Something feynman said

[TurboRotary]

He said that if you think in terms of classical physics then you need the uncertainty principle.. but if you think with a soley quantum mindset you then the uncertainty principle isnt needed.. what did he mean by this? why would it not be needed?

[ydoaPs]

could you use a quote?

[TurboRotary]

It comes from the his book QED.. "If you get rid of all the old-fashioned ideas and instead use the ideas that Im explaining in these lectures then there is no need for an uncertainty principle."

[Severian]

That is correct. In Quantum Field Theory the HUP becomes contorted into a statement about on-shell or off-shell.

 

A particle is on-shell if [math]E^2 = m^2c^4+p^2c^2[/math] and off-shell if this is violated. The more off-shell a particle is, the more likely it is to decay, so we see the HUP emerging: it appears that the particle has borrowed energy (ie its energy is not [math]\sqrt{m^2c^4+p^2c^2}[/math]) and then must decay quickly to 'pay it back'.

 

In actuality all particles are slightly off-shell since it is impossible to hit the mass-shell relation exactly (a point on a line is infinitely small).

[TurboRotary]

Im sorry but I dont know what that means.. Im very new to this and dont know what on-shell or off-shell is.. nor do I understand the mathematics. Maybe I also need a more thorough understanding of HUP.. When I read that it just sort of struck me as odd because from what Ive read HUP is one of the bases of QM.

[Perturbation]

Im sorry but I dont know what that means.. Im very new to this and dont know what on-shell or off-shell is.. nor do I understand the mathematics. Maybe I also need a more thorough understanding of HUP.. When I read that it just sort of struck me as odd because from what Ive read HUP is one of the bases of QM.

 

It's not the basis of, it's a result of the approach of QM. The more general relation is between two self-adjoint operators A and B such that [imath][A, B]=iC[/imath], where C is another self-adjoint operator, then their variances [imath]\Delta_A[/imath] and [imath]\Delta_B[/imath] satisfy

 

[math]\Delta_A\Delta_B\geq\tfrac{1}{2}|\langle C\rangle |[/math]

 

Here the angle brackets denote the mean of C in some state ([math]=Tr(\rho C)/Tr(\rho )[/math], for some state operator [imath]\rho[/imath]).

 

In the case of position and momentum the commutation relation between the two is [imath][\vec{P},\vec{Q}] =i\hbar[/imath] giving the familiar Heisenberg uncertainty relation between position and momentum

 

[math]\Delta\vec{P}\Delta\vec{Q}\geq\tfrac{1}{2}\hbar[/math]

 

For the uncertainty relation between energy and time, the derivation is not the same as this, for there is no "time operator" per se. The time in this uncertainty relation is the "characteristic time" associated with the variance of some dynamical variable, say that of some measuring apparatus.

 

[math]\Delta E\Delta\tau\geq\tfrac{1}{2}\hbar[/math]

 

The deal with on/off mass shell is to do with four-momentum, a four-dimensional vector that has the energy and spatial momentum as components. The definition of four-momentum is

 

[math]p^{\mu}p_{\mu}=E^2-|\vec{p}|^2=m_0^2[/math] (units with c=1 for simplicity)

 

Here the p's on the left are the four-momentum, E is the energy, [imath]\vec{p}[/imath] is the spatial momentum and [imath]m_0[/imath] is the rest mass/energy (rest mass if we weren't using units with c=1).

 

In quantum field theory it is feasable for this equality to be violated by virtual particles. If it is the case that the equality holds the particle is said to be on-mass-shell. If the identity does not hold then the particle is said to be off-mass-shell. As Severian said, the violation of the equation he gave for energy indicates a particle being off-shell. It's usually called being off-mass-shell so I gave a definition where the relation to mass is more explicit, the two are obviously equivalent.

 

What Severian was saying is that particles being of shell can be seen to imply Heisenberg uncertainty: a particle could have too much energy by being off-shell than it should have, so it has sort of borrowed energy from the uncertainty it is allowed in its energy from Heisenberg. So the norm [imath]p^{\mu}p_{\mu}[/imath] needn't be equal to the square of the mass because of this uncertainty.

[swansont]

If you think in quantum terms you are thinking about waves already. In classical thought, you are still thinking of a particle with a well-defined position and momentum, and you have to "smear" that out with the HUP.

[TurboRotary]

So basicially what hes saying is that the HUP really just helps people used to thinking in classical terms grasp QM right? And again... I do not know any mathematics so all of that is gibberish to me!

[Severian]

So basicially what hes saying is that the HUP really just helps people used to thinking in classical terms grasp QM right? And again... I do not know any mathematics so all of that is gibberish to me!

 

Not really. It is still there - your fields can't be in position and momentum eigenstates at the same time. But the idea of 'borrowing energy' is really just one way of looking at it, and probably not the best way.

[Perturbation]

So basicially what hes saying is that the HUP really just helps people used to thinking in classical terms grasp QM right? And again... I do not know any mathematics so all of that is gibberish to me!

 

I din't expect you'd get some of it, having already said you don't really know the maths. I was just bored and went for a bit of a ramble. You should be alright with the bit about on- and off-shell if you ignore the mention of four-momentum, which I needn't really have included, and take the equation I gave for the square of the mass as given. I'd edit the reply but it's too late to edit it now.