Question

5. (a) The uncertainty principle is $\Delta p \Delta x \ge h/2$. State the precise meaning of the quantities $\Delta p$ and $\Delta x$. State the general relationship between the wavefunction describing a particle and the particle's position and momentum, and briefly explain how this leads to the uncertainty principle. [4 marks] (b) A particle moves in a one-dimensional harmonic oscillator potential, so that its Hamiltonian is $H = \frac{p^2}{2m} + \frac{m\omega^2}{2}x^2$. Use the uncertainty principle to explain why the lowest-energy state has a non-zero $\Delta p$ and hence a non-zero energy. [2 marks] (c) Assuming the uncertainty principle is holds as an equality $\Delta p \Delta x = h/2$, write down an estimate of the energy of a eigenstate of the oscillator as a function of its uncertainty in position, $\Delta x$. Use this estimate to obtain an approximate formula for the size of the ground-state wavefunction and its energy. [Hint: the function $f(x) = A/x^2 + Bx^2$ has a minimum at $x = (A/B)^{1/4}$ for $A/B > 0$.] [4 marks]

          5.
(a) The uncertainty principle is $\Delta p \Delta x \ge h/2$. State the precise meaning of the
quantities $\Delta p$ and $\Delta x$. State the general relationship between the wavefunction
describing a particle and the particle's position and momentum, and briefly
explain how this leads to the uncertainty principle.
[4 marks]
(b) A particle moves in a one-dimensional harmonic oscillator potential, so that its
Hamiltonian is $H = \frac{p^2}{2m} + \frac{m\omega^2}{2}x^2$. Use the uncertainty principle to explain why the
lowest-energy state has a non-zero $\Delta p$ and hence a non-zero energy.
[2 marks]
(c) Assuming the uncertainty principle is holds as an equality $\Delta p \Delta x = h/2$, write down
an estimate of the energy of a eigenstate of the oscillator as a function of its
uncertainty in position, $\Delta x$. Use this estimate to obtain an approximate formula for
the size of the ground-state wavefunction and its energy. [Hint: the function
$f(x) = A/x^2 + Bx^2$ has a minimum at $x = (A/B)^{1/4}$ for $A/B > 0$.]
[4 marks]

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5.
(a) The uncertainty principle is Δ p Δ x ≥ h/2. State the precise meaning of the
quantities Δ p and Δ x. State the general relationship between the wavefunction
describing a particle and the particle's position and momentum, and briefly
explain how this leads to the uncertainty principle.
[4 marks]
(b) A particle moves in a one-dimensional harmonic oscillator potential, so that its
Hamiltonian is H = (p^2)/(2m) + (mω^2)/(2)x^2. Use the uncertainty principle to explain why the
lowest-energy state has a non-zero Δ p and hence a non-zero energy.
[2 marks]
(c) Assuming the uncertainty principle is holds as an equality Δ p Δ x = h/2, write down
an estimate of the energy of a eigenstate of the oscillator as a function of its
uncertainty in position, Δ x. Use this estimate to obtain an approximate formula for
the size of the ground-state wavefunction and its energy. [Hint: the function
f(x) = A/x^2 + Bx^2 has a minimum at x = (A/B)^1/4 for A/B > 0.]
[4 marks]

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University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Jacob Schulze

10/24/2022

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Step 1: The uncertainty principle states that the product of the uncertainties in momentum (Ap) and position (Ax) of a particle is greater than or equal to Planck's constant divided by 4π. Show more…

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5. (a) The uncertainty principle is Ap∆x ≥ h/2. State the precise meaning of the quantities Ap and Ax. State the general relationship between the wavefunction describing a particle and the particle's position and momentum, and briefly explain how this leads to the uncertainty principle. [4 marks] (b) A particle moves in a one-dimensional harmonic oscillator potential, so that its Hamiltonian is H = p² mw² + 2m 2 x². Use the uncertainty principle to explain why the lowest-energy state has a non-zero Ap and hence a non-zero energy. [2 marks] (c) Assuming the uncertainty principle is holds as an equality Ap∆x = h/2, write down an estimate of the energy of a eigenstate of the oscillator as a function of its uncertainty in position, Ax. Use this estimate to obtain an approximate formula for the size of the ground-state wavefunction and its energy. [Hint: the function f(x) = A/x² + Bx2 has a minimum at x = (A/B) 1/4 for A/B > 0.] [4 marks]

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Transcript

00:01 In this problem, we're going to work with a quantum spring, essentially.

00:05 So the spring we know has an energy as shown here.

00:10 It's going to be kinetic energy plus the spring energy.

00:13 And with the first problem asks us to do is use the uncertainty principle to put an inequality on this energy.

00:21 So the uncertainty principle tells us that momentum times position is greater than or equal to h bar over 2.

00:29 So this tells you that this is like essentially the certainty that you can have in momentum and position itself essentially tells you that it's impossible to know both momentum and position at the same time.

00:46 So what we can do here is essentially in a way solve this for x, right? x is greater than or equal to h bar over 2 px.

00:59 So if you plug in this value for x, right, that will be the lower bound to that energy, right? so plugging that in, you should find that this is going to be less than or equal to.

01:16 Let's see, k2 squared.

01:19 It's going to be four in the bottom times two is eight, and then px squared.

01:24 And that is exactly what they ask just to show.

01:29 Now in part b, where we're going to do is minimize this.

01:33 So this system is essentially the harmonic oscillator, and we know that the minimum of the harmonic oscillator is h bar omega over 2, right? because we know that the eigen values of the harmonic oscillator are this and plus one half times h bar omega.

01:55 And so when you plug n equals zero into this, you get a half of h bar omega.

02:00 But let's show that with the equation that they give us here.

02:05 So what we're going to do here is we're going to minimize this effectively.

02:12 So we're going to take a derivative of that with respect to px and set that equal to zero.

02:26 Okay.

02:29 So what will we get, we're going to get px over m plus, actually it will be minus over m...

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