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Vacuum energy and principle of uncertainty

  1. Apr 19, 2012 #1
    Let us assume that harmonic oscilator has the lowest energy level equals 0, (and next one [itex]\hbar \omega[/itex] and so on). Does this harmonic oscilator violates the principle of uncertainty?

    Let us assume that energy of vacuum equals 0 and not [itex]\hbar \omega/2[/itex]. Does such vacuum violates the principle of uncertainty?
     
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  3. Apr 19, 2012 #2

    vanhees71

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    Of course, this does not violate anything since as long as you do not consider general relativity, where absolute energy-momentum densities matter in the sense that they are sources of gravitational fields, you cannot observe the absolute value of energy levels. You can always ad a constant value to all energies without changing anything in the description of the system (except the absolute values of energy of course).

    The ground-state energy [itex]\hbar \omega/2[/itex] results from choosing the zero-energy level as the minimum of the potential energy. Of course, you can as well shift all energies down by this value. Then the minimum of the potential energy is at [itex]-\hbar \omega/2[/itex] and the ground-state energy is 0. This is the usual choice in quantum field theory, where you set the ground-state energy to 0.

    The uncertainty principle comes into play, when you ask whether a particle in the state of lowest energy is at rest. Then you'll easily find that [itex]\langle p^2 \rangle \geq 0[/itex] and thus on average the particle has a non-vanishing kinetic energy. You can easily figure out that the harmonic oscillator ground state is a state of "minimal uncertainty" in the sense that [itex]\Delta x \Delta p =\hbar/2[/itex] (according to the Heisenberg uncertainty relation for position and momentum, you must have [itex]\Delta x \Delta p \geq \hbar/2[/itex] of course!).
     
  4. Apr 19, 2012 #3
    A good thing to consider is where the harmonic oscillator potential comes from - fundamentally it comes from taking a taylor expansion about an equilibrium (thus harmonic). According to Zee's book this is also true in QFT. The point is, [itex]\omega[/itex], is not some arbitary frequency - it relates to the second derivative of the potential about an equilibrium, evaluated about that equilibrium (I think you also have to assume a stable equilibrium to get the right values and functions out)

    The point is, the Harmonic potential is an approximation that is liable to change. This is especially true in something like the vibrational spectra of a diatomic molecule, where the potential can change based on the binding, and the equilibrium position too! Then the functional dependence of the "zero" point of field can become something very physical. Of course, as one can only really measure transitions between energy levels, actually deciphering this effect is somewhat difficult - and clearly the approximation to a harmonic oscillator is what's at fault if anything particuarly interesting happens.

    For a "fundamental" oscillator like the Quantum field, it depends whether you think it's fundamentally an oscillator or not. If it really is and [itex]\omega[/itex] is a true constant, then it doesn't matter in the slightest. If it isn't, and [itex]\omega[/itex] is actually a function, then potentially it could. Someone may no more on this.

    The relationship with the uncertainty principle, specifically with the energy-time uncertainty, states that
    [itex]\Delta E \Delta t ≈\hbar[/itex] (or was that [itex]\hbar / 2 [/itex] ...)
    The question is, what the time actually means - turns out it's the lifetime of the state - for a state of well defined energy, like those that you get from treating the time independent Schrodinger equation), they don't evolve in time, and hence [itex] \Delta t[/itex] is infinite!, and therefore energy is arbitrarily well defined. Of course, this is telling you that states of well defined energy don't physically exist. Or more precisely, no real particle exists in a state of well defined energy, otherwise it can never evolve, and thus could never have got into that state in the first place! However, they are very useful, for dealing with a whole load of systems for which the energy is reasonably well defined (or long lived states, on the order of quantum oscillation of the state, which has scale [itex]\hbar / E [/itex], so is not necessarily very long at all!
     
  5. Apr 20, 2012 #4
    When we shift the ground state energy to zero, what happens to the energy-time uncertantity of the ground state?

    thanks
     
  6. Apr 20, 2012 #5
    Maybe I understand why harmonic oscillator should have not zero energy equal to zero.
    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html

    But, why photon should not have zero energy equal to zero? Because harmonic oscillator with energy zero still ever exists, but a photon with energy zero does not exist. This is one dillema. The mathematic aparatus of harmonic oscillator or photon is the same. Anyway, why, because of this undistinction exists?

