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Velocity operator inconsistency and discrete particle

  1. Oct 12, 2008 #1
    I'm going to mix a couple questions together instead of creating a new topic for each question.
    I hope you don't mind.
    I'm an electrical engineer(micro-electronics), so while I got the basics of QM in my studies I had to do most of my more 'in depth' learning on my own by reading books/ocwm from MIT.

    I took a course in Semiconductor Physics in my last year and came across some questions that seemed to be more of a coincidence rather than a rule. Thus here goes:

    1.H[tex]\Psi = E_{bloch}\Psi[/tex] for a bloch electron in a crystal lattice.
    with [tex]\Psi[/tex]= u(r,t)e([tex]\vec{k} \vec{r}[/tex])

    So far nothing out of the ordinary. Now we apply QM's to the wave function to find the average velocity by [tex]\int \Psi* \nabla_{\vec{r}}\Psi = \nabla_{\vec{k}}E(\vec{k}) [/tex]

    The last equation is said to be derivated from certain properties. Nothing wrong with that I am sure it's correct. But it's a strange coincidence that this happens to be equal to the group velocity of a packet of Bloch functions.
    By assuming the wave packet of the bloch functions and reworking the Hamiltonian we get

    [tex] E_{n}(\nabla_{\vec{r}}) + V(\vec{r}) [/tex]
    Now by going to the classical limit with [tex]\nabla_{\vec{r}} = \vec{p} [/tex] and [tex] \vec{r} = \vec{r} [/tex]
    we get [tex] E_{n}(\vec{p}) + V(\vec{r}) [/tex]

    In classical Mechanics we get
    [tex] \vec{v} = \frac{\partial H}{\partial \vec{p}} = \frac{\partial E_{n}(\vec{p})}{ \partial \vec{p}}= \nabla_{\vec{k}} E_{n}(\vec{k})[/tex]
    and
    [tex] \vec{F} = \frac{\partial H}{\partial \vec{r}} = \frac{\partial V(\vec{r})}{\partial \vec{r}}[/tex]

    it seems that just applying the velocity operator on the Bloch wave or finding the speed through the wave packet give the same speed indications. Which seems weird to me.

    I'm also not completely sold on the velocity operator = momentum operator / mass.
    Because the equality stems from the free particle where the frequency of the Broglie wave function = [tex] \frac{p^{2}}{2m}[/tex]. But this is not the case. The Bloch function is made up of several momentum waves(fourier) which propagate with the same speed since the the Bloch wave function acts with frequency [tex] E_{n}(\vec{k}) [/tex]

    Only when you use a wavepacket of Bloch waves do you get different Bloch waves move at different speed resulting in a different group velocity from the fase velocity.

    I'm very confused in that the velocity operator does not hold into account the time evolution. This is understandable in the case of the free wave particle where the frequency is in direct relation to it's momentum because of the lack of a potential function.
    But How can this hold when we get a potential function within the hamiltonian like for example with Bloch electrons. The momentum operator holds but I don't see how the velocity operator could hold in this case.


    2.I've gone through alot of threads on this forum to find a clear cut answer but I doubt there is any so just in case I'm just asking it again here.

    As we know there's alot of debate upon the interpretation of the wave function in QM.
    I was just wondering if there's any good theories or proof on the particles existing in discreet states. With that I mean (for example with an electron) when we detect it, do we really detect it as a physical little ball (discreet) or is the detection just a interaction with the particle that forces it to localise into a wavepacket and thus remain nonlocal. In other words is there any proof of the electron and other particles existing outside of these wavefunctions. I know what the wavefunctions mean so don't try to explain the copenhagen interpretation again. I'm just wondering if 'the chance you find the particle in this location' = 'the chance the wave functions crumbles up into a localised wavepacket in with the average position in this location'
    Both are very different.

    The second interpretation would seem to be the more logical one. Since else particles would teleport at random? Take for example a localised particle in a wavepacket. Now lets say we measure the momentum of the electron. QM states that this particle will collapse into 1 Broglie wavefunction and thus the electron can be anywhere even alot further into space than it was in the localised wavepacket. Thus the particle would teleport.

    I'm a big fan of QM. non locality is the only way for a particle to move in a continuous space since else it would have to pass through an infinite number of 'points' to move. That is if you consider space to be continuous.(I've considered space at a very small level to be discrete but it just doesn't feel right.
     
  2. jcsd
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