Do alpha particles travel faster than c?

[NeonBlack]

(I think this is my first thread in speculations)

This probably doesn't even count as a speculation - just something I noticed when I was helping out a friend with some intro QM.

 

Most people know special relativity's formula for momentum,

[math]p=\frac{mv}{\sqrt{1-v^2/c^2}}[/math]

and that if you put in a velocity greater than c, you will get an imaginary momentum. This is usually discarded as unphysical.

 

Physicists believe that alpha decay occurs when an alpha particle "quantum tunnels" out of a nucleus. I have included the typical simplified cartoon picture of what this looks like (yes, I drew it in paint). The dashed line in the energy of the alpha and the solid line is the potential. The alpha is held inside the nucleus's potential well by the strong force and outside the nucleus, there is a very strong Electro-static repulsion. Obviously, it does not have enough energy to escape classically, but according to QM, it has a very small chance of moving through the potential and out of the nucleus.

 

When the alpha particle is inside the classically forbidden region, it has an exponential wave function kind of like

[math]\Psi \propto e^{-x/a} [/math]

where a is a real constant with dimension of length.

 

Now nobody really knows what the particle is doing while it is tunneling. If we try to apply the momentum operator,

[math]\hat P = -i \hbar [/math]

we get an imaginary eigenvalue (imaginary momentum).

 

This is also regarded as unphysical, but what if it were a physical situation where the particle was traveling faster than c?

Maybe someone will volunteer to do some numbers and get an estimate of what superluminal speed would be required for this to be consistent.

 

I apologize if this is completely incoherent. The change to daylight saving time has wrecked my sleep schedule.

[YT2095]

the bottom line is simply, No they don`t travel faster than c

 

Nothing does, and certainly not a Helium Nucleus whose Mass is 4 grams for every 6.02 x10^23 nuclei.

 

IIRC, in order it goes something like (Fastest to slowest) Gamma, Beta, Alpha, and I think the Slowest is Thermalised Neutrons and they`re about 20 kilometers a second (Quite slow!).

[timo]

I think you have an incorrect view of QM, at least your statement

When the alpha particle is inside the classically forbidden region, it has an exponential wave function kind of like

[math]\Psi \propto e^{-x/a} [/math]

where a is a real constant with dimension of length.

seems to indicate that. The next best correct statement would be "when the wave-function is non-zero in a region R, then the particle can be found in that region". Finding it in a region, however, is a measurement process, meaning you change the system.

 

QM quick-guide how tunneling should be seen (several simplifications and non-stated assumptions implicit):

- Particles are in a state |state>, an element of a Hilbert space. States can be represented by wave-functions f(x).

- The Hilbert space is more or less (might depend on how you treat the normalization constraint) just a vector space.

- A vector space has a basis. A nice base for the Hilbert space is a base of states which are eigenstates to the Hamiltonian (Energy-eigenstates).

- If |n> is an EV to the Hamiltonian with energy En, then from the Schrödinger equation, the equation of motion of |n> is f(x,t) = exp(-i En t/hbar) f(x,t=0).

=> The probability density for finding a particle in some location x remains constant if it was in an energy eigenstate.

- For the potential you sketched, there is no energy eigenstate for which the whole wavefunction is constrained inside the middle sink (that's probably what you meant with the exp(-x/a)).

=> A wave-function f(x,t=0) for an initial state (at t=0) that is constrained to being inside the middle sink is not equal to the WF of a single energy-eigenstate but can be expressed as a linear combination of energy eigenstates, e.g. |state> = c1 |1> + c2 |2>. The coefficients (c1 and c2 here) add up in such a way that the probability for finding the particle outside =0.

- Due to |1> and |2> having different energies, the equation of motion allows that the probability of finding the particle outside does not need to remain 0 under time evolution.

=> Even though |state> initially had zero probability for finding the particle outside the nucleus, this does not have to hold true under time-evolution.

[thedarkshade]

Alphas are helium nuclei, so they have mass, hence NO c for them!