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

Stationary States Question

  1. Jan 29, 2005 #1
    I am trying to understand the nature of uncertainty relations in quantum mechanics. I am looking specifically at a relation between energy and position uncertainty... the book that I am reading hints that this relationship has no meaning in a stationary state. Why would that be?
  2. jcsd
  3. Jan 29, 2005 #2


    User Avatar

  4. Jan 29, 2005 #3

    Yes that did help a lot. However, I think I am still missing something. I understand that in the stationary state nothing changes with time. Yet, I do not quite see how this makes the uncertainty relationship between energy and position any different from the one between position and momentum. Maybe if you could elaborate on how the constant probability effects this uncertainty relation...

    Thanks for the link by the way. I didn't search through it entirely yet, but many questions that I have been having were certainly addressed in it.
  5. Jan 29, 2005 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In non-relativistic QM, there is difference in nature between the x-p uncertainty and the E-t uncertainty.
    x and p are observables, so they both are results of measurements in NR QM ; the uncertainty principle just gives you a property of their probability distributions (namely that the product of their standard variations has a lower bound).
    E is also an observable, but time (t) is a parameter. There is NO T - operator or T observable in QM. So you cannot talk about "an uncertainty in time" as the standard deviation of the probability distribution of t.

    But what then does the E - t "uncertainty" relation mean ?

    It actually means that if you want to have significant changes in the expectation values of ANY observable within a time lapse dt, then you need to have a superposition of several energy eigenstates such that the standard deviation of the energy measurement becomes at least dE, with
    dt . dE > hbar/2

  6. Feb 1, 2005 #5
    I would just like to add that your posts are very clear and helpful, Patrick!
  7. Feb 1, 2005 #6

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    Does your "RPI" stand for "Rennselaer Polytechnic Institute"?

    I went to RPI, and I now teach at HVCC. Hi! :smile:
  8. Feb 1, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    That's an inexact formulation of a property of the stationary states,not exactly the DEFINITION.Stationary states in the standard formulation are described through NORMALIZED EIGENSTATES OF THE TIME-INDEPENDENT HAMILTONIAN (Schroedinger picture) ASSOCIATED TO EIGENVALUES (denoted generally by [itex]E_{n} \in \sigma_{d}(\hat{H}) [/itex])...

    What wave function and what probability are u talking about...??

  9. Feb 1, 2005 #8
    Indeed the post of Vanesch explains everything and it is very well written. Now if you just realized that a stationary state is represented by a wavefunction with only ONE energy-value in it (not a superposition of several energy values [tex]E_{i}[/tex]), it is quite straightforward to see that the uncertainty-relations do not apply here. Just focus on the example that Vanesch gave as to explain the uncertainty between both E and t


    ps : indeed, the probability density of a stationary state is independent of time but NOT position and the expectation value of any observable (provided the observable is independent of t itself) is also independent of time BUT NOT position.
    Last edited: Feb 1, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Stationary States Question
  1. Stationary states (Replies: 10)

  2. Stationary state? (Replies: 5)