Question about total energy in quantum mechanics

[Johnny5]

If you start out with Schrodinger's equation, and then derive the total energy of a free particle, you get [math] \frac{mv^2}{2} [/math]. But suppose that instead of using the ordinary Laplacian, you use this instead:

 

[math] \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} [/math]

 

Then in this case you get:

 

[math] E^2 = (pc)^2 + (m_0 c^2)^2 = (hf+ m_0c^2)^2 [/math]

 

As the total energy of a free particle, unless I made a mathematical error somewhere. So if this is right then the total energy of a free particle is given by:

 

[math] E = hf + m_0 c^2 [/math]

 

rather than E=hf

 

Does this lead to the conclusion that quantum mechanics and special relativity are logically incompatible theories?

 

Thank you

[timo]

The mathematical error is that (pc)² + (mc²)² does not equal (pc + mc²)²

 

The correct result would not lead to any problems. Special relativistic quantum mechanics is actually very well established.

 

Another note: mv²/2 is not the total energy - even in classical mechanics. It´s the kinetic energy.

[5614]

Sadly I do not know the latex maths code thing... but

 

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

 

so yeah, p^2c^2 = (pc)^2... but the end bit is slightly different.

[Johnny5]

The mathematical error is that (pc)² + (mc²)² does not equal (pc + mc²)²

 

The correct result would not lead to any problems. Special relativistic quantum mechanics is actually very well established.

 

Another note: mv²/2 is not the total energy - even in classical mechanics. It´s the kinetic energy.

 

Can you derive the formula for total energy of a free particle for me, using the 'spatiotemporal' Laplacian, because I agree, I do think there's an error somewhere.

 

Thanks

[Johnny5]

Sadly I do not know the latex maths code thing... but

 

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

 

so yeah' date=' p^2c^2 = (pc)^2... but the end bit is slightly different.[/quote']

 

Example:

 

[something] E^2 = (pc)^2 + (m_0 c^2)^2 [/something]

 

Replace the word "something" by the word "math", and you get this:

 

[math] E^2 = (pc)^2 + (m_0 c^2)^2 [/math]

 

Lowercase omega

 

[something] \omega [/something]

 

Replace the word "something" by the word "math" and you get this:

 

[math] \omega [/math]

 

Uppercase Omega

 

[something] \Omega [/something]

 

Replace the word "something" by the word "math" and you get this:

 

[math] \Omega [/math]

 

All other greek letters work similarly.

 

Exponents

 

To write the speed of light squared you have:

 

[something] c^2 [/something]

 

replace the word "something" by the word "math" and you have

 

[math] c^2 [/math]

 

And fractions are easy too:

 

To write a/b you have this:

 

[something] \frac{a}{b} [/something]

 

[math] \frac{a}{b} [/math]

 

And you can do square roots too:

 

[something] \gamma = \frac{1}{\sqrt{1-v^2/c^2}} [/something]

 

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

 

J5

[5614]

OK, thanks man

[timo]

I rather think your problem is that you think the energy for particles was E = hf. That´s only true for photons. The energy is E = sqrt[ (pc)² + (mc²)² ] just like you got it.

 

Step-by-step derivation:

 

- Start with a plain wave: [math] \psi = \exp \left( -i(Et - px) \right) [/math] - I´ll leave it 1D to save some time typing. Also I set hbar to one. It would cancel out in the next equation anyways.

 

- Plug it into the Klein-Gordon wave-equation:

[math] \left[ \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} + m^2 \right] \psi = 0 [/math] (note that due to my lazyness in typing I also set c=1).

 

=> [math] \left[ \dots \right] \psi = \left[ -E^2 \psi + p^2 \psi + m^2 \psi \right] = \left[ -E^2 + p^2 + m^2 \right] \psi = 0[/math]

Since the wavefunction is nonzero the term in brackets must be zero

=> -E² + p² + m² = 0 => E² = p² + m²

 

Putting back the c´s to get the SI units thus gives:

E² = (pc)² + (mc²)²

or E = +- sqrt[ (pc)² + (mc²)² ]

 

Note that you also get negative total energies. If I had used the Dirac equation (which is a bit more complicated) in above those solutions with negative energies would be associated to anti-particles. To be honest I´m not completely sure if this can also be said in case of the Klein-Gordon equation.

[5614]

The energy is E = sqrt[ (pc)² + (mc²)² ] just like you got it.

 

now Atheist I will be the first to admit I do not know all the maths in that post, but isn't it:

[math]E = sqrt[(pc)^2 + (mp^2)^2][/math]?

[timo]

you mean mp² instead of mc²? No, a simple check of the units will show you it cannot be mp².

[5614]

oh yeah, sorry, my mistake... i dunno why, but i thought it was p^4 at the very end when its c^4... my mistake, appologies.

[Johnny5]

I rather think your problem is that you think the energy for particles was E = hf. That´s only true for photons. The energy is E = sqrt[ (pc)² + (mc²)² '] just like you got it.

 

For one I really don't like that derivation, isn't there a simpler one?

