# Massless Schrödinger equation

1. Jul 1, 2008

### Orion1

A massless particle situated in a 1D infinite square well with momentum only in the direction of quantum confinement (the x direction):

$$E_t \psi (x) = \hbar \omega \psi (x) = i \hbar \frac{\partial}{\partial t} \psi (x)$$

$$p(x) \psi(x) = \hbar k_x \psi (x) = -i \hbar \frac{\partial}{\partial x} \psi (x)$$

Integration by substitution:
$$E_k(x) \psi(x) = c p(x) \psi(x) = -i \hbar c \frac{\partial}{\partial x} \psi (x)$$

$$E_t \psi (x) = E_k(x) \psi(x) + V_u(x) \psi (x)$$

Integration by substitution:
$$i \hbar \frac{\partial}{\partial t} \psi (x) = -i \hbar c \frac{\partial}{\partial x} \psi (x) + V_u(x) \psi (x)$$

Massless Schrödinger equation:
$$\boxed{i \hbar \frac{\partial}{\partial t} \psi (x) = -i \hbar c \frac{\partial}{\partial x} \psi (x) + V_u(x) \psi (x)}$$

Mass Schrödinger equation:
$$i \hbar \frac{\partial}{\partial t} \psi (x) = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} + V(x) \psi(x)$$

Is my solution for the massless Schrödinger equation correct?

Reference:
http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation" [Broken]
http://en.wikipedia.org/wiki/Particle_in_a_box" [Broken]

Last edited by a moderator: May 3, 2017
2. Jul 1, 2008

### malawi_glenn

Massless particles travles at the speed of light, hence you need relativistic QM - Klein Gordon eq.

3. Jul 1, 2008

4. Jul 1, 2008

### Orion1

A relativistic massless particle situated in a 1D infinite square well with momentum only in the direction of quantum confinement.

Relativistic mass Klein-Gordon equation:
$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi + \frac {m^2 c^2}{\hbar^2} \psi = 0$$

$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi + \left( \frac {m c}{\hbar} \right)^2 \psi = 0$$

$$\mathbf{p} = -i \hbar \mathbf{\nabla}$$

$$\mathbf{\nabla}^2 = - \left( \frac{p}{\hbar} \right)^2 = - \left( \frac{\hbar}{ \overline{\lambda} \hbar} \right)^2 = - \frac{1}{\overline{\lambda}^2}$$

$$\overline{\lambda} = \frac{\hbar}{m c} = \frac{\hbar c}{m c^2} = \frac{\hbar c}{E}$$

Relativistic massless Klein-Gordon equation:
$$\boxed{\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi + \frac{1}{\overline{\lambda}^2} \psi + \left( \frac{E}{\hbar c} \right)^2 \psi = 0}$$

Is this the correct equation solution for a relativistic massless Klein-Gordon equation?

Reference:
http://en.wikipedia.org/wiki/Klein-Gordon_equation" [Broken]
http://en.wikipedia.org/wiki/Compton_wavelength" [Broken]

Last edited by a moderator: May 3, 2017
5. Jul 1, 2008

6. Jul 1, 2008

### arunma

Maybe I'm mistaken, but it seems that there's no need for a massless Schrodinger equation. The Klein-Gordon equation is derived from the energy-momentum relation for relativistic (i.e. massive) particles. Massless particles always travel at the speed of light, and thus obey Maxwell's Equations. Maxwell's Equations yield the wave equation similar to that for a wave on a string, pressure waves, etc., as Mentz114 has written down above. The Schrodinger Equation was constructed specifically because massive particles have a different dispersion relation than the simple one obeyed by photons. But if the particle in question has no mass, then there's no need for any Schrodinger Equation, and one can simply use the classical wave equation.

Again, I could be mistaken, since I've never encountered a problem such as a massless particle bound in a potential (is this even possible?).

7. Jul 1, 2008

### muppet

Maxwell's equations describe the behaviour of a field, not a particle. So in order to describe the behaviour of particles you need a QFT...
As far as I know no particles are massless apart from photons (neutrino oscillations and all that ) so you probably wouldn't have encountered such a problem. But whatever you use certainly wouldn't be a Schroedinger equation of any description:
"So we multiply that by h-bar squared over 2m to give ... bugger."

8. Jul 1, 2008

### ismaili

BTW, just supplemantarily mention that, actually, the gluons which are the intermediate gauge bosons in QCD are massless. Moreover, the graviton is a spin 2 massless particle.

9. Jul 2, 2008

### malawi_glenn

Orion1: in your post #4, just let m=0, and you'll get the correct result (Mentz114 post #5)

muppet: Gluons are massless =D

10. Jul 2, 2008

### Orion1

A relativistic massless particle situated in a 1D infinite square well with momentum only in the direction of quantum confinement.