    Or it not exist? In this case this is clear.
     
  7. Apr 20, 2012 #6
    I think it was precisely clear by what both I and vanhees said - energy time uncertainty relates to the length or time scale by which the state exists - not some arbitary time by which the state evolves by - in this Hyperphysics has a needless ambigious implementation of time-energy uncertainty - for imaginary states of well defined energy, no such principle exists!

    Furthermore, any ambigious result regarding the divergeness of the harmonic model should always be referred back to the origin of such a potential - clearly a low momentum / energy limit model (i.e. the simple harmonic oscillator model) has no reason to extend to arbitrary high energy levels. Any problems can be comfortably be dealted with by renormlisation, again, I would recommend that you read Anthony Zee's book on Quantum Field Theory in a Nutshell to realise why no ambiguity of zero point energies exists.

    I would also note, for the amount of scotch I've consumed, I cannot see my own keyboard. Shameful.
     
    Last edited: Apr 20, 2012
  8. Apr 21, 2012 #7
    If you answered to me, I do not problematize the divergeness here.
    At an ordinary harmonic oscillator, zero momentum and zero location is at contradiction with uncertainty principle. So it needs ℏω/2 "zero" level.
    But, at the photon zero momentum and zero location means that the photon does not exist. (Ok it is an analogy for momentum and location, this means B, H, exp(i..) and so on...).

    So where is contradiction of a photon without ℏω/2 level?
     
  9. Apr 21, 2012 #8
    The energy-time uncertainty relation has a completely different meaning than momentum-position uncertainty relations.

    Namely, the latter is between two non-commuting observables, and it reflects the principle that they cannot be simultaneously measured to an arbitrary precision.

    The former, on the other hand, involves time, which is not an observable in Quantum Mechanics. The derivation of this relation comes from time-dependent perturbation theory.

    Nevertheless, how does adding a constant to the energy affect [itex]\Delta E[/itex]?
     
  10. Apr 21, 2012 #9
    Not at all, of course. But let's stay with the specific case of the harmonic oscillator. How do we get the groundstate from the time-independent Schrödinger when the lowest energy is zero? What is the eigenvector for E=0, instead of E= (hbar/2 times frequency)?

    To get the time evolution for an eigenstate of the time-independent Schrödinger equation, we normally append exp(energy x time/hbar). But with E set to zero, how do we get a time evolution?

    And also, if energy is set to zero, does uncertainty in energy does not imply negative energy?

    I assume since only energy differences matter, not absolute energy, the word negative energy does not mean much, just a convention how we count. But then the groundstate of fermionic harmonic oscillators has negative energy and that certaintly has some meaning. But which?

    Last but not least, can we say that the quantum groundstate energy always differs from the lowest classical energy value by (hbar/2 times frequeny)? And that quantum groundstate does not insist to be (hbar/2 times frequency) or zero or something else, but only to differ from the classical value by (hbar/2 times), minus for the fermionic case and plus for the bosonic case?

    sorry for this long and bit confusing post, any clarifications are very welcome
    thanks!
     
  11. Apr 21, 2012 #10
    A constant shift in energy by an amount [itex]\Delta E[/itex], [itex]E_n \rightarrow E_n + \Delta E[/itex] would lead to a phase factor
    [tex]
    \exp \left( -\frac{i}{\hbar} \, \Delta E \, t \right)
    [/tex]
    in front of all stationary kets, so it may be pulled in front of the sum [itex]\vert \Psi, t \rangle = \sum_{n}{c_n \, \exp \left( -\frac{i}{\hbar} \, E_n \, t \right) \, \vert n \rangle}[/itex], leading to
    [tex]
    \vert \Psi, t \rangle \rightarrow e^{-i \, \Delta E \, t/\hbar} \, \vert \Psi, t \rangle
    [/tex]

    But, a state ket is determined up to an arbitrary phase, so this does not change the quantum state of the system at any time.
     
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