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Does a particle have negative kinetic energy in delta function wall

  1. Sep 24, 2013 #1
    E is finite and V is infinite, So T should be negative infinite
     
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  3. Sep 24, 2013 #2

    DrClaude

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    Can you please give more details about the situation you are contemplating? "delta function wall" is not very clear.
     
  4. Sep 24, 2013 #3

    stevendaryl

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    Well his question about negative kinetic energy applies to any kind of bound particle in the region with [itex]V(x) > E[/itex]. He's specifically talking about the case where [itex]V[/itex] is a delta-function potential, but the issue comes up for any binding potential.

    For example, in the case of the potential

    [itex]V(x) = 0[/itex] for [itex]-L < x < +L[/itex]
    [itex]V(x) = V_0[/itex] for [itex]|x| > L[/itex]

    the energy eigenfunctions look like this:

    [itex]\Psi(x) = A cos(k x) + B sin(k x)[/itex] for [itex]-L < x < +L[/itex]
    [itex]\Psi(x) = C e^{-K |x|}[/itex] for [itex]|x| > L[/itex]

    where
    [itex]k = \sqrt{2m E}/\hbar[/itex]
    [itex]K = \sqrt{2m (V_0 - E)}/\hbar[/itex]

    So in the region [itex]|x| > L[/itex], the kinetic energy operator [itex]-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}[/itex] gives a negative number, [itex]-\frac{\hbar^2}{2m} K^2[/itex]

    My first inclination is to say that the operator [itex]-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}[/itex] doesn't actually mean kinetic energy except for unbound particles, but maybe there's something more to be said.
     
  5. Sep 24, 2013 #4
    Yes, what stevendaryl said is exactly what i'm curious about.
     
  6. Sep 24, 2013 #5

    ZapperZ

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    But the question also doesn't make any sense. A "delta function" is something that is infinitely thin, meaning it doesn't have a width. So there's no "inside" for it to be in.

    A delta function potential barrier is to illustrate the condition where a wavefunction can be continuous, but its derivative isn't! There is a derivative jump across the delta function. If you try to "rationalize" and try to analyze this problem beyond what is given, then you run into absurdities such as this.

    Zz.
     
  7. Sep 24, 2013 #6
    Yes, Kinetic energy can be negative in quantum mechanics which leads to penetration into a region that would be forbiden in classical mechanics allowing for instance for tunnel effect.
     
  8. Sep 24, 2013 #7
    No. Kinetic energy is not a position dependent property. To be precise, the only sensible definition of kinetic energy is the quadratic form <psi|T|psi> for normalized |psi>. And such a state |psi> which is entirely contained within the region where you'd classically expect "negative kinetic energy" has an localization induced momentum uncertainty that makes the total kinetic energy positive.

    There's a theorem which is not hard to prove that <psi|T|psi> is always >0 for states on the hilbert space of square integrable functions, because otherwise they could not be normalized.

    Cheers,

    Jazz
     
  9. Sep 24, 2013 #8
    We're not talking about exactly the same thing. The expected value will be positive. But the question is about the value of the function before the integration is performed to find the expected value. At least that's how a read it.
     
  10. Sep 24, 2013 #9
    What function? There is no kinetic energy function is quantum theory. And the expectation value IS the kinetic energy.

    Cheers,

    Jazz
     
  11. Sep 24, 2013 #10

    stevendaryl

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    Well, I would say that kinetic energy is an operator, not an expectation value.

    As for a kinetic energy function, there isn't one in standard quantum mechanics. But we can define one (I only know of one case where this was ever done):

    For any operator [itex]\hat{O}[/itex] and any wave function [itex]\Psi(x)[/itex], we can define a position-dependent (and time-dependent, if [itex]\Psi[/itex] is time-dependent) function [itex]O_Q(x)[/itex] via:

    [itex]O_Q(x) = \dfrac{\hat{O} \Psi(x)}{\Psi(x)}[/itex]

    If [itex]\Psi(x)[/itex] is an eigenstate of [itex]\hat{O}[/itex] then [itex]O_Q(x)[/itex] is the eigenvalue. Otherwise, [itex]O(x)[/itex] is in general a complex-valued function.

    Defining such functions can be useful in seeing how QM approaches the classical limit. For example, for the special case of momentum, we have:

    [itex]p_Q(x) = -i \hbar \frac{d}{dx} ln(\Psi(x))[/itex]

    The corresponding classical "momentum function" is:

    [itex]p_C(x) = \dfrac{\sqrt{2m (E - V(x))}}{\hbar}[/itex]

    They both become pure imaginary in the "forbidden" region where [itex]E < V(x)[/itex]
     
  12. Sep 24, 2013 #11
    Kinetic energy is a property of a state. Operators are not state dependent, so kinetic energy is the quadratic form of the operator. That's the quantity that enters conservations laws too.