Relativistic spinless massless Klein-Gordon equation:
$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi = 0$$

Relativistic massless Bosons:
Photon - Spin-1
Gluon - Spin-1
Graviton - Spin-2

1D infinite square well boundary condition:
$$n \frac{\lambda}{2} = L$$

1D infinite square well momentum:
$$p =\frac{\hbar}{\overline{\lambda}} = \frac{\pi \hbar n}{L}$$

Bosonic energy:
$$E_n = pc$$

Integration by substitution:
$$E_n = pc = \left( \frac{\pi \hbar n}{L} \right) c$$

Eigenenergy for a relativistic massless particle situated in a 1D infinite square well:
$$\boxed{E_n = \frac{ \pi \hbar c n}{L}}$$

According to Wikipedia, the Klein–Gordon equation is described as a relativistic version of the Schrödinger equation for 'spinless' particle models.

Is there really no relativistic wavefunction equation in quantum field theory (QFT) that describes relativistic massless particles with spin?

Reference:
http://en.wikipedia.org/wiki/Particle_in_a_box" [Broken]

Last edited by a moderator: May 3, 2017
11. Jul 2, 2008

### malawi_glenn

Gravitons has not been discovered yet, so no need for it in that list.

Why are you still talking about 'integration' when you don't integrate?

Yes, Klein Gordon is for spinless particles, but you should also be careful here, in Relativistic QM we don't have 'wavefunctions', we have fields instead.

So you might wanna pick up a text about Quantum electrodynamics (QED), to study the quantum theory of photons.

12. Jul 2, 2008

### Orion1

Massless Euler-Lagrange equation...

Relativistic spinless massless Klein-Gordon equation:
$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi = 0$$

(QED) Mass Euler-Lagrange equation (spin-1/2):
$$i \gamma^\mu \partial_\mu \psi - m \psi = e \gamma_\mu A^\mu \psi$$

(QED) Massless Euler-Lagrange equation (spin-1/2):
$$\boxed{i \gamma^\mu \partial_\mu \psi = e \gamma_\mu A^\mu \psi} \; \; \; m = 0$$

$$\boxed{i \gamma^\mu \partial_\mu \psi - e \gamma_\mu A^\mu \psi = 0}$$

Can the relativistic massless Klein-Gordon equation be used to describe fields the same way that the (QED) massless Euler-Lagrange equation does?

Reference:
http://en.wikipedia.org/wiki/Quantum_electrodynamics" [Broken]

Last edited by a moderator: May 3, 2017
13. Jul 2, 2008

### malawi_glenn

Well since the Euler-Lagrange here contains gamma-matrices and 4-derivatives, what do YOU think? =)

I think you need a better source than wiki, try the textbook of Mandl, very good introduction.

14. Jul 2, 2008

### Orion1

$$\boxed{\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi = i \gamma^\mu \partial_\mu \psi}$$

$$\boxed{\mathbf{\nabla}^2 \psi = e \gamma_\mu A^\mu \psi}$$

It seems possible for the massless Euler-Lagrange (QED) equation to 'emulate' the relativistic massless Klein-Gordon equation as a spin-1/2 particle solution.

Reference:
http://en.wikipedia.org/wiki/Emulator" [Broken]

Last edited by a moderator: May 3, 2017
15. Jul 2, 2008

### malawi_glenn

where did

$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi = i \gamma^\mu \partial_\mu \psi$$

and

$$\mathbf{\nabla}^2 \psi = e \gamma_\mu A^\mu \psi$$

Come from?

Do you for example know what $$\partial_\mu$$ is?

16. Jul 2, 2008

### muppet

I stand corrected (twice!)
Particle theory next year... =D

17. Jul 2, 2008

### Orion1

Relativistic massless Klein-Gordon equation equivalent to massless Euler-Lagrange equation:
$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi = i \gamma^\mu \partial_\mu \psi - e \gamma_\mu A^\mu \psi = 0$$

$$\boxed{\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi = i \gamma^\mu \partial_\mu \psi}$$

$$\boxed{\mathbf{\nabla}^2 \psi = e \gamma_\mu A^\mu \psi}$$

$$\partial_\mu$$ is a vector of derivatives in the Euler-Lagrange equation:
$$\partial_\mu = \left(\frac{1}{c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)$$

And is one of the derivative variables for second order in time in the massless Klein-Gordon equation:
$$\boxed{\partial_\mu = \frac{1}{c} \frac{\partial}{\partial t}}$$

Reference:
http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equations" [Broken]

Last edited by a moderator: May 3, 2017
18. Jul 2, 2008

### malawi_glenn

1) Klein-gordon is for spin-0, the other ones are for fermions.. so there is no equivalence

2) Klein-gordon is for free particles, the other ones are for particles in potential (hence the A-mu)

Last edited by a moderator: May 3, 2017
19. Jul 4, 2008

### Orion1

Degenerate Bosonic pressure...

