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Homework Help: Position expectation value of a particle in a box

  1. Jun 10, 2006 #1
    I have calculated the expectation value of a particle in a box of width a to be a/2. The wavefunction of the particle is:

    [tex]N Sin(k_n x) Exp[-i \frac{E_n t}{\hbar}][/tex]

    Now, in the first excited state with [tex]k_n[/tex] equal to [tex]2\pi / a[/tex] the position probability density peaks at a/4 and 3a/4 but is zero at a/2! But the expectation value is still given by a/2 because it is still the average value of the position. How can it be that the expectation value of the position is illegal (position probability equal to zero)? I thought the position expectation value was the position where the particle was most likely to be found?

  2. jcsd
  3. Jun 10, 2006 #2
    The expectation value of an observable in a QM state is the weighted average of all the values you could obtain when measuring the variable. It is not, itself, necessarily a value that you could measure that observable to be.

    Let's take a simpler example to illustrate this. If we call the eigenstates that you've listed [tex]\Psi_n[/tex], we can see that any linear combination of these will also satisfy the Schroedinger Equation.

    In particular, we can choose a state like [tex]\Psi = \frac{1}{\sqrt{2}} (\Psi_1 + \Psi_2)[/tex]. (You can check to make sure this is properly normalized.)

    If we take the energy expectation value of [tex]\Psi[/tex], we find that [tex]\langle E \rangle = \frac{E_1 + E_2}{2}[/tex]. This is certainly not an energy eigenvalue of the system; so it isn't an energy you would ever measure the system to have. Instead, what this represents is that, were the system in the state I describe, half of the times you measured the energy you would get [tex]E_1[/tex] and the other half you would get [tex]E_2[/tex].

    Taking this kind of reasoning back to your example, we see that the natural interpretation for the expectation value of position is that it is the point at which there is equal probability of the particle being observed to either side. That is, half the times you measure the particle's position and don't find it to be at [tex]\langle x \rangle[/tex] it will be to the left of [tex]\langle x \rangle[/tex], and the other half it will be to the right.
  4. Jun 10, 2006 #3
    Expectations are really what the average measure value would be if you made measurements on an ensemble of many, many identically prepared systems.
  5. Jun 11, 2006 #4


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    This has nothing to do with quantum theory per se:
    What is the expectation value of the value of rolling a fair dice (with faces 1,2,3,4,5, and 6) ?

  6. Jun 11, 2006 #5


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    No, the expectation value is NOT the most likely outcome. It is simply the *average* outcome.

    A striking is example is the following: consider tossing a coin and let's assign a value of 1 to the head side and a value of 2 to the tail side. Assume that the coin is fair (both sides are equally probable). The expectation value after N tosses is (N_1 + 2 N_2)/(N_1 +N_2) where N_1 is the number of heads and N_2 is the number of tails. Unless all tosses are heads or they are all tails, it is clear that the expectation value is somewhere between 1 and 2, and it is clear that as N goes to infinity, the expectation value approaches 1.5. All this is true even though th eonly possible outcomes are 1 or 2!

    This is a striking example because the distribution is discrete (50% of the time the result is 1 and 50% of the time it is 2). But even with continuous distributions, one can end up with the result that the expectation value is none of the possible measurements, as in your example where the expectation value correspond to a result which is never obtained.

    Even if a probability distribution is nonzero everywhere so that all results are possible (think for example of the Maxwell-Boltzmann distribution), the expectation value is not necessarily equal to the most likely outcome. That is not the case for example for the MB distribution or even for the measurement of the radial position of an electron in the gorund state of hydrogen. That's because those distributions are not symmetric.

    Hope this makes sense
  7. Jun 11, 2006 #6


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    Hi Pat,

    That's why I gave the example of the expectation value of a fair dice:
    it is 3.5, which is not one of the obtainable values...
    As such, I didn't need to assign a value to heads and tails.
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