    That's only a sensible construction in the context of approximations (like WKB), but not as part of the structure of quantum theory itself. Specifically you cannot formulate conservation laws or frame changes coherently with such a function.

    Cheers,

    Jazz
     
  13. Sep 26, 2013 #12
    Let's do a thought experiment. How might you measure the kinetic energy of an electron? Well, it's hard to measure a single electron, but you could measure the average energy of a stream of electrons. You could set up a probe with a retarding potential in front of a detector, and measure the rate of detections. Decrease the potential until the counts start to drop off exponentially, and that gives a measurement. Consider some electrons trapped in a potential well, and we stick the end of the probe in a region which is classically energetically inaccessible to the electrons. If we set the potential in the probe to a positive potential relative to the probe tip, then some electrons can tunnel through the wall into the probe into the detector. If we label the kinetic energy by the potential difference in the probe, then it seems like the kinetic energy is negative. Of course, you could say that this is an incorrect analysis of the probe. Or this is a peculiar definition of kinetic energy. Anyways, the physics are understood, it's just a matter of terminology.
     
  14. Sep 26, 2013 #13

    Jano L.

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    The confusion of negative kinetic energy comes about in the following way. First, most physicists believe that the energy of the system described by the Hamiltonian eigenfunction ##\phi_n## corresponding to negative value ##E_n## is a definite number that has to be equal to ##E_n##. Then, since

    $$
    (\hat T + \hat U) \phi_n = E_n \phi_n
    $$

    one may be lead to believe on the grounds of energy conservation that

    $$
    \hat T \phi_n = [ E_n - \hat U(q) ] \phi_n
    $$

    implies that kinetic energy of the system at ##q## is ##E_n - \hat U(q)##. As U goes to zero at large distances, there is always region of configuration space where this number is negative.

    However, from the definitino it is clear that the kinetic energy can have only positive values, so something is wrong.

    The fallacy of the above derivation is that the kinetic energy of the system at ##q## is not necessarily ##E_n - \hat U(q)##, because the total energy of the system is not necessarily ##E_n##. ##E_n## is just an eigenvalue of the Hamiltonian.

    If we believe in the validity of the momentum representation and derived probability distributions for momentum, it can be easily shown that the energy of the system can be larger than this value.

    Take harmonic oscillator and let ##\psi## be the "ground state" function with eigenvalue ##E_0 = \frac{1}{2}\hbar \omega##. The kinetic energy alone can be larger than the eigenvalue ##E_0##, due to the following reason. The kinetic energy is a function of momentum, and in momentum representation the average value of the kinetic energy is

    $$
    \langle T \rangle = \int_{-\infty}^{\infty} \frac{p^2}{2m} |\psi(p)|^2 dp.
    $$

    Momentum ##p## takes only real values, so the kinetic energy assumes only values from ##\langle 0,\infty)##.
    The probability distribution ##|\psi(p)|^2## is a Gaussian function non-zero even for ##p## so large that the corresponding kinetic energy ##p^2/2m## is greater than the eigenvalue ##E_0##. Since the potential energy is positive for all ##q##, the total energy may be larger than ##E_0##.

    Since the kinetic energy cannot be negative, the total energy clearly is greater than ##E_0## for all ##q## for which ##U > E_0##, i.e. whenever the particle is far enough from the minimum of the potential.
     
  15. Sep 27, 2013 #14

    stevendaryl

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    I don't think that makes any sense, any more than saying that the position is a property of the state. In non-relativistic quantum mechanics, the position operator has the same sort of status as the kinetic energy.

    The expectation value of kinetic energy, or position, or momentum, or whatever, is a property of the state. But those quantities themselves are not properties of the state.

    My point is that there is a way to make sense of a position-dependent kinetic energy (or momentum, or whatever). It's useful for understanding the relationship between quantum mechanics and the classical limit.
     
  16. Sep 27, 2013 #15
    Kinetic energy cannot be state independent (like being just the operator you suggested), instead, it must be a function of the state of the system. We know that kinetic energy is a sensible concept, even in quantum mechanics, because of the conservations laws that involve it. That makes it a property of the state. It doesn't mean that a state necessarily has to be in a kinetic energy eigenstate. It's the "expectation value" (which is really a misnomer, because it automatically implies you perform a measurement) for which the conservation law holds, so that is the proper definition of kinetic energy in a quantum context.

