# Why is the de Broglie equation as well as the Schrodinger equation is correct for massive particle?

Starting from special relativity, here I see the de Broglie approximation is valid only if $m_0=0$.

Derivation:

$E^2=P^2C^2+m_0^2C^4$. Here we put Plank-Einstein relation $E=h\nu=h\frac{C}{\lambda}$. Finally,

$\lambda=\frac{h}{\sqrt{P^2+m_0^2C^2}} \hspace{2cm} (1)$.

If $m_0=0$ then $\lambda=\frac{h}{p}$ (de Broglie approximation).

Furthermore we know the Schrodinger equation was derived by assuming that the de Broglie approximation is true for all particles, even if $m_0 \neq 0$. But if we take special relativity very strictly then this approximation looks incorrect.

In addition, if we try to derive the Schrodinger equation from the exact relation found in '1', we find completely different equation. For checking it out, lets take a wave function-

$\Psi=Ae^{i(\frac{2\pi}{\lambda}x-\omega t)}=Ae^{i(\frac{\sqrt{P^2+m_0^2C^2}}{\hbar}x-\frac{E}{\hbar} t)} \hspace{2cm}$ (putting $\lambda$ from '1', $\frac{h}{2\pi}=\hbar$ and $E=\hbar\omega$).

Then, $\frac{\partial^2 \Psi}{\partial x^2}=-\frac{P^2+m_0^2C^2}{\hbar^2}\Psi=-\frac{E^2}{C^2\hbar^2}\Psi$

$\implies E^2\Psi=-C^2\hbar^2\frac{\partial^2 \Psi}{\partial x^2} \hspace{4cm} (2)$

Again, $\frac{\partial \Psi}{\partial t}=-i\frac{E}{\hbar}\Psi \implies E\Psi=-i\hbar\frac{\partial \Psi}{\partial t}$.

Here we see operator $E=-i\hbar\frac{\partial}{\partial t} \implies E^2=-\hbar^2\frac{\partial^2}{\partial t^2}$.

$\implies E^2\Psi=-\hbar^2\frac{\partial^2 \Psi}{\partial t^2} \hspace{4cm} (3)$

Combining (2) and (3) we find the differential equation:

$\frac{\partial^2 \Psi}{\partial x^2}=\frac{1}{C^2}\frac{\partial^2 \Psi}{\partial t^2}$

It is the Maxwell's equation, not the well known Schrodiner equation!

Therefore for the Schrodinger equation to exist, the de Broglie approximation must hold for $m_0 \neq 0.$ I see a clear contradiction here. Then why is the Schrodinger equation correct after all?

• In your step $E=h\nu=\frac{hc}{\lambda}$, you use the relation $c=\nu\lambda$, which is not valid for massive particles. – probably_someone Jan 31 '18 at 23:52
• Also, $E\Psi=-i\hbar\frac{\partial\Psi}{\partial t}$ is the Schrodinger equation. Why are you assuming the Schrodinger equation in your derivation of the Schrodinger equation? – probably_someone Jan 31 '18 at 23:59
• Phase is a 4-scalar, the product of 2 4-vectors: $\phi = k_{\mu}x^{\mu} = (\vec{p}\cdot\vec{x} - Et)/\hbar = \vec{k}\cdot\vec{x}-\omega t$; moreover, the dispersion relation $\omega^2 = (ck)^2-(mc^2/\hbar)^2$ is fine--it means that there is finite frequency at $0$ wave-number--just like a waveguide. – JEB Feb 1 '18 at 0:03

Your relation in (1) is wrong, because for massive particles $\lambda\nu=c$ does not hold. I will show that if you insert the correct relation between $\lambda$ and $\nu$, you recover the de Broglie relation, whether the particle is massive or massless.

To see why this is true, take the usual expression for energy,

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

and substitute the usual Planck-Einstein relations, $E=\hbar\omega$ and $p=\hbar k$. From this we get

$$\hbar^2\omega^2=\hbar^2k^2c^2+m^2c^4$$

which reduces to

$$k=\sqrt{\left(\frac{\omega}{c}\right)^2-\left(\frac{mc}{\hbar}\right)^2}$$

When we take the product $\lambda\nu=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}$, we get

$$\lambda\nu=\frac{\omega}{k}=\frac{\omega}{\sqrt{\left(\frac{\omega}{c}\right)^2-\left(\frac{mc}{\hbar}\right)^2}}=\frac{1}{\sqrt{\left(\frac{1}{c}\right)^2-\left(\frac{mc}{\hbar\omega}\right)^2}}=\frac{c}{\sqrt{1-\left(\frac{mc^2}{\hbar\omega}\right)^2}}$$

Now note that

$$\frac{mc^2}{\hbar\omega}=\frac{mc^2}{E}=\frac{mc^2}{\gamma mc^2}=\frac{1}{\gamma}=\sqrt{1-\beta^2}$$

which, when we substitute in the above, gives us

$$\lambda\nu=\frac{c}{\beta}$$

Now that we know how to substitute $\lambda$ for $\nu$, let's proceed with your derivation. Starting with $E^2=p^2c^2+m^2c^4$ and taking the Planck-Einstein relation $E=h\nu=\frac{hc}{\beta\lambda}$, we get the following:

$$\frac{h^2c^2}{\beta^2\lambda^2}=p^2c^2+m^2c^4$$

which reduces to:

$$\lambda=\frac{h}{\beta\sqrt{p^2+m^2c^2}}$$

Inserting the standard definition again, $\frac{E}{c}=\sqrt{p^2+m^2c^2}$, we get

$$\lambda=\frac{hc}{\beta E}$$

and inserting $\frac{E}{p}=\frac{\gamma mc^2}{\gamma\beta mc}=\frac{c}{\beta}$, we end up with

$$\lambda=\frac{h}{p}$$

which is the de Broglie relation. It's not an approximation; it's exact. There were no approximations in the above derivation.

• Dear probably, you started taking $P=\hbar k$, which is in fact the de Broglie approximation. At the end of the calculation you are getting the same relation back which is quite expected. Here is my question, why is this relation be taken to be true at first place? – Alam Feb 1 '18 at 16:15
• @Alan Where in the world are you getting that $p=\hbar k$ is an approximation? Do you have a source that agrees with you? Because I've been searching everywhere in the past 24 hours and I haven't seen anyone ever say this. The relation is taken to be true because it has been experimentally verified (see e.g. spiedigitallibrary.org/conference-proceedings-of-spie/3767/0000/…). – probably_someone Feb 1 '18 at 16:26