Wavefunction of relativistic free particle

In summary, according to Zacku, for any relativistic free particle, the frequency of the wavefunction must be equal to the square of the relativistic energy times the frequency of the wave. However, this relation is dependent on the spin of the particle.
  • #1
I_wonder
9
0
Could anyone please help with the following, rather unusual, query?

I know that for spin 0 bosons, the Klein Gordon equation gives solutions that are similar to the solutions of the Schrodinger equation for a non-relativistic free particle, the only difference being that the energy used when calculating the wave frequency (E/h-bar) is the relativistic energy m*c-square*gamma.

I don't know how the wavefunctions look for spin 1 or spin 1/2 particles. What I would like to know is if there is any kind of particle for which one gets a wavefunction where the frequency is different then m*c-square*gamma/h-bar? Also, is it possible to verify these things experimentaly? I mean, can an experiment get an eigenstate and measure both the energy and the wave frequency? Has that been done for relativistic particles? Have any discrepancies been found? Where could I find such data?

Thanks!
 
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  • #2
I_wonder said:
Could anyone please help with the following, rather unusual, query?

I know that for spin 0 bosons, the Klein Gordon equation gives solutions that are similar to the solutions of the Schrodinger equation for a non-relativistic free particle, the only difference being that the energy used when calculating the wave frequency (E/h-bar) is the relativistic energy m*c-square*gamma.

I don't know how the wavefunctions look for spin 1 or spin 1/2 particles. What I would like to know is if there is any kind of particle for which one gets a wavefunction where the frequency is different then m*c-square*gamma/h-bar? Also, is it possible to verify these things experimentaly? I mean, can an experiment get an eigenstate and measure both the energy and the wave frequency? Has that been done for relativistic particles? Have any discrepancies been found? Where could I find such data?

Thanks!

As far as I know, for any relativistic free particle, the frequency [tex]\omega[/tex] that appears in [tex]e^{-i \omega t}[/tex] when regarding the evolution of a steady state (i.e. an eigenstate of the Hamiltonian operator) must be equal to [tex] \sqrt{m^2c^4+p^2c^4}/\hbar [/tex] because the hamiltonian is the generator of time translations.
However, the relation between [tex]\omega[/tex] and the wave vector [tex]\vec{k}[/tex] will be spin dependent.

P.S1 : Unfortunately the word "wave function" is no more appropriate to qualify a state in relativistic quantum theory. Results can be obtained with this point of vue only for the Klein-Gordon equation and for the Dirac equation (i.e. for a spin zero particle and a spin 1/2 particle) but they lead to fondamental problems that necessite a new approach of quantum mechanics ; that is quantum field theory.

P.S2 : Excuse me for the english, I live in France and as everybody knows french people are pretty bad in english :) .
 
  • #3
Zacku said:
As far as I know, for any relativistic free particle, the frequency [tex]\omega[/tex] that appears in [tex]e^{-i \omega t}[/tex] when regarding the evolution of a steady state (i.e. an eigenstate of the Hamiltonian operator) must be equal to [tex] \sqrt{m^2c^4+p^2c^4}/\hbar [/tex]


Thankyou Zacku! Your english is just fine. It's not my language, either.

I realize of course that I will not be able to say anything inteligent about this subject before I study QFT, which I hope to do next year. What I'm trying to find now, for my own peculiar reasons, is if there are any experimental results that might suggest that some correction to the frequency may be required in the case of a free relativistic particle, that is, that the dependence of the frequency on the energy may be more complicated.

Thanks.
 
  • #4
I_wonder said:
I realize of course that I will not be able to say anything inteligent about this subject before I study QFT, which I hope to do next year. What I'm trying to find now, for my own peculiar reasons, is if there are any experimental results that might suggest that some correction to the frequency may be required in the case of a free relativistic particle, that is, that the dependence of the frequency on the energy may be more complicated.
Thanks.