    And my point is that this is highly misleading. Suggesting that there is any such property as kinetic energy that you can associate with position implies an underlying structure of some sort where the particle is actually located somewhere, in form of a single point, but we just don't know where. This is exactly the kind of comparison to classical physics that confuses the heck out of students and results in weird debates about the real meaning of quantum physics. It also implies a much too simplified and superficial transition from quantum to classical. But most importantly, there is no way to make sense of any such definition in terms of conservation laws.

    Cheers,

    Jazz
     
  17. Sep 27, 2013 #16

    stevendaryl

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    I don't understand what you're saying. Why is kinetic energy more of a "property of state" than position is?

    How does having a position-dependent kinetic energy function imply such a thing, any more than having a position-dependent wave function [itex]\Psi(x)[/itex]?

    I think avoiding confusing students is way over-rated. Confusion is an important stage in reaching understanding. I think for any given topic, such as quantum mechanics or relativity, or whatever, the more different ways you know how to think about it, the better.

    What's the proof that there is no way to "make sense of any such definition in terms of conservation laws"?

    Actually, understanding the classical limit using "momentum functions" was the topic of research done by my advisor years ago. I thought it was very interesting, even though not much came of it.
     
  18. Sep 28, 2013 #17
    There is a conservation law for energy(-momentum), and that law holds independently from the system being in a corresponding eigenstate. The law specifically refers to the "expectation value" of the involved quantities, which makes them physically meaningful attributes of the state of the system. That's what I call a property.

    For position you have no such conservation law, and it makes much less sense to talk about the expectation of position as a physically relevant quantity. In some situations however, it might make sense, but that's not really my point, as we're discussing energy and not position.

    Specifically the operator of (kinetic) energy is not what you would call energy as such, because it does not depend on the state, which energy does. The operator is important for the mathematical formulation of energy, but it itself is not energy, it's quadratic form together with a state is.



    The wave function is a state that you can expand in any basis that you like. The kinetic energy scalar you proposed does not have this property, it's not a vector in a linear space in any physically meaningful way.


    But this is not a good or meaningful way to think about it, so it doesn't really help you to get a different perspective on things. But it does suggest a relationship between quantum physics and classical physics that does not exist on this level and it also suggests a classical hidden variable interpretation where the particle is at a single position, which you can use to construct classical energy quantities. This is misleading and undermining the very idea of having a theory that is agnostic of the choice of a basis and that, as we know today, does not have any classical explanation.


    The proof is simply that the conservation law does not involve these quantities, and the formulation of energy conservation in quantum theory is unique.

    Anyone can research whatever he wants. But whatever comes out, it will certainly not revolutionize our understanding of energy conservation in quantum theory or the quantum to classical transition.


    Cheers,

    Jazz
     
  19. Sep 28, 2013 #18

    stevendaryl

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    But kinetic energy by itself isn't conserved, so I'm not sure how it's relevant that total energy is conserved. They are two different operators.

    We have: [itex]\hat{H} = \hat{K} + \hat{V}[/itex] where [itex]\hat{K}[/itex] is kinetic energy, and [itex]\hat{V}[/itex] is potential energy. It doesn't make any sense to me for you to say that the kinetic energy is a function of state, and the potential is not. But the potential is explicitly position-dependent, so it can't be a function of state.

    Well, I think that what you are saying is not a good or meaningful way to think about it. So there.

    You're making claims that are beyond what you actually know.

    On the other hand, I DO know what I'm talking about, and I agree that nothing much came of Dr. Leacock's research. So you're right, but a guess doesn't count for much.
     
    Last edited: Sep 28, 2013
  20. Sep 28, 2013 #19

    stevendaryl

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    Hamilton-Jacobi Formulation of Quantum Mechanics

    Robert Leacock's approach to quantum mechanics is a quantum version of the Hamilton-Jacobi formulation of classical mechanics. It's mentioned as one of the nine formulations of quantum mechanics in this review article:

    https://www-physique.u-strasbg.fr/cours/l3/divers/meca_q_hervieux/Articles/Nine_form.pdf [Broken]

    The technique itself is described here:
    http://siba.unipv.it/fisica/articoli/P/Physical%20Review%20D_vol.28_no.10_1983_pp.2491-2502.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  21. Sep 28, 2013 #20
    Going from total energy to kinetic energy is only a matter of fixing a frame, which I assumed to have happened.

    That logic is flawed. Only because it doesn't usually make much sense to speak of the position of a state, the actual potential energy associated with a superposition thereof can very well be a valid and physically meaningful concept. The potential energy does NOT depend on the expectation value of position, which is what you're arguing about, it depends on the actual state expansion in position.


    That's not just me, it's everyone who is investigating the structure of quantum theory.

    Now that's a claim that is beyond your knowledge.

    So what are you arguing about?

    Cheers,

    Jazz
     
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