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Probability of measuring specific energy of particle

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    How do you calculate the probability of measuring a specific energy value of a particle (eg. what's the prob that you'll measure exactly E=h^2/8ml^2 in a given particle)?

    3. The attempt at a solution
    When I saw the word probability, I just thought: Probability=Integral(psi* H psi)... where psi is the wavefunction of the particle and H is the Hamiltonian operator. But how would you set the range of that integral? I've done problems where I had to find the probability of finding the particle in a certain space (like from x=0 to x=length/2) but I'm not sure as to solving this problem... Also, how would you include the energy value specified by the problem into that integral?
  2. jcsd
  3. Sep 16, 2007 #2
    Suppose that [tex]|\phi\rangle[/tex] is an eigenstate of the operator A with the eigenvalue a. The probability that a measurement of A on the state [tex]|\psi\rangle[/tex] will result in a is [tex]\left| \langle \phi | A | \psi \rangle \right|^2[/tex].

    In the case that states are represented as functions over an interval [tex]\left[x_0, x_1\right][/tex], the inner product is defined as:

    [tex]\langle f | g \rangle = \int_{x_0}^{x_1} f^* g\,dx[/tex]

    Hope that makes sense...
  4. Sep 16, 2007 #3


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    watch out! This expression is not a probability! It's an expectation value!!
    the probability of obtaining the eigenvalue corresponding to the eigenstate [itex]|\phi\rangle[/itex] is [itex]\left| \langle \phi | \psi \rangle \right|^2[/itex].
  5. Sep 16, 2007 #4
    Indeed -- my bad. It's always good to have more than one person checking! In fact, I left it as an exercise for physgirl! Ahem... :blush:
  6. Sep 16, 2007 #5

    Can you tell me what the difference between what you mean by phi vs. psi? :(

    I would guess that psi is the actual wavefunction of the particle (which is given in the problem)? So how do you determine phi? I would guess that has something to do with the specified E value...
    Last edited: Sep 16, 2007
  7. Sep 16, 2007 #6
    Oh I feel bad now -- I've managed to confuse you by my mistake. The probability is, as nrqed has said:

    [tex]| \langle \phi | \psi \rangle | ^ 2[/tex]

    [tex]\psi[/tex] is the state of the system before measurement. [tex]\phi[/tex] is the state of the system after measurement -- it is the eigenstate corresponding to the eigenvalue a -- the measured value.
  8. Sep 16, 2007 #7
    Oh, correct me if I'm completely off-track... but do I have to worry about the time-dependent schrodinger equation in this case? The original wavefunction given is at time 0... but I guess that doens't mean I can ignore time? (do I just need to add on the e^(-iEt/h.bar?? and plug in the E expression that I'm interested in?)
  9. Sep 16, 2007 #8
    In quantum mechanics, the measurement process instantaneously. So if you know the state just before measurement, then you don't need the time evolution. However, if you do need to use it, and if your state is not an energy eigenstate, the evolution will not be given by multiplying [tex]e^{-\frac{iEt}{\hbar}}[/tex]
  10. Sep 16, 2007 #9
    [tex]| \langle \phi | \psi \rangle | ^ 2[/tex]

    I think I have a problem tying this in with the information I have to get what I want... I am given the wavefunction of the particle psi(x,0) that is in terms of x and L (length of box). So is that the psi you want to put into the above expression? So then what is phi?
  11. Sep 16, 2007 #10
    You need to know the energy eigenstates of the system. Perhaps you could reproduce the entire question?
  12. Sep 16, 2007 #11
    Actually, genneth, I think you might have helped me in the calculus forum also, about integrating the step function. This is the same problem, different parts though.

    where A is the normalization factor that I have already found. What's the probability of measuring a particle's energy=h^2/(4ml^2).... (particle is in a 1-D box of length l)
  13. Sep 16, 2007 #12
    I guessed as much. Have you covered the concept of energy eigenstates or eigenstates in general? Do you know how to find them? Assuming that your problem involves no potential, and the energy of the particle is just kinetic energy, then the energy eigenstates are just sinusoidal waves, with nodes at the ends of the boxes. Each additional half-wavelength (and so node) gives you a little more energy.
  14. Sep 16, 2007 #13
    The concept was covered, but I guess not necessarily understood by me :( I think qualitatively I almost view eigenstate=eigenvalue... I mean, the eigenvalue of a hamiltonian operator is the quantized energy values, right? How is eigenstate different? Isn't it also supposed to describe the "state" (with respect to energy values) in which the particle is in?

