Massless Schrödinger equation

[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"
http://en.wikipedia.org/wiki/Particle_in_a_box"

 

[malawi_glenn]

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

 

[malawi_glenn]

http://psi.phys.wits.ac.za/teaching/Connell/phys284/2005/lecture-01/lecture_01/node16.html

http://en.wikipedia.org/wiki/Klein-Gordon_equation

 

[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"
http://en.wikipedia.org/wiki/Compton_wavelength"

 

[Mentz114]

Have a look here -

https://www.physicsforums.com/archive/index.php/t-227790.html

I think this

$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi = 0$$

is what you are looking for.

 

[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?).

 

[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."

 

[ismaili]

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 )

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.

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."

 

[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

 

[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"

 

[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 want to pick up a text about Quantum electrodynamics (QED), to study the quantum theory of photons.

 

[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"

 

[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.

 

[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"

 

[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?

 

[muppet]

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

 

[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"

 

[malawi_glenn]

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"

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)

 

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

 

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

 

[per.sundqvist]

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"
http://en.wikipedia.org/wiki/Particle_in_a_box"

Yes, you are perfectly right! It follows from the relativistic hamiltonian, when lim m->0.
$$H=\frac{\lim}{m_0\rightarrow 0} \left(V+\sqrt{c^2p^2+m_0^2c^4}\right)\rightarrow c\cdot \hat{p}+V$$

but, that's just "classically", To do the correct thing you should use the Dirac equation, which acts on a four component spinor, and substitute the lower two-component spinor \ksi into psi and you obtain the Klein-Gordon, as has been described before here.

Ok, this is the time-independent solution for the Hamiltonian:
$$H\Psi=V\Psi-c^2\hbar^2\nabla\cdot\left(\frac{1}{E-V(\vec{r})}\nabla\Psi\right)+ \frac{ic^2\hbar^2e^2}{(E-V)^2}\vec{\sigma}\cdot (\vec{E}\times\nabla\Psi)$$
where $$\vec{E}=-\nabla U(r)$$, the electric field.

 

[Orion1]

Relativistic mass Dirac equation for spin-1/2 free particle:
$$\left(\mathbf{\nabla}^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) \psi = \frac{m^2c^2}{\hbar^2} \psi$$

Relativistic massless Dirac equation for a spin-1/2 free particle:
$$\boxed{\left(\mathbf{\nabla}^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) \psi = 0} \; \; \; m = 0$$

Relativistic mass Klein–Gordon equation for spinless free particle:
$$\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi + \frac {m^2 c^2}{\hbar^2} \psi = 0$$

Relativistic massless Klein–Gordon equation for spinless free particle:
$$\boxed{\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi = 0} \; \; \; m = 0$$

I noticed that the difference in the massless Dirac equation and the massless Klein–Gordon equation for a free particle is only a difference in signs. Does the particle spin account for the difference in signs?

Reference:
http://en.wikipedia.org/wiki/Dirac_equation"
http://en.wikipedia.org/wiki/Klein-Gordon_equation"

 

[per.sundqvist]

Relativistic mass Dirac equation for spin-1/2 free particle:
$$\left(\mathbf{\nabla}^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) \psi = \frac{m^2c^2}{\hbar^2} \psi$$

Relativistic massless Dirac equation for a spin-1/2 free particle:
$$\boxed{\left(\mathbf{\nabla}^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) \psi = 0} \; \; \; m = 0$$

Relativistic mass Klein–Gordon equation for spinless free particle:
$$\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi + \frac {m^2 c^2}{\hbar^2} \psi = 0$$

Relativistic massless Klein–Gordon equation for spinless free particle:
$$\boxed{\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi = 0} \; \; \; m = 0$$

I noticed that the difference in the massless Dirac equation and the massless Klein–Gordon equation for a free particle is only a difference in signs. Does the particle spin account for the difference in signs?

Reference:
http://en.wikipedia.org/wiki/Dirac_equation"
http://en.wikipedia.org/wiki/Klein-Gordon_equation"

There is no sign difference. Let V=0, and E\Psi=H\Psi, and E=-i\hbar*d/dt. Then move up the E in the denominator and you get i^2=-1 and you got Klein-Gordon.

PS.I wonder what kind of particle that has no mass that could feel a potential though? Maby my derivation was a bit silly...

 

[Orion1]

Why is the Dirac equation capable of describing spin-1/2 free particles and the Klein–Gordon equation can only describe spinless free particles?

According to the relativistic classical perspective, they are the same equation.

Relativistic mass Dirac equation for spin-1/2 free particle:
$$\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi - \frac {m^2 c^2}{\hbar^2} \psi = 0$$

Relativistic mass Klein–Gordon equation for spinless free particle:
$$\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi + \frac {m^2 c^2}{\hbar^2} \psi = 0$$

Does the particle spin account for the difference in signs?

 

[masudr]

The 4-component Dirac spinor transforms under the (1/2, 1/2) representation of the Poincaré algebra. This is essentially 2 spin-1/2 particles, i.e. the left handed and right handed Weyl spinors.

The scalar Klein-Gordon field transforms under the (0, 0) rep., and so describes a spin-0 particle.

 

[reilly]

Look up the physics of neutrinos; consider the ultra-relativistic form of QED.
Regards,
Reilly Atkinson

 

[Orion1]

Dirac equation for free-particles...



Dirac equation for free-particles plane-wave:
$$\psi = \omega e^{-i p \cdot x} \,$$

Normalization for the four-spinor:
$$\omega^\dagger \omega = 2 E \,$$

Four-component Dirac spinor for free particles:
$$\omega = \begin{bmatrix} \phi^{(s)} \\ \frac{\mathbf{\sigma \cdot p}}{E + m} \phi^{(s)} \end{bmatrix} \,$$

$$u(\mathbf{p}, s) = \sqrt{E+m} \begin{bmatrix} \phi^{(s)} \\ \frac{\mathbf{\sigma \cdot p}}{E + m} \phi^{(s)} \end{bmatrix} \,$$

s = 1 or 2 (spin "up" or "down")

Can the 4-component Dirac spinor transform under any spin representation?

example:
S = 0, 1/2, 1, 2

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

Fundamental particle:
electron - Spin-1/2
neutrino - Spin-1/2

Composite particle:
pion - Spin-0 (spinless)

Reference:
http://en.wikipedia.org/wiki/Dirac_spinor#Two-spinors"

 

[per.sundqvist]

Why is the Dirac equation capable of describing spin-1/2 free particles and the Klein–Gordon equation can only describe spinless free particles?

According to the relativistic classical perspective, they are the same equation.

Relativistic mass Dirac equation for spin-1/2 free particle:
$$\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi - \frac {m^2 c^2}{\hbar^2} \psi = 0$$

Relativistic mass Klein–Gordon equation for spinless free particle:
$$\frac {1}{c^2} \frac{\partial^2}{(\partial t)^2} \psi - \mathbf{\nabla}^2 \psi + \frac {m^2 c^2}{\hbar^2} \psi = 0$$

Does the particle spin account for the difference in signs?

Honestly, the trick with E-> -i\hbar d/dt is awful physics, when going from the time-independent Dirac equ to time-dependent. But when m=0, you don't need four components to obey the relativistic energy-momentum dispersion, so in this case, you don't need to do this substitution between spinor components, and a scalar field is enough, but with m included you need it. Note that Dirac equ itself does not tells you anything about anti-symmetry, which is the rule for fermions. You have to apply this "extra" condition on top of Dirac's equ.