 

Next, you just blatantly say that it is only true for photons, but if the formula for total energy is reference frame independent, then it applies to all particles even photons, so that even for a photon we have:

 

[math] E^2 = (pc)^2 + (m_0c^2)^2 [/math]

 

Where m0 is rest mass of a photon, in the rest frame of a photon. So if the rest mass of a photon isn't zero, then isn't your formula E=hf is wrong for photons?

 

Regards

[5614]

So if the rest mass of a photon isn't zero

But as photons dont really have a rest frame its hard to say since photons have no rest frame the concept of rest mass for a photon is not defined.

 

However in the equation:

 

if m = 0 that cancels out all of the last brackets leaving , which could go to [math]e=pc[/math]is that correct?

[timo]

a) A frame where photons are at rest does not exist.

b) The (rest)mass of photons is zero.

c) I made a mistake. E=hf applies to all particles. But pc = hf only applies to massless particles. Simple reason: (hf)² = E² = (pc)² + (mc²)² => [math] pc = \sqrt{(hf)^2 - m^2c^4} [/math]. So pc = hf exactly if m=0.

[Johnny5]

But as photons dont really have a rest frame its hard to say since photons have no rest frame the concept of rest mass for a photon is not defined.

 

However in the equation:

 

if m = 0 that cancels out all of the last brackets leaving ' date=' which could go to [math']e=pc[/math]is that correct?

 

Of course photons have rest frames, and if m0=0 then for a photon you will get E=pc, but as I said, photons do have rest frames.

 

Regards

[Johnny5]

a) A frame where photons are at rest does not exist.

b) The (rest)mass of photons is zero.

c) I made a mistake. E=hf applies to all particles. But pc = hf only applies to massless particles. Simple reason: (hf)² = E² = (pc)² + (mc²)² => [math] pc = \sqrt{(hf)^2 - m^2c^4} [/math]. So pc = hf exactly if m=0.

 

Your answer to A is incorrect.

Your answer to B depends upon whether or not photons can be accelerated in their frame does it not? Or to come at the question from another angle, according to the principle of equivalence, the inertial mass of a photon is equivalent to its gravitational mass, its gravitational mass is nonzero, which means that its inertial mass is nonzero.

 

Would you mind deriving the formula for total quantum mechanical energy for me in a simpler way, then maybe we can get this right.

 

Regards

[5614]

photons do not have a rest frame.

 

However we still take m=0 so then what type of mass is the m referring to? (relativistic, inertial or what?)

[timo]

No, I´m obviously no match for your knowledge in physics if I don´t even know that there´s a rest frame for photons.

[Johnny5]

photons do not have a rest frame.

 

However we still take m=0 so then what type of mass is the m referring to? (relativistic' date=' inertial or what?)[/quote']

 

I insist that photons do have a rest frame. Well you have to derive the formula somehow. In general, rest mass is the mass that something would have in a frame in which it is at rest.

 

If the principle of equivalence is true, then I can respond by either saying rest inertial mass, or rest gravitational mass, as they are equivalent. And there is no need to say relativistic mass, since in the rest frame of an object, its relativistic mass is equivalent to its rest mass, so it makes the most sense to respond to you by saying this:

 

If the principle of equivalence is true, then m0= rest inertial mass

If the principle of equivalence is false, then the answer depends upon the derivation of the formula for [math] E^2 [/math].

 

If the derivation assumes that the mass M is measured in a gravitational field, then the appropriate response would be... rest gravitational mass, but if the derivations assumes that the mass M is measured on a free particle, then the appropriate response would be... rest inertial mass.

 

Regards

 

PS Energy is not conserved in quantum field theory. Virtual photons violate the principle of conservation of energy don't they?

[Johnny5]

No, I´m obviously no match for your knowledge in physics if I don´t even know that there´s a rest frame for photons.

 

Start with the definition of rest frame, and go from there. You will see that photon's must have rest frames.

 

Regards

[5614]

if photons had a rest frame then they wouldn't be travelling at c in that frame and photons travel at c in all frames.

[Johnny5]

if photons had a rest frame then they wouldn't be travelling at c in that frame and photons travel at c in all frames.

 

Where on earth did you come up with that??? In a photonic frame, a photon's speed is zero by fiat. But, where did you get the idea that photons travel at c in all frames? That's just wrong.

 

Regards

[swansont]

An inertial observer can never be in a frame where the photon is at rest.

[5614]

The rest mass is the mass of a particle (in our case the photon) as measured by an observer who sees the particle still and with zero speed. In other words, the particle is at rest as far as this observer is concerned. Thus comes the term REST mass. But according to special relativity, light ALWAYS travels with the light speed c, and is NEVER at rest. And so it has zero REST mass.

[Johnny5]

An inertial observer can never be in a frame where the photon is at rest.

 

Prove your statement please.

 

Regards

[Johnny5]

The rest mass is the mass of a particle (in our case the photon) as measured by an observer who sees the particle still and with zero speed. In other words, the particle is at rest as far as this observer is concerned. Thus comes the term REST mass. But according to special relativity, light ALWAYS travels with the light speed c, and is NEVER at rest. And so it has zero REST mass.

 

Your statement is incorrect. The fundamental postulate of the theory of special relativity is that in any inertial reference frame, the speed of a photon must be measured to be exactly 299792458 meters per second.

 

It does not postulate that the speed of light in ANY frame is 299792458 m/s.