 Alright, so let's try to solve Schrodinger's equation for an actual problem that we could run across in the real world. So we know Schrodinger's equation, written in operator form here, the Hamiltonian acting on psi is equal to e times psi, that's an eigenvalue problem, we're looking for wave functions that are the eigenfunctions of this operator, the Hamiltonian. So to solve Schrodinger's equation we need to get a little more specific about exactly what problem we're trying to solve. So first, because this is our first real example, let's keep it simple and solve a problem in one dimension rather than three dimensions. So if I write Schrodinger's equation in one dimension, my derivative, so the function itself is just going to be a one-dimensional function of one variable, x. The Hamiltonian contains a kinetic energy term, which is just d squared dx squared and some constants and a potential energy term. To proceed any further, I have to tell you what the potential energy term is. So Schrodinger's equation is not an equation where we just solve it once and we're done and we never talk about Schrodinger's equation again. To solve Schrodinger's equation for a particular problem we need the potential energy of that particular system. So every different system has its own potential energy, has its own solutions to Schrodinger's equation. So in order to solve it, I have to tell you what is the potential energy and again because we're keeping things simple, the simplest I can make the problem to start with is to make the potential energy equal to zero. So first I'll write down what the equation becomes, then I'll tell you what real world system might correspond to. So the kinetic energy term is still h squared over 8 pi squared mass second derivative of the wave function. The potential energy is zero so the P e term goes away and that should be equal to energy times the wave function. So what actual system would have a potential energy of zero? If the system were multiple molecules interacting with each other then that potential energy of their interaction would be included in this potential energy term. So if there's no potential energy it means particles are not interacting with each other. So a single particle all by itself whether it's an atom or an electron or whatever it is not interacting with anything else. There's no other particles nearby that it's interacting with. There's no gravity that feels, there's no planet earth for it to be attracted to, there's no other charged particles nearby for it to have electrostatic interactions with. Essentially this is a particle floating all by itself in vacuum with nothing else for it to interact with. So that's why it has a potential energy of zero. So the problem that we're solving is the free particle, it's a free particle because it's floating around free in vacuum or in space not interacting with anything else. So the free particle is another way of saying potential energy is equal to exactly zero. So what are the solutions to this differential equation? If we were solving this equation properly like you might do in a differential equations class we would proceed formally and find a characteristic equation and find general solutions and particular solutions that obey boundary conditions and so on. And we're not going to go through all that formal process so that may be good news to you or bad news to you depending on whether you didn't or did like differential equations. All we're going to do is propose some solutions since we know the form of the Schrodinger equation and see whether those solutions are or are not solutions to Schrodinger's equation. So to start with we're looking for, we know we're looking for wave functions whose second derivative is related to the original function. That's the eigenvalue problem we have to solve. Second derivative of the wave function gives us back the original function but notice the negative sign here. I need a function that when I take the second derivative of it and multiply by a negative number I get back the original function with some constants. So in particular what we're looking for is a function whose second derivative looks like negative the original function with maybe some other constants thrown in there as well. So we can think of functions whose second derivative give back the original function and one function like that is exponentials. So e to the kx, first derivative of e to the kx kicks out of k and we get k e to the kx second derivative. We'll kick out another k and I've got k squared e to the kx. So that looks half good. I've taken the second derivative of the function. I've got constants times the original function which is what I want but I wanted negative constants. The only way k squared is going to be a positive number unless perhaps k is imaginary. So if I back up and I say okay maybe if k is an imaginary number then a square is negative and that's what I want but that's a complication we don't need to deal with just yet. So if I wanted k to be a real number if I'm assuming k is a real number then this is not a good solution because k squared will be positive and I want it to be negative. So as another possibility a function whose second derivative is negative is trigonometric functions like sine or cosine. So sine of x if I take the in fact not just sine of x but sine of any constant times x sine of kx and in fact let me go ahead and keep things as general as possible stick another constant out front. So some constant a times the sine of k times x is a candidate that sounds good because derivatives of trig functions keep giving us more trig functions. First derivative of a sine kx is going to pull out a k so I'll get k times a times derivative of sine is cosine and then when I move on to the second derivative pulls out another factor of k so I've got k squared times a derivative cosine is negative sine and that's good news that's pulled out the negative sine that I'm looking for so the second derivative looks like minus k squared a sine of kx so remembering that a sine of kx is the original function that's good now I've got negative constants times the original function. So let's plug those back into the original Schrodinger equation. So if I take the wave function and plug it into the right side of Schrodinger's equation second derivative plug it in the left side what I've got is minus h squared over pi squared mass second derivative is minus k squared a sine x and I'm asking is that equal to e times the function a sine of kx so I can simplify the negative sines cancel h squared over 8 pi squared mass k squared I've got an a sine x on both sides that I can cancel and after I've done that the right side just looks like an e so and now stop and ask yourself that was it legit to do algebra I told you not to treat this like an algebraic equation and yet I just used algebra to cancel an a sine x on both sides a sine kx I've got a slight typo here this should be a sine kx but using algebra is not a problem it's the difficulty is we want to make sure that the stuff on the left can be always equal to the stuff on the right so in fact that can be true some constants times sine of kx can be equal to some constants times sine of kx and I'm just canceling the sine of kx that appears on both sides and these constants must be equal to this constant on the right so what we found here is that yes this function a sine kx does solve Schrodinger's equation a sine of kx is an eigenfunction of this one dimensional Hamiltonian with the potential energy zero and the eigen value e that pairs with that eigenfunction is this collection of constants h squared over 8 pi squared mass times k squared so what we say is that a wave function a sine of kx does sine of kx does solve Schrodinger's equation this version of Schrodinger's equation and it does so with an energy h squared over 8 pi squared m all times k squared so notice a few things about this expression number one this isn't just one solution this is a whole family of solutions k equals 1 k equals 2 k equals 3 and a half k equals pi k equals any value at all will solve Schrodinger's equation so I've got a whole family of solutions with any value of k that will solve Schrodinger's equation each one of those solutions with its own value of k is paired with a particular value of the energy that value of k changes the value of energy each wave function has its own eigen as its own energy each eigenfunction has its own eigenvalue the other thing we could point out is that this is not the only solution to this Schrodinger equation that's the downside of just having proposed some solutions and double-check them doesn't give us all of the solutions that are possible the way it would if we solved it formally as in a differential equations class for example you can take a look at these and convince yourself that they must also be solutions not just sine but also cosine would be a valid solution and if we want to get a little more exotic like we mentioned up here if the exponent was imaginary rather than real e to the i times some real constant times x is also going to solve Schrodinger's equation because the second derivative of this function is going to pull down an i twice which will get the negative sign that we're interested in so there's several different families of solutions that are somewhat related to one another that solve Schrodinger's equation again each one of them paired with its own individual energy the next thing that we're going to do is move on and extend this problem a little bit to make it a little more realistic so that we don't have to just consider a particle floating all by itself out in vacuum so that's what we'll do next