Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uncertainty question

  1. Mar 1, 2010 #1
    If the energy Eigen values for hydrogen atom can be precisely determined, does that not violate uncertainty?
    Since DeltaE is 0 in this case as energy (and therefore momentum) is known exactly?
  2. jcsd
  3. Mar 1, 2010 #2


    User Avatar
    Science Advisor
    Gold Member

    What is delta t in this case? The energy eigen-value solutions are called "stationary states" because they supposedly never evolve. Therefore, delta t in these cases are infinite. Remember that the energy, time uncertainty relation deals with the time it takes to evolve from one energy state to another (more precisely, it is the time it takes the expectation value of some observable Q to change by one standard deviation, which in this case is infinite since you're in a stationary state).

    At least this is true in a non-relativistic quantum theory wherein time is treated not as a dynamic observable variable and position is.
  4. Mar 1, 2010 #3
    You are correct! The uncertainty in energy is zero. This is always true when the state vector is an eigenstate of energy. But the momentum does not commute with energy in this case. Therefore, the momentum is uncertain because the energy eigenstate is not an eigenstate of momentum. The energy uncertainty is zero, but the momentum uncertainty is infinite. The uncertainty principle is not violated.
  5. Mar 1, 2010 #4

    If momentum uncertainty is infinite, uncertainty of kinetic energy becomes infinite! Electron in hydrogen atom has finite momentum uncertainty and position uncertainty satisfying uncertainty relation.

  6. Mar 1, 2010 #5
    You are correct. I have not done the calculation and I should not have said the momentum uncertainty is infinite. However, the momentum is still uncertain, as you point out. For the Hydrogen atom, energy does not commute with momentum.

    In a Hydrogen energy eigenstate, if we were to repeatedly measure the energy, we would always get the same value. If we repeatedly measured the momentum for that same energy eigenstate, we would get a spectrum of values. Thus, there is an energy-momentum uncertainty relation for the Hydrogen atom.

    The position-momentum uncertainty relation is not violated in an energy eigenstate. In fact it is always valid because position and momentum never commute.
  7. Mar 1, 2010 #6


    User Avatar
    Science Advisor

    That is correct, even in the case of the particle in an infinite square well, where the momentum and energy operators *do* commute. As in the OP's example, in this case the fact that the uncertainty of the energy is zero means that the uncertainty in the *magnitude* of the momentum [tex]\left(\left|p\right|=\sqrt{2mE}\right)[/tex] is also zero. However, the *sign* of the momentum is not known, and thus it has a finite uncertainty, as required by the HUP. In fact, you can actually determine the ground state energy for this case directly from the HUP (with some straightforward assumptions) ...
  8. Mar 2, 2010 #7
    Just a short calculation.

    In the case of square well potential, both finite and infinite depth, the momentum and energy operators do not commute.
    H=P^2/2m + V(x), V(x)=-V for -a<x<a, 0 for otherwise. Then [P,H]=[P,V(x)]=-ih'[d/dx,V(x)]=ih' V{δ(x+a)-δ(x-a)}, not zero. Applet http://www.falstad.com/qm1d/index.html shows us distribution of momentum. Particle is not in simple to and fro motion of the constant speed.

    Last edited: Mar 2, 2010
  9. Mar 2, 2010 #8
    Please allow me to muddle things further. The infinite square well is a very special case. It is not the simple example taught to us in Quantum I.

    We know that observables must be Hermitian operators. It turns out that the momentum operator is not Hermitian for the infinite square well! Any wave function in the infinite well must vanish at the boundaries. But the momentum eigenfunctions exp(ikx) do not vanish anywhere.

    I am not sure that we can even discuss momentum in such a case.