The eigenenergy of a particle of mass m in a one dimensional infinite square well of length L is equivalent to the Fermi energy of a degenerate neutron, which is used to determine neutron degeneracy pressure in a neutron star Fermi sphere. Therefore, the infinite square well is the basis of all models of neutron stars. The Fermi energy is the Fermionic energy of fermions, in the case of bosons, it is the Bosonic energy of a Boson.

Fermi energy = eigenenergy:
$$\boxed{E_f = E_n}$$

Fermionic eigenenergy:
$$E_f =\frac{\hbar^2 \pi^2}{2m L^2} n_f^2$$

1D infinite square well eigenenergy:
$$E_n = \frac{\hbar^2 \pi^2}{2 m L^2} n^2 \,$$

Eigenenergy for a relativistic massless particle situated in a 1D infinite square well:
$$E_n = \frac{\pi \hbar c n}{L}$$

$$n_b = \left( \frac{3 N}{\pi} \right)^{\frac{1}{3}}$$

Integration by substitution:
$$E_n(N) = \frac{\pi \hbar c}{L} \left( \frac{3 N}{\pi} \right)^{\frac{1}{3}}$$

Integration by substitution:
$$E_t = {\int_0}^{N_0} E_n(N) dN = {\int_0}^{N_0} \frac{\pi \hbar c}{L} \left( \frac{3 N}{\pi} \right)^{\frac{1}{3}} dN$$

$$E_t = \frac{3^{\frac{1}{3}} \pi^{\frac{2}{3}} \hbar c}{L} {\int_0}^{N_0} N^{\frac{1}{3}} dN = \frac{3^{\frac{1}{3}} \pi^{\frac{2}{3}} \hbar c}{L} \left( \frac{3 N_0^{\frac{4}{3}}}{4} \right) = \frac{3^{\frac{4}{3}} \pi^{\frac{2}{3}} \hbar c}{4 L} N_0^{\frac{4}{3}}$$

Total Bosonic eigenenergy:
$$\boxed{E_t = \frac{3^{\frac{4}{3}} \pi^{\frac{2}{3}} \hbar c}{4 L} N_0^{\frac{4}{3}}}$$

The elimination of L in favor of V:
$$\boxed{L = V^{\frac{1}{3}}}$$

$$\boxed{E_t = \frac{3^{\frac{4}{3}} \pi^{\frac{2}{3}} \hbar c N_0^{\frac{4}{3}}}{4 V_0^{\frac{1}{3}}}}$$

Degenerate Bosonic pressure:
$$P_b = - \frac{\partial E_t}{\partial V}$$

Integration by differentiation substitution:
$$P_b = - \frac{3^{\frac{4}{3}} \pi^{\frac{2}{3}} \hbar c N_0^{\frac{4}{3}}}{4} \left( \frac{d V_0^{- \frac{1}{3}}}{dV} \right) = - \frac{3^{\frac{4}{3}} \pi^{\frac{2}{3}} \hbar c N_0^{\frac{4}{3}}}{4} \left( - \frac{1}{3 V_0^{\frac{4}{3}}} \right) = \frac{3^{\frac{1}{3}} \pi^{\frac{2}{3}} \hbar c N_0^{\frac{4}{3}}}{4 V_0^{\frac{4}{3}}}}$$

Degenerate Bosonic pressure:
$$\boxed{P_b = \frac{3^{\frac{1}{3}} \pi^{\frac{2}{3}} \hbar c N_0^{\frac{4}{3}}}{4 V_0^{\frac{4}{3}}}}}$$

Degenerate Bosonic density:
$$\rho_0 = \frac{M_0}{V_0}$$

Degenerate Bosonic pressure:
$$\boxed{P_b = \frac{3^{\frac{1}{3}} \pi^{\frac{2}{3}} \hbar c \rho_0^{\frac{4}{3}}}{4}}$$

Reference:
http://en.wikipedia.org/wiki/Particle_in_a_box" [Broken]
http://en.wikipedia.org/wiki/Fermi_energy" [Broken]
https://www.physicsforums.com/showpost.php?p=1787244&postcount=11"
https://www.physicsforums.com/showpost.php?p=1787309&postcount=10"

Last edited by a moderator: May 3, 2017
20. Jul 4, 2008

### Orion1

Total Bosonic eigenenergy:
$$E_t = \frac{3^{\frac{4}{3}} \pi^{\frac{2}{3}} \hbar c}{4 L} N_0^{\frac{4}{3}}$$

$$E_t = \left( \frac{3 N_0}{4} \right) \frac{\pi \hbar c}{L} \left( \frac{3 N_0}{\pi} \right)^{\frac{1}{3}} = \frac{3^{\frac{4}{3}} \pi^{\frac{2}{3}} \hbar c}{4 L} N_0^{\frac{4}{3}}$$

$$\boxed{E_t = {\int_0}^{N_0} E_n(N) dN = \left( \frac{3 N_0}{4} \right) E_n}$$

Total Bosonic eigenenergy:
$$\boxed{E_t = \left( \frac{3 N_0}{4} \right) E_n}$$

Last edited: Jul 4, 2008