +\frac{\partial}{\partial x}\right) \left(\frac{\partial}{\partial t} - \frac{\partial}{\partial x}\right) f(x,t) = 0$$,
so any function $$a(x+t) + b(x-t)$$ satisfies the equation. The trigonometric/exponential functions you mention satisfy this, but they also form a complete set, meaning that any normalizable solution to the wave equation can be expressed in that basis. Furthermore, the exponential functions are eigenfunctions of the momentum operator, in other words, if you work in that basis, your momentum operator is diagonal. Thus, if you start quantizing the theory, your particles are states with definite momentum.

[dooh! I thing all of us except jtb answered the wrong question.]

The wave equation can be written as
$$\left(\frac{\partial}{\partial t} +\frac{\partial}{\partial x}\right) \left(\frac{\partial}{\partial t} - \frac{\partial}{\partial x}\right) f(x,t) = 0$$,
so any function $$a(x+t) + b(x-t)$$ satisfies the equation. The trigonometric/exponential functions you mention satisfy this, but they also form a complete set, meaning that any normalizable solution to the wave equation can be expressed in that basis. Furthermore, the exponential functions are eigenfunctions of the momentum operator, in other words, if you work in that basis, your momentum operator is diagonal. Thus, if you start quantizing the theory, your particles are states with definite momentum.
You've got me curious, lbrits. Why include the cross terms when they disappear?

[dooh! I thing all of us except jtb answered the wrong question.]
You've got me curious, lbrits. Why include the cross terms when they disappear?
I think every answer is constructive. In any case, the reason for including cross-terms that disappear is due to d'Alembert. Essentially, the wave equation as given is second order in each variable. If you change variables to $$u = t + x$$ and $$v = t - x$$, then the equation becomes

$$\frac{\partial^2}{\partial u \partial v} f(u,v) =0$$,

which is first order in each variable, and has general solution found by simple integration $$f(u,v) = g(u) + h(v)$$.

The reason for all this mess is that if you know the initial conditions of the wave, you can very easily write down the exact solution for all time, and deal with reflections and stuff.

I guess the best answer is that the problem simplifies in some coordinate systems. If you complexify the problem (i.e., Wick rotate), then you'll find that any solution is a sum of a holomorphic and antiholomorphic parts, so then you have all of the tools of complex analysis at your disposal. It makes conformal field theory in 2D also very interesting.

I didn't understand all that, but thanks, lbrits. It would be nice to see the big picture in terms of tensors, if that's at all possible. I done very little (just some algebra) with this (u,v) substitution, but as I recall, lorentz transforms in this coordinate system preserve both area and angles.

Well, tensors aren't really necessary... i'm just using rules of changing variables in partial differentiation. Let
$$t = \frac{1}{2}(u + v)$$
$$x = \frac{1}{2}(u - v)$$
Then
$$\frac{\partial}{\partial t} = \frac{\partial u}{\partial t} \frac{\partial}{\partial u} + \frac{\partial v}{\partial t} \frac{\partial}{\partial v} = \frac{\partial}{\partial u} + \frac{\partial}{\partial v}$$

$$\frac{\partial}{\partial x} = \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} = \frac{\partial}{\partial u} -\frac{\partial}{\partial v}$$

Now, tensors transform in the same way... Specifically, upstairs indices transform the same way as $$dx$$, while downstairs indices transform the same way as $$\tfrac{\partial}{\partial x}$$. Let me demonstrate.

$$T_x = \frac{\partial u}{\partial x} T_u + \frac{\partial v}{\partial x} T_v$$
$$T_t = \frac{\partial u}{\partial t} T_u + \frac{\partial v}{\partial t} T_v$$

$$T^x = \frac{\partial x}{\partial u} T^u + \frac{\partial x}{\partial v} T^v$$
$$T^t = \frac{\partial t}{\partial u} T^u + \frac{\partial t}{\partial v} T^v$$

You can keep going with more indices, this doesn't really have anything to do with wave equations anymore =)

Here's an exercise for you... Express $$T^x,T^t$$ in terms of $$T^u,t^v$$, which you can in turn express in terms of $$T^x,T^t$$ again... When you're done, you'll have something like $$T^x = \dots = T^x$$, and the stuff in between will give you an identity between the partial derivatives. It is this identity that makes $$\frac{\partial}{\partial x^\mu} \frac{\partial}{\partial x_\mu} = \partial_\mu \partial^\mu$$ a Lorentz invariant.

lbrits, Thanks! That's a lot of work in latex. I want to get back to this in a day if you're still around. I've had a lot on my plate.