    Best wishes
  10. Mar 2, 2010 #9


    User Avatar
    Science Advisor

    I am not sure that your derivation is correct for an infinite square well, although it is certainly correct for a finite V. The fact is that the infinite square well is not a physically realistic case, but rather a "toy" problem used for instructive purposes. As such, I think it is reasonable to claim that the momentum and energy operators commute over the range of positions where the problem is well-defined. However, the (non-physical) discontinuity of the slope of the wavefunction at the boundaries of the box does present a problem with my (admittedly cursory) analysis. The ground state wf is only one half-period of the plane-waves representing the momentum eigenstates corresponding to |p|=sqrt(2mE) (I believe this is part of eaglelake's argument in his latest post) ... therefore my statement that the "uncertainty in the absolute value of the momentum is zero" was incorrect. Actually, this should have been obvious, since the uncertainty has a precise mathematical definition, namely:

    [tex]\Delta p^{2} = \left\langle\hat{p}^{2}\right\rangle - \left\langle\hat{p}\right\rangle^{2}[/tex], but here [tex]\left\langle\hat{p}\right\rangle=0[/tex], so we have
    [tex]\Delta p = \sqrt{\left\langle\hat{p}^{2}\right\rangle} = \sqrt{2mE}=\frac{h}{2L}[/tex],
    where L is the width of the box.

    A similar calculation for the uncertainty in position gives:

    [tex]\Delta x=\sqrt{\left\langle\hat{x}^{2}\right\rangle - \left\langle\hat{x}\right\rangle^{2}}=\frac{L}{2\pi}\sqrt{\frac{\pi^2}{3}-2}[/tex],
    and so
    [tex]\Delta x\Delta p=\frac{\hbar}{2}\sqrt{\frac{\pi^2}{3}-2}\approx1.136\frac{\hbar}{2}>\frac{\hbar}{2}[/tex]

    as required by HUP.

    One minor point is that, although I did make an incorrect statement about the "uncertainty of the absolute value of momentum", I definitely did not say anything about a particle "moving with constant speed", as you implied in your post. That is a classical description that is not meaningful in a QM context.
  11. Mar 2, 2010 #10

    Take it easy. Let ψ(x) and its Fourier transform ψ(p) be wave functions of a energy eigenstate of a finite square well potential of depth V. We get their formula easily. Then we enlarge V in formula of ψ(x) and ψ(p) to infinity and get the wave functions for infinite square well potential.
    Last edited: Mar 2, 2010
  12. Mar 4, 2010 #11
    If we want to measure the momentum of a particle known to be in the energy eigenstate [tex]\psi _n (x)[/tex] of the infinitely deep square well, then we must write [tex]\psi _n (x)[/tex] in the momentum representation. Here, the basis functions are the momentum eigenfunctions [tex]\varphi (x)[/tex] which satisfy the momentum eigenvalue equation
    & - i\hbar {{d\varphi (x)} \over {dx}} = p\varphi (x)[/tex]

    subject to the boundary conditions [tex]\cr
    & \varphi ( - a) = \varphi (a) = 0 \cr}
    But the solutions of the differential equation [tex]e^{ipx/\hbar }[/tex] cannot satisfy the boundary conditions! The only solution of the momentum eigenvalue equation is [tex]\varphi (x) = 0[/tex] everywhere. The Fourier integral doesn't exist. The probability of getting a value, any value, of momentum is zero! Does this mean that we cannot measure the momentum? Is this result simply because the infinite square well is unrealistic? I welcome everyone’s response.

    It is true that the momentum eigenvalue equation does have solutions for the finite well, but when we let [tex]V \to \infty [/tex], I think we still have the difficulty with the boundary conditions.

