The relativistic de Broglie equation

[mjda]

I'm having trouble understanding how the 4-wave vector is derived, and also how it is then used alongside the 4-momentum vector to formulate the relativistic de Broglie equation.

The inner product of the 4-momentum vector with itself, is an invariant quantity. If we define the 4-momentum vector, P, as:

P = (p''' , iE/c) ---- where p''' is just the 3 dimensional momentum vector.

This then leads to finding:

P.P = p² - (E/c)² <-- this is invariant

The 4-wave vector, N, is defined as:

N = f (c/w , 1) <-- how does one derive this?

From what I have read, you should be able to derive the de Broglie equation by considering the inner product of P and N, with P.P?

The result being:

cP = hN

Has anyone seen this notation before and able to show this result is true? I can't find it anywhere in this form!

 

[Orodruin]

First of all, you have to realize that the de Broglie relation is an empirical relation. It was historically convenient to consider particles having a wavelength proportional to its momentum. For the same reason, you should not expect to ”prove” the relativistic version of the relation.

What you can do is to look at plane wave solutions to the Klein-Gordon or Dirac equations. For a plane wave, you can easily derive the 4-frequency as the gradient of the phase function and relate it to how the solution behaves under space-time translations, which are generated by the momentum and energy operators.

 

[Mister T]

From what I have read, you should be able to derive the de Broglie equation by considering the inner product of P and N, with P.P?

Where did you read this?

 

[mjda]

Where did you read this?

It is in a question from a past paper I found. Maybe my wording is a bit misleading, but the exact phrasing is:

By considering the scalar product of P both with itself, and with N, deduce the deBroglie equation.

 

[PeterDonis]

It is in a question from a past paper I found.

What paper? Please give a reference.

 

[sweet springs]

Relativistic relation

$$E^2-p^2c^2=m^2c^4$$

replacing E and p by QM operators

$$[-\frac{\partial^2}{c^2\partial t^2}+\mathbf{\nabla}^2-\frac{m^2c^2}{\hbar^2}]\psi=0$$

$$(\mathbf{\square}^2+\frac{m^2c^2}{\hbar^2})\psi=0$$

Are you looking for something like this?

 

[Sorcerer]

Relativistic relation

$$E^2-p^2c^2=m^2c^4$$

replacing E and p by QM operators

$$[-\frac{\partial^2}{c^2\partial t^2}+\mathbf{\nabla}^2-\frac{m^2c^2}{\hbar^2}]\psi=0$$

$$(\mathbf{\square}^2+\frac{m^2c^2}{\hbar^2})\psi=0$$

Are you looking for something like this?

That is pretty sexy looking. Would you mind telling me what the square symbol is called so I can look it up? Wild guess is it has something to do with a wave since there is a second order partial derivative there.



Edit- Nevermind, I found it. It is indeed related to waves, specifically the four dimensional, Minkowski space variety.


https://en.m.wikipedia.org/wiki/D'Alembert_operator

 

[Orodruin]

The $\square$ is a second order differential operator, the d'Alembert operator $\square = g^{\mu\nu}\partial_\mu \partial_\nu$. Note that there should not be a square on the $\square$, i.e., the equation should contain $\square$, not $\square^2$.

 

[haushofer]

It's the d'Alembertian operator,

https://en.m.wikipedia.org/wiki/D'Alembert_operator

 

[Sorcerer]

The $\square$ is a second order differential operator, the d'Alembert operator $\square = g^{\mu\nu}\partial_\mu \partial_\nu$. Note that there should not be a square on the $\square$, i.e., the equation should contain $\square$, not $\square^2$.

I really need to take the time to learn Einstein summation convention. But four dimensional partial differential equations and matrices are hard enough without bringing weird dot product rules and tensor stuff into it, let alone short hand for it. ;) *sigh* So much work to be done...

But thanks for the insight! Slowly but surely I’ll learn, until I can get back into school.

 

[mjda]

Relativistic relation

$$E^2-p^2c^2=m^2c^4$$

replacing E and p by QM operators

$$[-\frac{\partial^2}{c^2\partial t^2}+\mathbf{\nabla}^2-\frac{m^2c^2}{\hbar^2}]\psi=0$$

$$(\mathbf{\square}^2+\frac{m^2c^2}{\hbar^2})\psi=0$$

Are you looking for something like this?

I am trying to find it in the form of:

cP = hN

where P and N are both the 4-Momentum, and 4-Wave vectors.

It can also be written in the form:

cm( u , ic ) = hf( cn / w , 1 )

c - speed of light
m - mass
u - 3-D velocity vector
i - imaginary component
f - frequency
h - Planck's const.
w - ang. freq.
n - unit vector

 

[sweet springs]

I suppose you would say
4-momentum
$$mc^2u^\mu=mc^2(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}},\frac{v_x/c}{\sqrt{1-\frac{v^2}{c^2}}},\frac{v_y/c}{\sqrt{1-\frac{v^2}{c^2}}},\frac{v_z/c}{\sqrt{1-\frac{v^2}{c^2}}})$$
equals 4-wave vector
$$c\hbar k^\mu =c\hbar (\omega/c,k_x,k_y,k_z)$$
in de Broglie relation.