I found the coordinate transform interesting for a Lorentz boost. The coordinates (u,v) come out of finding the eigenvalue of the matrix,
$$[(\gamma\ \ \beta\gamma)(\beta\gamma\ \ \gamma)]$$
the metric becomes
$$ds^{2} = dudv + dy^{2} + dz^{2}$$
and a Lorentz transform in the (u,v) coordinate basis is
$$[(D\ \ 0)(0\ 1/D)]$$
Taking the limit
$$D\rightarrow0$$
is my answer to "what happens if you ride along on a beam of light?," or on a plane wave, actually. Space-time is reduced to a three dimensional plane normal to v where all of time happens at once. Not all events are included, but only those that happen to exist on the wave crest you're ridding. The rest are taken beyond the horizon.

Sticking to a Minkowski space, where things are nice and flat all around, you might think of stacking a lot of these 3d planes corresponding to wave crests and those in-between, so that they are reassembled into 4 dimenisons. I suppose I've have to take some first and second derivatives in u and v, before taking D to zero, and see what happens.

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Yeah, the coordinates happen to be called "lightcone coordinates", for what it's worth.

Yeah, the coordinates happen to be called "lightcone coordinates", for what it's worth.
Yeah, well, that was some time ago. I'm more interested in how your de Alembertian is related to the Laplace-de Raham operator,
$$(d+\delta)^{2}$$

Yeah, well, that was some time ago. I'm more interested in how your de Alembertian is related to the Laplace-de Raham operator,
$$(d+\delta)^{2}$$
Mmm... that really is overkill. my d'Alembertian operator might as well be a Laplacian (the signature of the metric isn't really important). If you want to know how to show $$\nabla^2 f = (d+\delta)^2 f$$ (i.e., when acting on 0-forms, then look here: http://en.wikipedia.org/wiki/Laplace-Beltrami_operator/Proofs

reilly
siresmith -- You might consider small oscillation problems -- connected springs. This problem is covered in almost all texts on mechanics. If you work through this type of problem you will see clearly how the exponential functions arise.

In other than rectangular coordinates you may not get straightforward exponentials -- check out wave equations in parabolic or elliptic coordinates.

For many problems in rectangular coords (or spherical) plane wave solutions are the easiest to work with, and often lead naturally to Fourier series or integrals. No magic here, just practical math for physics.
Regards,
Reilly Atkinson

I've been taking an informal survey of harmonic k-forms with complex entries in the lorentz metric in four and five dimensions, thus my interest. Thanks for the feedback!

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I've been taking an informal survey of harmonic k-forms with complex entries in the lorentz metric in four and five dimensions, thus my interest. Thanks for the feedback!
I believe there is a one-to-one map between these and objects built out of spinors. If you are interested...

I believe there is a one-to-one map between these and objects built out of spinors. If you are interested...
I dont really know. My math skills are pretty weak. I'm not using the full complex space, but a slice of it,
$$\textbf{R}^{4}\otimes exp(i\textbf{R})$$
so it may not relate.

Quite by accident, I was looking for a pair of type(3,1) 2-forms to find system of equations to do what is commonly said could not be done. In hindsight turn out to be a regauging of the two forms. In fact, they regauge each other. Very peculiar.

One interesting effect is this,
$$i\xi = *\xi$$

I stumbled across the de Rham Laplacian in similar manner. It took me a week to find the name for it. Hope I haven't bored you.