    However, there is a way to give meaning to the Fourier transform: Given the particle in energy state [tex]\psi _n (x)[/tex], imagine that we suddenly remove the walls of the box containing the particle. Now we have a free particle in the energy state [tex]\psi _n (x)[/tex]. The momentum eigenvalue problem has non-zero solutions for a free particle and we can now proceed in the usual way.
    Best wishes
  13. Mar 4, 2010 #12


    User Avatar
    Science Advisor

    your analysis is rather flawed ... the ground state of a particle in an infinite square well can be written as a linear combination of momentum eigenstates. What's hard about that? Then the probability of finding the system in a particular eigenstate when a measurement is taken is given by the square modulus of the expansion coefficient for that state. Now, since the momentum eigenstates are continuous, we represent the "linear combination" as an integral rather than a sum, but the meaning is analogous. So, with this continuous distribution the probability of measuring a momentum between p and dp is given by:


    where [tex]\tilde{\psi}\left(p\right)[/tex] is the fourier transform of the position representation of the wavefunction. Now, the fact that there are discontinuities in the slope of the wavefunction at the boundaries of the well means that the fourier transform can have non-zero components across the entire range of possible values (i.e. from [tex]\infty[/tex] to [tex]-\infty[/tex]). However, this is not a problem, since we know from Parseval's theorem that the momentum space and position space representations of the wavefunction must have the same norm. The position space eigenstate is certainly normalizable, so the FT in momentum space must also be.

    I don't have time to derive the precise form of the momentum space wavefunction right now, but I know how to do it ... you just create a step function that is 1 inside the well and zero outside it by adding together two Heaviside functions, and take the product with the ground-state wavefunction (just a sine function). You then take the Fourier transform of this product, using the convolution theorem. The FT's of the step functions and the sine function are well-known, so it should be simple enough to write the solution ...
  14. Mar 5, 2010 #13
    Here I write down formula of wave functions.
    Ground state of a particle in a square well potential of finite depth

    m: mass of the particle
    hbar: Planck constant / 2π
    D>0 :depth of the square well potential
    W: width of the well
    Potential V(x)= 0 for |x|> W/2, -D for |x|< W/2.
    Hamiltonian H=P^2/2m+V(x)
    E : enrgy of the ground state measured from the bottom line of the well

    I. Coordinate representation

    The wave function of the ground state in coordinate representation is
    ψ(x)= C cos (kx) ,sinusoidal wave inside the well
    ψ(x)= B exp (κ(x+W/2)) , B exp (-κ(x-W/2)) exponentially dumping function outside the well
    k = sqrt(2mE)/hbar
    κ= sqrt{2m(D-E)}/hbar
    κ= k tan(kW/2)
    C = sqrt{ 1/(W/2 + 1/κ)} ,amplitude at center of the well
    B = cos(kW/2) * C ,amplitude at the boundary of the well

    Introducing non dimensional constant θ=kw/2, they are expressed as its function, i.e.
    k = 2/w * θ
    κ= 2/w * θ tan θ
    C^-2 = w/2 ( 1 + 1 / (θtanθ) )
    E = hbar^2 / (2m) * (2/w)^2 * θ^2
    D = hbar^2 / (2m) * (2/w)^2 * θ^2 (1 + (tanθ)^2 )
    B^-2 = (C cosθ)^-2 = w/2 ( 1 + 1 / (θtanθ) ) ) *(cosθ)^-2.
    From formula of D we know that when m,w and D are given all the parameters are determined through θ.

    Infinite square potential D→+∞ are realized by θ→π/2 -0. Then,
    k = π/w
    κ= +∞
    C = sqrt(2/w)
    E = hbar^2 / (2m) * (π/w)^2
    D = +∞
    B = 0
    The coordinate wave function outside the well disappears.

    II. Momentum representation

    The wave function of the ground state in momentum representation is Fourier transform of the coordinate wave function, thus
    φ(p)= 1/√(2πhbar) * ∫ψ(x)e^-ipx/hbar dx
    = C/sqrt(2πhbar)* [ sin{(k-p/hbar)W/2} /(k-p/hbar) + sin{(k+p/hbar)W/2} /(k+p/hbar) ]
    + B/sqrt(2πhbar)* 2* 1/(κ^2 + p^2/hbar^2) * {κcos(p/hbar*W/2) - p/hbar*sin(p/hbar*W/2)}