Actually wathever the formalism (i mean the "classical" one or the "field" one) any states in QM has to satisfy the equation of evolution :
[tex]i \hbar \frac{d}{dt} |\psi(t) \rangle = \hat{H} |\psi(t)\rangle [/tex]
You know that, if [tex]|\psi(t)\rangle [/tex] is an eigenket of [tex]\hat{H}[/tex] so that :
[tex]\hat{H} |\psi(t)\rangle = \hbar \omega |\psi(t) \rangle [/tex]
you get obviously :
[tex] |\psi(t) \rangle = e^{-i \omega t} |\phi \rangle [/tex]
Thus [tex]\omega[/tex] is directly linked to the eigenvalue of the hamiltonian in QM, in non relativistic QM but also in realtivistic QM.

So, if there exists an experiment which result is that [tex] \omega [/tex] is different from the energy, then QM is false and we have to find something else...
 
  • #5
I_wonder said:
Could anyone please help with the following, rather unusual, query?

I know that for spin 0 bosons, the Klein Gordon equation gives solutions that are similar to the solutions of the Schrodinger equation for a non-relativistic free particle, the only difference being that the energy used when calculating the wave frequency (E/h-bar) is the relativistic energy m*c-square*gamma.

I don't know how the wavefunctions look for spin 1 or spin 1/2 particles. What I would like to know is if there is any kind of particle for which one gets a wavefunction where the frequency is different then m*c-square*gamma/h-bar? Also, is it possible to verify these things experimentaly? I mean, can an experiment get an eigenstate and measure both the energy and the wave frequency? Has that been done for relativistic particles? Have any discrepancies been found? Where could I find such data?

Thanks!

For what it's worth, http://www.ensmp.fr/aflb/AFLB-301/aflb301m416.pdf [Broken] . The experiment was performed with weakly relativistic electrons (80 MeV). The article is unusual, as it was published in 2005 in a rather obscure journal, whereas the experiment seems to have been performed in 1988, and at that time the authors published some of its results in Phys. Rev. B (I have not read that article). So use your judgement.
 
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  • #6
For somewhat unrelated reasons, I invite you to write the Klein-Gordon equation in lightcone coordinates. The result is interesting
 
  • #7
Thanks to everyone!

I_wonder...
 
  • #8
I_wonder said:
I don't know how the wavefunctions look for spin 1 or spin 1/2 particles.

Peskin and Schroeder, An Introduction to Quantum Field Theory, Chap. 3, give a very accessible description of the free-particle solutions of the Dirac equation (the governing equation for relativistic spin-1/2 particles). It's one of the classic texts for relativistic quantum mechanics and field theory, very well written, and I recommend it wholeheartedly!

One of the subtleties of the Dirac equation (and other relativistic equations) is that it admits both positive and negative-frequency solutions, which correspond to electron and positron wavefunctions respectively.
 

1. What is the wavefunction of a relativistic free particle?

The wavefunction of a relativistic free particle is a mathematical function that describes the probability amplitude of a particle in a quantum system. It is a solution to the Schrödinger equation and is used to determine the behavior and properties of the particle.

2. How does the wavefunction of a relativistic free particle differ from that of a non-relativistic particle?

The main difference between the wavefunction of a relativistic free particle and a non-relativistic particle is that the former takes into account the effects of special relativity, such as time dilation and length contraction. This means that the wavefunction for a relativistic particle is more complex and can change over time, while the wavefunction for a non-relativistic particle remains constant.

3. What is the role of the wavefunction in quantum mechanics?

The wavefunction plays a central role in quantum mechanics as it describes the probability of a particle's position, momentum, and other physical properties. It is used to calculate the expected outcomes of measurements and to understand the behavior of quantum systems.

4. Can the wavefunction of a relativistic free particle be directly measured?

No, the wavefunction of a relativistic free particle cannot be directly measured. This is because the wavefunction is a mathematical representation of the particle's state and does not have a physical form. However, the effects of the wavefunction can be observed through experiments and measurements.

5. How does the wavefunction evolve over time for a relativistic free particle?

The wavefunction of a relativistic free particle evolves over time according to the Schrödinger equation, which describes how the wavefunction changes in response to the particle's energy and potential. This evolution is also affected by the particle's motion and interactions with other particles in the system.

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