thanks for a step-by-step solution!!! What is the average potential energy? Expr
ID: 822443 • Letter: T
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thanks for a step-by-step solution!!!
What is the average potential energy? Express it with Delta x. (Hint: ) What is the average kinetic energy? Express it with Delta p. (Hint: ) Express the total average energy with the Delta x, Delta p (Hint: ) If the fluctuations in position Delta x are somehow forced to be small, then the Heisenberg uncertainty principle requires large momentum fluctuations Delta p. This implies that there exist optimal values of Delta x and Delta p that lead to a minimum in energy. If so, then we would expect the QM system to find these optimum values naturally if the energy is minimized. Heisenberg estimate of Delta p = /2 Delta x can give quite reasonable results. Using it leads answer c) to: Find the optimum value for Delta x which leads minimum energy (Hint: Find Delta x which satisfies d/d Delta x H^ = 0) What is the corresponding spread in momentum (Delta p) in d)? What is the expectation value of the total energy for the minimum uncertainty situation? Is it the same as ground state energy in the QM solution?Explanation / Answer
Well, one way to think about the problem in relation to classical mechanics is to examine the expected ("average") values of certain physical quantities. In a sense, as the quantum number increases, things tend to look "more classical," so one can examine the expected values for the highly excited states of the quantum harmonic oscillator.Start with the Hamiltonian operator for the quantum 1-dimensional harmonic oscillator,
H = T + V = (p^2)/(2m) + (1/2)m w^2 x^2,
where T is the kinetic energy of the particle, V is the potential energy of the particle, p is the momentum of the particle, x is the displacement of the particle from its equilibrium position at x = 0, m is the mass of the particle, and w is the angular frequency of the oscillation.
If you look at the expected value of the energy of the oscillator when it is in the nth stationary state (n = 1, 2, 3, ...), you find
E = <H> = hbar w (n + 1/2).
Now, take a look at the expected value of the kinetic energy and the potential energy of the oscillator when it is in the nth stationary state:
<T> = <p^2>/(2m) = (hbar w / 2) (n + 1/2)
<V> = (1/2) m w^2 <x^2> = (hbar w / 2) (n + 1/2).
In other words, <T> = <V> = E/2 for the quantum harmonic oscillator. In a larger sense, though, you can see that both the average kinetic energy and average potential energy grow as you go to states of the oscillator that have higher total energies.
How is this relevant to your question? Well, first of all, note that the equations for position and momentum uncertainty are
Delta x = sqrt(<x^2> - <x>^2)
Delta p = sqrt(<p^2> - <p>^2) .
Since we are working with a potential energy function that is symmetric about x = 0, that means that there is no preference for the particle to be found left of the center versus right of the center, or to be traveling to the left versus to the right. That means that the expected values of position and momentum are zero, since position and momentum are signed quantities (in three dimensions, they are vector quantities). (To be more exact, the expected value of the momentum is zero, and the expected value of the position is at the center of the potential, which we have chosen to be x = 0.) We find that
<x> = 0,
<p> = 0.
So, for the special case of a symmetric potential, <T> = <p^2>/(2m). Since the harmonic oscillator potential energy operator is proportional to x^2, we find that
Delta x = sqrt[2<V> / (m w^2)] (harmonic oscillator)
= sqrt[hbar / (m w)] sqrt(n + 1/2),
Delta p = sqrt[2 m <T>] (harmonic oscillator)
= sqrt[hbar m w] sqrt(n + 1/2).
You can see that both (Delta x) and (Delta p) increase as the square root of (n + 1/2), which goes like simply sqrt(n) for large values of n.
So far, this has just summarized what you already said -- that the uncertainty increases with increasing energy of the oscillator. Now that we've established a few relations, we can try to see why.
First of all, the wavefunction of the oscillator (technically, the position-space wavefunction) gives the probability amplitude for the oscillator to be found at a certain position (the absolute square of this amplitude gives the probability density). The ground state of the harmonic oscillator has a wavefunction that is shaped like a Gaussian function -- a high probability of finding the oscillator near its equilibrium state (x = 0), and a low probability of finding the oscillator stretched (x > 0) or compressed (x < 0).
Now, the momentum-space wavefunction is just the "Fourier transform" of the position-space wavefunction. In general, that means that a single narrow probability distribution in position space corresponds to a single wide probability distribution in momentum space, and vice-versa. The width of the probability distribution is related to the measurement uncertainty -- if the distribution is narrow, then the particle is confined to a small region, and the uncertainty in its location is small. It turns out that the Fourier transform of a Gaussian function is another Gaussian function, so the distributions in position space and momentum space are the same -- centered around zero, not too narrow, not too wide. This actually corresponds to the case where the lower bound of the uncertainty relation is saturated, (Delta x) (Delta p) = hbar / 2.
As you go to higher energy states of the harmonic oscillator, you find that the oscillator can be found displaced farther and farther from its equilibrium position. This is congruent with our finding that the average potential energy increases, because the potential energy increases when the oscillator is displaced from its equilibrium position. In fact, at high quantum numbers, the oscillator is mosty likely to be found near the "turning points" of its oscillation (maximum stretching and maximum compression). This is the classically-expected behavior, because the oscillator spends most of its time where the momentum is the lowest (which is when the oscillation is switching from stretching to compressing or vice-versa). This results in a position-space wavefunction that has high-amplitude "wings" far from zero displacement. One can see that in this case, the position uncertainty is large, because the oscillator is likely to be found at large displacements from x = 0. In fact, as n goes to infinity, one can find the oscillator stretched almost to infinity or near full compression. Compare this to the case of the ground state in which the particle is likely to be found in relatively narrow region around the equilibrium displacement.
At the same time, the momentum uncertainty is going up because the oscillator's kinetic energy is going up. One can find the oscillator moving to the left at a high rate or moving to the right at a high rate, which results in a large spread of possible momentum values. Compare this to the ground state where the particle has less energy and is likely to be found with a momentum relatively close to zero.
So, you can see that there are some physical reasons why the uncertainty goes up with increasing energy -- the amplitude of the oscillation increases, so the distribution of oscillator positions widens, which increases the position uncertainty. The kinetic energy of the oscillator also increases (although the frequency of the oscillation stays the same) -- the oscillator is moving faster, so distribution of oscillator momenta widens, which increases the momentum uncertainty. In fact, one finds,
(Delta x) (Delta p) = (hbar / 2) (2n + 1),
which is much greater than the lower limit of the uncertainty relation (Delta x) (Delta p) = (hbar / 2) when the quantum number n (and hence the total energy of the oscillator) increases toward infinity.
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