    Introducing non-dimension parameter τ=p/hbar * w/2, and θ as in I,
    φ(p)= 1/√(2π)* √(w/2hbar)* 2θ^2 secθ / √(1 + cotθ/θ)*
    (τ sinτ - θtanθcosτ) * (τ-θ)^-1 * (τ+θ)^-1 * (τ^2 + (θtanθ)^2 )^-1
    Further introducing t=τ/θ,
    φ(p)= 1/√(2π)* √(w/2hbar)* 2θ^(-1) secθ / √(1 + cotθ/θ)*
    * (t sin(tθ)cosθ - cos(tθ)sinθ ) * (t-1)^-1 * (t+1)^-1 * ( t^2 + (tanθ)^2 )^-1 , dp= (2 hbar/w) θdt.
    Here (t sin(tθ)cosθ - cos(tθ)sinθ )^2 = ( t^2 + (tanθ)^2 ) /2 - Re[ ( t - i tanθ)^2 /2 * e^2tθi ] in alternative expression.

    Momentum distribution function is
    |φ(p)|^2 dp= 1/(2π)* 4 (secθ)^2 /(θ+cotθ) * (t sin(tθ)cosθ - cos(tθ)sinθ )^2 * (t-1)^-2 * (t+1)^-2 * ( t^2 + (tanθ)^2 )^-2*dt, where t=p/hbar * w/(2θ)

    Momentum distribution of the ground state of infinite depth square well is given by θ→π/2 - 0,
    |φ(p)|^2 dp = 4π^-2 * {cos (πt/2)}^2 * (t-1)^-2 * (t+1)^-2 dt, where t=p/hbar * w/π


    There seems to be OK.
    Last edited: Mar 6, 2010
  15. Mar 6, 2010 #14

    You misunderstood what is bothering me! I apologize for the confusion. Let me try again.

    We want to determine the possible results of a measurement and the probability distribution of those results. These are the fundamental calculations that we make in quantum mechanics. The postulates tell us how to do this. Among other things, we are told to solve the eigenvalue equation for the observable to be measured.

    For a particle in an infinitely deep square well, the energy eigenvalue equation (Schrodinger’s time independent equation) is

    [tex] - {{\hbar ^2 } \over {2m}}{{d^2 \psi (x)} \over {dx^2 }} = E\psi (x)[/tex]
    where [tex]\psi ( - a) = \psi (a) = 0[/tex]

    This is a boundary value problem; the boundary conditions cannot be ignored. We solve for [tex]\psi (x)[/tex] without difficulty. Since we have written [tex]\psi (x)[/tex] in position space, [tex]\psi (x) = \left\langle {x}\mathrel{\left | {\vphantom {x \psi }}\right. \kern-\nulldelimiterspace}{\psi } \right\rangle [/tex] is the amplitude for finding the particle, known to be in state [tex]\psi (x)[/tex], at position x.

    Now we want to measure the momentum for a particle known to be in energy eigenstate [tex]\psi _n (x)[/tex]. We proceed in the same way. We need to write [tex]\psi _n (x)[/tex] in momentum space, where the basis functions are the eigenfunctions of the momentum operator. You state, “the ground state of a particle in an infinite square well can be written as a linear combination of momentum eigenstates. What's hard about that? “ It is not “hard”, rather I think, it’s impossible! Let me try to explain------

    By definition, the eigenfunctions are solutions of the eigenvalue equation. (I apologize for lecturing, but I want to be clear on this.) We must solve the momentum eigenvalue equation for a particle in the infinitely deep well, [tex]
    - i\hbar {{d\varphi } \over {dx}} = p\varphi
    where [tex]
    \varphi ( - a) = \varphi (a) = 0
    There is no solution because the boundary conditions cannot be satisfied! You assume incorrectly that [tex]
    [/tex] is the momentum eigenfunction. You cannot write the Fourier transform [tex]
    \left\langle {k}
    \mathrel{\left | {\vphantom {k {\psi _n }}}
    \right. \kern-\nulldelimiterspace}
    {{\psi _n }} \right\rangle = \int_{ - a}^a {\varphi ^* (k,x)\psi _n (x)dx}
    [/tex] because [tex]
    \varphi (k,x) = 0
    [/tex] everywhere inside the well. What am I missing?
  16. Mar 6, 2010 #15


