 Okay, so let's consider what to do in a case where we can write down the Schrodinger equation for a problem, but we can't solve it necessarily, like the helium atom is a Schrodinger equation that we can write down, but we can't solve on paper or any larger atom or any more complicated system. So even though we can't solve the Schrodinger equation, that doesn't mean that solutions don't exist. Solutions certainly do exist. Any helium atom have a particular energy. They occupy particular wave functions, even if we can't solve the Schrodinger equation and figure out what the wave functions are that they obey. So it's certainly still true that the Schrodinger equation is true and not just for a single solution, but there's ground state, excited states for electrons and helium atoms. So any particular wave function that they occupy, if I act with that wave function with the Hamiltonian operator, I'll get the energy of that particular wave function multiplied by the wave function again. For the specific case, let's say of the ground state, if I put a helium atom in the ground state, an electron in the ground state of a helium atom, we could perhaps experimentally measure that energy, even if we can't solve Schrodinger's equation to find it out. The Schrodinger's equation is still valid. One thing we can do with this equation, let's pre-multiply on both sides of this equation. So I'm going to multiply on the left side of the equation by psi-naught complex conjugate. So I'm pre-multiplying by psi-naught star on both sides of the equation. And notice that I said pre-multiply, it's important to distinguish between whether I'm multiplying on the left or multiplying on the right, because remember this Hamiltonian includes some derivatives. This is a function of r's and theta's and phi's. H includes derivatives of r's and theta's and phi's. If I put it after the Hamiltonian that it would, those derivatives would act on the wave function by saying pre-multiply, I'm putting it out here where I don't want to take the derivatives of those r's and theta's and phi's. So if I've written this expression down, the reason I've written it down in this way is because if I then integrate this expression on both sides, then it turns out we can do something with the right-hand side. It doesn't matter whether I'm pre-multiplying or post-multiplying on this side, because E, the energy, that's just a number. That's not an operator. So that's just a particular constant. I can pull that energy out of the integral. So this is E-naught times the integral of a wave function complex conjugate times itself. So this looks familiar. We know if the wave function itself is normalized, if we have normalized wave functions, then this term is going to be equal to one. In general, we may have functions, either wave functions or non-wave functions, that are not normalized. So I'm going to leave this written this way. If we happen to know that our wave functions are normalized, then this simplifies and becomes one. But if what we're interested in is knowing the energy, even if we can't solve the wave function, I can rearrange this expression to say that the energy E-naught is equal to the integral on the left-hand side. Integral of psi-naught star times h acting on psi-naught, all divided by integral of psi-naught star times psi-naught. So, so far, all I've done, I need to know what psi-naught is in order to evaluate the energy. So I haven't really gained anything. What I've gained is a more complicated way of learning what the ground state energy is. The easy way is take the wave function, act with it on the Hamiltonian. And what I find is I get back the same wave function times an energy that number is called the ground state energy. Here, I have another recipe for finding E-naught. It's just a little bit more difficult. I have to still act with the Hamiltonian on the wave function, then multiply by the wave function, then take an integral. And when I'm done, either divide by one or divide by a different integral. So if I know psi-naught, I can do this procedure. It hasn't saved me any work. But it is an alternate way of finding the energy. The same thing would be true if I replace all these zeros for the ground state with a different index for a different wave function. I could find the energy of any particular wave function I want via this difficult technique. There's not yet any reason why I should do that if I know the wave functions. But what it does suggest is that even if I don't know the wave functions, remember we're hypothetically considering a case where we can't solve Schrodinger's equation and we don't know the wave functions. But let's say we can make a guess, and even though we can't solve the Schrodinger equation, we can write down a function that might or might not be the correct wave function for the system. What this suggests is I can calculate something that looks like an energy of any function in this variable phi. So I'm not writing psi because I don't want to imply that it's the actual wave function. The psi is something we'll call a trial function. If I compute the energy like this, if my function happens to be the correct wave function, and what I compute is my trial function complex conjugated times h acting on the trial function, all that integrated over phi star times phi. So I've just defined this new function. Any function I want, I can obtain this expression e of that function by doing the integral in the numerator divided by the integral in the denominator. This whole thing is something we're going to call the variational energy. And the reason we call it variational will become a little more clear as we talk more about it. But the variational energy is just an expression for finding a quantity that we're naming the energy for any function that we want. Notice even if we don't know whether it's a wave function or not, I can take derivatives of a function, I can multiply by itself, I can integrate it, I can divide by another integral. The function doesn't have to be normalized because I'm doing this integral in the denominator. If I perform this somewhat complicated procedure what I get is some quantity that I can call the energy and I'm calling it an energy because if I get lucky and plug in the actual true wave function what I've discovered is the energy of that wave function. So this variational energy is going to be useful for us in trying to figure out the energy of problems where we don't know the exact wave function.