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Hbar or hbar/2?

  1. Aug 12, 2004 #1
    hbar or hbar/2??

    Regarding the uncertainty principle, some books say

    (delta p) (delta x) >= hbar

    and others say

    (delta p) (delta x) >= hbar/2.

    Which is right? This matters because I get different results when I
    let p x=hbar(or hbar/2), plug in to the expression for energy, and
    minimize it to get the ground state energy of the system.
    Last edited: Aug 12, 2004
  2. jcsd
  3. Aug 12, 2004 #2
    Sigh... unless exact definitions are given for those deltas, the statement of the Uncertainty Principle should be taken as an order-of-magnitude estimate only. I believe that when one employs standard deviations, which is the correct way to go about it, the answer is hbar/2.
  4. Aug 12, 2004 #3

    zefram is correct
  5. Aug 12, 2004 #4
    Cohen-Tannoudji's "Quantum mechanics" writes the HUP as [tex]\Delta[/tex]x[tex]\Delta[/tex]p(>~)hbar
    Hameka's "Quantum mechanics" as [tex]\Delta[/tex]x[tex]\Delta[/tex]p>hbar
    Last edited: Aug 12, 2004
  6. Aug 12, 2004 #5
    There is a simple way to settle the issue. The inequality becomes equality for the SHO in its ground state. So if anyone wants to calculate the the standard deviations [itex]\sigma_x=\sqrt{ <x^2>-(<x>)^2} [/itex] and similar for p, then multiply them out, we have the answer. Actually it's easy:
    [tex]<x>=<p>=0 \; so \; \sigma_x \sigma_p = \sqrt { <x^2> <p^2> } [/tex]

    By the virial theorem:
    [tex] \frac{<p^2>}{2m} = \frac{1}{2} k<x^2> = \frac{E_0}{2} = \frac{\hbar \omega(0+1/2)}{2} [/tex]

    [tex] <p^2>=\frac{m \hbar \omega}{2}, <x^2>=\frac{\hbar \omega}{2k} [/tex]

    [tex]\sigma_x \sigma_p = \frac{\hbar}{2} \sqrt {\frac{m \omega ^2}{k}} \; but \; \omega^2 = k/m [/tex]

    Hence [tex] \sigma_x \sigma_p = \hbar /2 [/tex] is the final answer. Note to self: learn Latex. It took me far longer to compose this than to actually solve the problem.
  7. Aug 13, 2004 #6
    That is how the HUP appears in Griffith's "Introduction to Quantum mechanics"
    Last edited: Aug 13, 2004
  8. Aug 13, 2004 #7
    :rofl: :rofl: :rofl:

    Maybe not appearing in your derivation is the fact that the equality holds when the wavefunction has the same "shape" in postion and momentum variables. That is, when the Fourier transform of the wavefunction is analogous to the initial wavefunction : in the gaussian case, which applies to Harm. Oscill. you used.

    Maybe I'll give a try to latex too... :wink: :tongue2:
  9. Aug 13, 2004 #8
    OK, the all story is simple : the Heisenberg undeterminacy principle simply follows from the Schwarz inequality. Let us see how. Consider a state [tex]\psi[/tex] and two observables [tex]\hat{A}[/tex] and [tex]\hat{B}[/tex].
    Now the standard deviation is given by :
    ( \Delta a )^2 = \langle \psi|(\hat{A} - \langle a\rangle )^2 |\psi\rangle = \langle (a - \langle a\rangle )^2 \rangle
    This seems natural. Why bother an overall factor at this stage ?
    Let [tex] \hat{A'} = \hat{A} - \langle a\rangle[/tex]
    Then [tex] ( {\Delta}a )^2 = \langle\psi|\hat{A'}^2|\psi\rangle [/tex]
    Likewise for [tex]\hat{B}[/tex] one gets
    [tex] ( {\Delta}b )^2 = \langle\psi|\hat{B'}^2|\psi\rangle [/tex]

    Now the real argument : Schwarz inequality. I redemonstrate.
    Consider the norm of the vector
    [tex](\hat{A'} + i\lambda \hat{B'} )|\psi\rangle[/tex]
    This vector has positive norm :
    \langle\psi|(\hat{A'} - i\lambda \hat{B'} )(\hat{A'} + i\lambda \hat{B'} )|\psi\rangle\geq 0
    From this follows simply :
    (\Delta a)^2 + \lambda^2 (\Delta b)^2 + i \lambda \langle \psi |[\hat{A'},\hat{B'}]|\psi\rangle\geq 0

    As you can see, a 2nd order polynomial in [tex]\lambda[/tex] which is always positive will lead to :
    (\Delta a)(\Delta b) \geq \frac{1}{2}\langle \psi |[\hat{A},\hat{B}]|\psi\rangle
    and I did not bother about the primes, since the commutators are equal :
    This is the general way of deriving the [tex]\frac{1}{2}[/tex] factor.
    Let me add the HO argument's origin : let us see how gaussian functions appear. The inequality becomes an equality iff the second order polynomial vanishes, that is when
    [tex]\lambda = \lambda_0 = \frac{\hbar}{2(\Delta b)^2}=\frac{2(\Delta a)^2}{\hbar}[/tex]
    in which case the vector has vanishing norm, so :
    [tex][\hat{A}-\langle \hat{A}\rangle+i\lambda_0(\hat{B}-\langle \hat{B}\rangle)]|\psi\rangle = 0[/tex]
    Therefore, the condition for the inequality to become an equality is that the vectors [tex][\hat{A}-\langle \hat{A}\rangle]|\psi\rangle = 0[/tex] and [tex][\hat{B}-\langle \hat{B}\rangle]|\psi\rangle = 0[/tex] be proportional to each other (linearly dependent).
    Let us take [tex]\hat{A}=\hat{x}[/tex] (position) and

    We collect the equation :
    \left[ x + \hbar\lambda_0\frac{d}{dx} -\langle \hat{A}\rangle - i \lambda_0 \langle \hat{B} \rangle \right] \psi(x)
    with [tex]\langle\hat{x}|\psi\rangle[/tex].
    We furthermore eliminate mean values :
    \psi(x) = e^{i\langle\hat{B}\rangle x/\hbar}\phi(x- {\langle \hat{A}\rangle} )
    in order to get :
    [tex]\left[ x + \lambda_0\hbar\frac{d}{dx}\right]\phi(x)=0[/tex]
    whose solution is :
    [tex]\phi(x) = C e^{-x^2/2\lambda_0\hbar}[/tex]
    C is an arbitrary compex constant.
    Finally :
    [tex]\psi(x) = \left[2\pi(\Delta x)^2\right]^{-\frac{1}{4}}e^{i\langle p\rangle x/\hbar}e{-\left[ \frac{x-\langle x\rangle}{2\Delta x} \right]^2}[/tex]
    We note that the same lines can be carried out in the momentum representation, where one gets :
    [tex]\bar\psi(p) = \left[2\pi(\Delta p)^2\right]^{-\frac{1}{4}}e^{i\langle x\rangle p/\hbar}e{-\left[ \frac{p-\langle p\rangle}{2\Delta p} \right]^2}[/tex]
    credit : Jean-Louis Basdevant "Mecanique quantique, cours de l'X"
    Last edited: Aug 13, 2004
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