    User Avatar
    Science Advisor

    You are missing that the fact that the eigenstates of the momentum operator always refer to the case of a free-particle, i.e. where the potential is zero everywhere in space. So, exp{ikx} *always* describes a momentum eigenstate, irrespective of the problem at hand. These eigenstates form a complete basis, so we can always expand *any* continuous function in terms of them. For the finite square well, the position space representation is continuous everywhere, so it can be described as an expansion in the basis of momentum eigenstates, as I said.

    In short, once the completeness of a basis is established for a given space (and the 1-D momentum eigenstates are a complete set for 1-D problems), then you can use that basis to expand any function in that space ... you are not forced to define new basis states every time the problem changes.
  17. Mar 6, 2010 #16


    User Avatar
    Science Advisor
    Gold Member

    This doesn't make sense because although the Inf square well cuts off the position of the particle to a set of values within -a and a, it does nothing to cut off the momentum of the particles.

    The below boundary condition:

    Doesn't make sense. "a" is a position, whereas phi is the probability of such momentum arising ("a" is NOT a momentum). Even if we just took the value of "a" without looking at what it is, why would the probability of the momentum having some value "a" be zero? The momentum can assume any range of values.

    EDIT: I may have gotten momentum space wavefunctions confused with eigenfunctions of momentum...please hold as I recheck this...

    EDIT 2: I was indeed thinking of momentum-space wavefunctions. However, I'm not convinced that the condition [tex]\varphi(a)=0[/tex] is a requirement. You certainty CAN construct the wavefunctions in the inf square well using the momentum operator's eigenfunctions (they are complete). Also, finding the eigenfunctions of momentum did not require us to impose any potentials, and therefore I have no reason to believe that the eigenfunctions of momentum depend on the potential we are dealing with. If you have a source that confirms this, I'd be glad to read it, I don't claim to be a QM guru haha. Lastly, I may point out that the Eigenfunctions of the momentum operator don't even exist within Hilbert Space, yet we still talk about a particle's "momentum" all the time.
    Last edited: Mar 6, 2010
  18. Mar 6, 2010 #17
    Hi, eaglelake.

    For a particle in a square well of depth D, Hamiltonian is H=P^2/2m+V(x) where potential energy V(x)= 0 for |x|> a, -D for |x|< a. Energy eigenstate for eingenvalue e is classified to bound states, discrete e<0, and to non binding states, continuous e>0. For |x|→∞, bound state ψ(x)→0 and non-binding state ψ(x) remains not zero as momentum eigenstates do. This is also true for the case D→+∞ i.e. infinite well. One finds the wall is too high to climb up, but the other can jump over the hole ahead how deep it it is and walk away. The boundary condition [tex]\psi ( - a) = \psi (a) = 0[/tex] does not apply to all the states but applies, in the limit of D→+∞, only to part of bound states that have finite energy measured from the bottom line of the potential well.

    Not each momentum eigenstate but their superposition or wave packet is satisfying the boundary condition [tex]
    \varphi ( - a) = \varphi (a) = 0
    [/tex] and [tex]
    \varphi (x) = 0 [/tex] outside the well. When you observe momentum of the ground state, one eigenstate of momentum should appear, that does not satisfy [tex]
    \varphi ( - a) = \varphi (a) = 0
    [/tex] and [tex]
    \varphi (x) = 0 [/tex] outside the well. From the former discussion of this post it means that observation process of momentum requires infinite energy to pump up the state so that momentum eigenstate realize.
    Last edited: Mar 7, 2010
  19. Mar 12, 2010 #18
    Although I am still uncomfortable with momentum as discussed here, I accept the wisdom of the group. Apparently I am making an issue where none exists! A wise man once said, "Listen to your critics. They are your friends and they will make you better." Thanks to all who responded to my query.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Uncertainty question