For the ground level of a harmonic oscillator,

Question

increment x increment p sub x equals h with stroke slash 2

. Do a similar analysis for an excited level that has quantum number

n

. How does the uncertainty product

∆ x ∆ p_{x}

depend on

n

?

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with the uncertainty principle. This problem says Heisenberg's uncertainty principle states that there is always some uncertainty when measuring any variable of a particle in a harmonic oscillator. The product of particle position and momentum uncertainties is delta X multiplied by delta P is equal to H bar divided by two for the ground state. What should be the product of uncertainty for the nth excited state? And how does the uncertainty vary with the quantum number in we four possible choices as our answers. For choice A we have delta X multiplied by delta P is equal to the quantity of N plus one divided by two N quantity multiplied by H bar. And this increases within, for choice B, we have delta X multiplied by delta P is equal to the quantity of N plus one plus one divided by two N quantity multiplied by H bar. And this increases within, for choice C we have delta X multiplied by delta P is equal to the quantity of N minus one N quantity multiplied by H bar. And this decreases within. And for choice D, we have delta X multiplied by delta P is equal to the quantity of N minus one divided by four N quantity multiplied by H bar. And this decreases with N. So we need to calculate this product of our uncertainties of delta X and delta p. So recall the relationships for the uncertainties in our position and um momentum for a harmonic oscillator. So for delta X, it's gonna be equal to a divided by the square root of two N. For delta P, it's gonna be equal to P max divided by the square root of two where a here is the amplitude of motion and P max is the maximum momentum that the oscillator undergoes. Now, we can actually write P max in terms of our amplitude. And so we call your definition of momentum. We're gonna have P max is equal to the mass M multiplied by its speed, which is V max. However, V max can be written in terms of the amplitude for harmonic oscillator. So we'll have P max is equal to M multiplied by. And here for V max, we'll have omega multiplied by a where omega here is the angular frequency of our harmonic oscillation. So now if I look at my uncertainty of relationship, we have delta X multiplied by delta P is equal to. And here for X, we're going to have a divided by the square root of two, that's gonna be multiplied by the quantity of M multiplied by Omega multiplied by A for P max. And that's going to be also divided by the square root of two. So simplifying we'll have delta X multiplied by delta P is equal to one half multiplied by M multiplied by omega multiplied by A squared. Now, what I ultimately want to do is write our amplitude A in terms of our quantum number and, and H bar. So to do that, we're going to look at the energy. So if we call your formula for the energy or a given state for a quantum oscillator, so that's gonna be en is equal to the quantity of N plus one divided by two N quantity multiplied by H bar omega. Now this is also equal to the total energy of a harmonic oscillator. So recall for that formula, we're gonna have one half K A squared we're in here. It's gonna be your quantum number H bar is a constant omega is again, our angular frequency for the oscillation K is our spring constant for our oscillator. And A again is our amplitude. So we can actually solve this for A. So when we do that, we'll get A is equal to the square root of the quantity of two multiplied by the quantity of N plus one divided by two N quantity multiplied by H bar multiplied by omega divided by K. So to solve for A, I simply multiplied both sides of the equation by two divided both sides by K and then took the square root. So we can now plug this formula in to our uncertainty relationship. And in doing so, we'll get delta X multiply by data P is equal to one half multiplied by M multiplied by omega multiplying the quantity of two multiplied by the quantity of N plus one divided by two in quantity multiplied by each bar multiplied by omega. And that is all divided by K. So we can simplify this expression a little bit. And to do that, we need to recall a relationship between our angular frequency, our mass and our spring constant. So we call for a harmonic oscillator that we have omega is equal to the square root of K divided by M. So using that relationship, we can actually cancel out a lot of terms. So we'll have delta X multiplied by delta P is equal to the factor of two and one half will cancel. And then I'll be left with the quantity of N plus one divided by two in quantity multiplied by H bar. That's gonna be multiplied by omega squared, which we can replace with K divided by M. And then that could be multiplied by M and divided by K. So the MS will cancel, the cases will cancel. And what I'm left with is delta X multiplied by delta P is equal to the quantity of N plus one divided by two N quantity multiplied by each bar. And so Now, I need to figure out how this varies with quantum number in. And for the right hand side, if I increase in that right hand side will actually increase as well. So this means that this is going to be actually increasing within. So the answer to the second half of that question is to increase within. So that is the answer that I'm actually looking for. And if I look at my answer choices, this actually corresponds to answer choice A. So just to recap of what we've done here, we started off by writing our uncertainty in X and our uncertainty in our momentum delta P in terms of the amplitude of our oscillation A, we then use our energy equation right, our amplitude A in terms of the quantum number and H bar. And then we just plug that into our uncertainty relationship and simplified our expression until we got the one that we were looking for. So I hope that this has been useful and I'll see you in the next video.

For the ground level of a harmonic oscillator, $increment x increment p sub x equals h with stroke slash 2$ . Do a similar analysis for an excited level that has quantum number $n$ . How does the uncertainty product $∆ x ∆ p_{x}$ depend on $n$ ?

Key Concepts

Harmonic Oscillator

Heisenberg Uncertainty Principle

Quantum Number n

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