    Given a wavefunction that is known to be an eigenfunction of the operator, you use the schrodinger equation to solve for the eigenvalues, yes?

    And as for the sinusoidal waves, are you referring to this sort of thing: http://www.lightandmatter.com/html_books/0sn/ch12/figs/particle-in-a-box.png
  15. Sep 16, 2007 #14
    That is indeed what I mean by sinusoidal waves.

    Given an eigenstate of an operator, you can find the eigenvalue by the equation

    [tex]A|\phi\rangle = a|\phi\rangle[/tex]

    In other words, an operator will simply "stretch" its eigenstates, though by different amounts per eigenstate. It's fairly important to realise this distinction. So even though given an operator, you can get the eigenvalue or eigenstate from the other one, they represent different things about the system (and even this is not technically true, but most likely is for the problems you will do).

    If possible, review your notes, or consult a textbook. They will be much more systematic (and less likely to be wrong!) than us random people on a web forum.
  16. Sep 17, 2007 #15
    So I tried looking back at notes, textbook, internet, and whatnot, and I think I have a better theoretical sense of what's going on but I'm still not sure how the pieces really connect together when given a real wavefunction and an operator. So if a system is in initial state |A>, and makes transition to eigenstate |H'> of the observable H because a measurement was made, the probability of a system making that transition into that specific eigenstate is |<A|H'>|^2. Ok, but what does that really mean involving everything that I am given? How do I tie it all in in a "real" problem? do you think you could expand on this explanation or maybe even give an example using a really simple wavefunction and a simple operator??
  17. Sep 17, 2007 #16
    Let's do it for your case of a particle in a 1D box. Let the box have length L, and the Hamiltonian is just [tex]H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}.[/tex] We'll be cheeky, and use our pre-knowledge of the eigenstates of H, rather than solving Schroedinger's equation directly. So we know that the states are [tex]\psi_n = A_n \sin \left( \frac{n \pi x}{L} \right)[/tex] for [tex]n \ge 1.[/tex] The energies are [tex]E_n = \frac{\hbar^2 n^2 \pi^2}{2mL^2}[/tex] which you can verify by applying H to [tex]\psi_n.[/tex] The normalisation coefficients [tex]A_n[/tex] will have to be worked out later, but I think you already know how to do that. Now, suppose we have the particle in some state, [tex]\phi.[/tex] The expectation energy is then [tex]\langle \phi | H | \phi \rangle[/tex]. If we actually do the measurement, then we will collapse the system to one of the [tex]\psi_n[/tex] states, and we will get a energy value of [tex]E_n.[/tex] The likelihood of getting this state is [tex]\left| \langle \phi | \psi_n \rangle \right|^2.[/tex]

    I'll leave the actual mess of doing the integrals to you :wink:
  18. Sep 17, 2007 #17
    Thanks so much for the help! I'm still not getting what I want though :( So the normalization factor is square root of "2/L". I took [tex]\phi[/tex] to equal [tex]\varphi[/tex]* (the complex conjugate) but in this case I took [tex]\varphi[/tex]*[tex]\varphi[/tex] to equal [tex]\varphi[/tex]^2 since all parts of [tex]\varphi[/tex]_n was real... is that wrong? I took the integral from 0 to L of "[tex]\varphi[/tex]^2"... and squared that value. I got values I totally didn't expect and doesn't make sense either :(

    (by [tex]\varphi[/tex] I really just mean psi)
  19. Sep 18, 2007 #18
    If you could just reproduce the entire question, verbatim, that would help us to have a concrete example to talk about. But what you say you've done is correct -- most likely, you've just got an algebraic mistake in there somewhere.
  20. Sep 18, 2007 #19
    Yes, indeed... I did make a stupid mistake. It's good now though! Thanks so much for the help!!!
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