CALC A particle of mass m has the wave function ψ(x) = Ax exp ...

Question

CALC A particle of mass m has the wave function ψ(x) = Ax exp (−x²/a²) when it is in an allowed energy level with E = 0. Find and graph the potential-energy function U(x).

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with quantum mechanics. This problem says an oscillator of mass M oscillates according to the equation I as a function of Y is equal to a multiplied by Y multiplied by E raised to the minus Y squared divided by A squared or permissible states where E is equal to zero represent the potential energy function on a graph. Below the question we're given four choices for answering. And each one contains a graph that has the potential energy U on the Y axis and the quantity of Y divided by A on the X axis. For answer choice A we have a problem with a maximum of three H bar squared divided by MA squared. And the two roots for the parabola are minus 1.22 and 1.22. For choice B, we again have a parabola that has a minimum at minus three H bar squared divided by MA squared. And it again has two roots one at minus 1.22 and another at 1.22. For choice C we have a prola where the minimum of the pro prola is at the origin and it opens upwards. And for choice D, we again have a problem that has a maximum at the origin and it opens downwards. So with this problem, we're given our wave functions I and so we can actually use um Schenscher equation in order to calculate our potential for this oscillator. So we call for Schrodinger's equation, we have minus H bar squared divided by two M multiplied by the second derivative of SI with respect to why? Plus U multiplied by P SI is equal to E multiplied by Phi where M here is our mass of our oscillator H bar is a constant P I is our wave function U is our potential energy function and E is the energy of the state. Now we're told that their energy E is equal to zero. That means that the whole right hand side is equal to zero. And I need to solve this for you, which is our potential energy function, which we're trying to graph. So I'll move the first term on the left hand side over to the right hand side and then divide by Psy. So I'll get you, it's equal to H bar squared divided by two M multiplied by the second derivative of SI with respect to Y. And that whole quantity is divided by SI. So at this point, I need to take derivatives of SI in order to get that second derivative. So we're gonna start off by taking the first derivative of si with respect to Y. And that's gonna be equal to when I take the derivative of our function that we were given in the problem, I'll have a multiplied by, eat the minus Y squared divided by A squared. Since I took a derivative of Y, which just gives me one, then I'll have, have plus because I'm using the product rule, I again have a multiplied by Y. But now I need to take a derivative of our exponential. And so when I take a derivative, our exponential, I'll get back that exponential. And that's gonna be multiplied also be multiplied by the derivative of what's in its exponent, which in this case is gonna be minus two Y divided by A squared. And this could be multiplied by E to the minus Y squared divided by A squared. I can simplify this a little bit. So I have my first derivative of Psy with respect to Y equal to, I'll have a multiplying the quantity of one minus two Y squared divided by A squared. And that quantity is multiplied by E to the minus Y squared divided by A squared. So now I can take another derivative to get my second derivative. So I have the second derivative of si with respect to why? And this could be equal to. And again, I'm gonna have to use the product rule. So I take a derivative of my first term of A. And when I take a derivative of that first term, I'll have minus four Y divided by A squared. That's gonna be multiplied by E to the minus Y squared divided by A squared. And they'll have plus again, have to take a derivative of the exponential. Let's all have a multiplied by the quantity of one minus two Y squared divided by A squared. And when I take the derived of the exponential, I'll get what I had originally back in the first derivative when I took that derivative of the exponential. So it's gonna be the quantity of minus two Y divided by A squared multiplied by E to the minus Y squared divided by A squared. So to simplify this, I'm gonna factor out an A ye to the minus Y squared divided by A squared. So I have my second derivative of P si with respect to why that could be equal to a multiplied by Y multiplied by E to the minus Y squared divided by A squared. And this could be multiplying the quantity of minus four divided by A squared. They don't have minus two divided by A squared and they'll have a plus four Y squared divided by eight to the fourth. Now, the reason why we factor out the quantity that we did that is equal to our original wavefunction si, so that means our second derivative of si with respect to why is equal to P si multiplying the quantity. And here, um I'm gonna go ahead and factor out another one over A squared. So I have actually out in front I divided by a squared multiplying the quantity of or Y squared divided by A squared minus six. So we can take this plug that back into our formula or the potential energy. That means that you, it's gonna be equal to H bar squared divided by two M multiplying the quantity of P si divided by A squared, multiplying the quantity of four Y squared divided by A squared minus six. And all of that is divided by P si the sites will cancel. And we can just simplify our expression a bit. We'll have U is equal to, we'll have H bar squared divided by ma squared and that's gonna be multiplying the quantity. And here I'm gonna bring in the factor of one half. So I'll have two multiplying the quantity of ad or sorry uh Y divided by a in quantity squared minus three. So when I have it in this form, I see that I actually do get a Puebla. So Y divided by A, it is going to be our variable that's gonna be on our X axis. And so this is quadratic in that variable. And so we can actually look at um the Y intercept. And so when U is, uh when Y divided by A is equal to zero, we'll have U of zero is going to be equal to minus three H bar squared divided by ma squared. So that would give me the white intercept and I can also find the roots. So U is going to be equal to zero when that term in the bracket is actually equal to zero. And so that means that Y divided by a going to be equal to plus or minus the square root of three divided by two, which is equal to plus or minus 1.22. So those are the um parameters that we want to use to look at for our graph. So that means we're going to have a problem that is opened upwards that has a minimum at minus three H bar squared divided by MA squared and has roots of plus or minus 1.22. And if I look at my answer choices, this actually corresponds to answer choice B because that um pre has a minimum of minus three H bar squared divided by MA squared and it has roots of plus or minus 1.22. So just to recap of what we've done, we started off with our schrodinger equation and solved it for our potential energy function. You but in order to calculate that using our wavefunction, we had to take two derivatives of our wavefunction and plug that back into our schrodinger equation. And once we did that, we can do a lot of cancellation and simplifying to get our expression for you, which showed to be a para. So I hope that this has been useful and I'll see you in the next video.

CALC A particle of mass m has the wave function ψ(x) = Ax exp (−x²/a²) when it is in an allowed energy level with E = 0. Find and graph the potential-energy function U(x).

Key Concepts

Wave Function

Potential Energy Function

Energy Levels in Quantum Mechanics

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