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Proof That the Two-Electron Ground State of a Spin-Independent Hamiltonian Is a Singlet(a) The mean energy of a two-electron system with Hamiltonian (32.3) in the s...

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Proof That the Two-Electron Ground State of a Spin-Independent Hamiltonian Is a Singlet(a) The mean energy of a two-electron system with Hamiltonian (32.3) in the state $psi$ can be written (after an integration by parts in the kinetic energy term) in the form:$$E=int d mathbf{r}_{1} d mathbf{r}_{2}left[frac{h^{2}}{2 m}left{left|mathbf{abla}_{1} psiight|^{2}+left|abla_{2} psiight|^{2}ight}+Vleft(mathbf{r}_{1}, mathbf{r}_{2}ight)|psi|^{2}ight]$$Show that the lowest value ( $32.28$ ) assumes over

Proof That the Two-Electron Ground State of a Spin-Independent Hamiltonian Is a Singlet (a) The mean energy of a two-electron system with Hamiltonian (32.3) in the state $psi$ can be written (after an integration by parts in the kinetic energy term) in the form: $$ E=int d mathbf{r}_{1} d mathbf{r}_{2}left[frac{h^{2}}{2 m}left{left|mathbf{ abla}_{1} psi ight|^{2}+left| abla_{2} psi ight|^{2} ight}+Vleft(mathbf{r}_{1}, mathbf{r}_{2} ight)|psi|^{2} ight] $$ Show that the lowest value ( $32.28$ ) assumes over all normalized antisymmetric differentiable wave functions $psi$ that vanish at infinity is the triplet ground-state energy $E_{t}$, and that when symmetric functions are used the lowest value is the singlet ground-state energy $E_{mathrm{s}}$ (b) Using (i) the result of (a), (ii) the fact that the triplet ground state $psi_{t}$, can be taken to be real when $V$ is real, and (iii) the fact that $left|psi_{2} ight|$ is symmetric, deduce that $E_{s} leqslant E_{T}$.



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$\textbf{Effective Magnetic Field.}$
An electron in a hydrogen atom is in the $2p$ state. In a simple model of the atom, assume that the electron circles the proton in an orbit with radius $r$ equal to the
Bohr-model radius for $n$ = 2. Assume that the speed $v$
of the orbiting electron can be calculated by setting $L$ = $mvr$ and taking $L$ to have the quantum-mechanical value for a $2p$ state. In the frame of the electron, the proton orbits with radius $r$ and speed $v$. Model the orbiting proton as a circular current loop, and calculate the magnetic field it produces at the location of the electron. also absorbed when the initial state is still the ground state. What is the value of $n^2$ for the final state in the transition for which this wavelength is absorbed, where $n^2$ = $n_x^2$ + $n_Y^2$ + $n_Z^2$ ? What is the degeneracy of this energy level (including the degeneracy due to electron spin)?

Question number 22. Old problem number 22 Do Do it, Spain Degeneracy we have m s is equal to plus minus Wanna work too? This means each electron. This means each state's can be occupied by two electrons. So each state each state each state two electrons. This is due to the spinning degeneracy. So in the ground state, we have a two electrons, two electrons Having an energy he won is equal to four times four times one constant square over in L. A square in Triple State, we have the energy e to that will be equal to six. This knowledge got to square eight m l square. This will contain six electrons while in ah, our energy equal to 33 This will contain the three electrons. So for the total off 11 we have a three year a six year and the two electrons here. So seven edge square or it Mm, Hill square. So the ground state energy, the ground state will be two electrons having energy one six electrons having energy too, and three electrons having energy. Three. If we plug all these values about mentioned here, if you plug these rallies here. So I plugged these values here. Then the ground state energy I get here here the 65 times our this factor h got square were head M l Square. Then the first excited state energy. I will get here first. Excited state he will be Two electrons will be in having energy one. Then the five silicones will be having energy to And then we have four electrons will be having energy tree. This will be the total energy off our first excited states. If we add up all these energies, then we end up with the number 66. It's square who are eight m l square.

Given the way functions, including the mixed eight wave function, we're going to start with the ground state, and we're gonna take the expectation value for the position operator and proceeding. We get the following integral, and we see that this is going to be an integral over and even or symmetric interval, meaning that it's symmetric about the vertical axis, the Y axis. In other words, it's an even interval or symmetric interval about the origin from minus infinity to infinity. But this is an integral of an odd function. So from that observation, we can already conclude that this integral should be zero. You could also look this up in an integral table or real quick. You can. You can actually do the integral Um, by making a quick change of variables a Z I'll show you. So there's a real quick way to do that. You can write. This is to in a girls so an integral from minus infinity to zero and an integral from zero to infinity, and you'll perform a change of integral, a change of variables on just the first integral. So we basically changed X two minus X. That's the change of variables. And when we do that, we get an integral from infinity to zero. We leave the second integral alone, and this is the former that we now have. So in doing that we come to this next form. We flipped the integral on the first integral itself and we have a minus sign. Now notice that the minus Sign on Negative X right here has cancelled with the minus sign in in de negative X and the minus Sign in negative X up here has gone away because it's negative X squared. So that leaves us with X dx in the next line. And we just have e to the negative a X squared. So we just have a minus sign right here simply because we flipped this integral right here. So now this integral will cancel with the integral that we left alone on the rights and we therefore have zero. So the expectation value for X in the ground state of zero. Next we continue and we calculate the expectation value for X and the first excited state. But again, we have the integral of an odd function over a symmetric interval, and we get a similar result, we should get zero, but we go ahead and just write out the same trick, the same procedure right here, and it works the same way, except that it's X cubed. Had this been an even function then, because you're squaring or cute or taking to the fourth power, etcetera. This method would not work, and you would not have any girls that cancel out. But if you have 135 etcetera for odd functions, they do cancel. And so you see once again that these entered rules will vanish because they add together, the minus sign causes them to cancel. So now we finally come to the expectation value for X in the mix state, and we get three terms when you when you substitute in the mix state in the expectation value formula, you get, ah, an expansion and their three terms that result in that expansion and one of them is across term, which gives us this to in the cross term, and the other two are square terms. The square terms will look familiar. We just calculated them and notice that we have the two there in the cross term because you had really you had four terms since it's ah wasn't expansion. So they should add together to give you the two. We just calculated the square terms. They vanished. So this leaves us with just what's left over two times the cross terms. But the two with 1/2 are going to cancel. And so now we just have that our expectation value equals the integral of X times, the to wave functions. And the only reason I have absolute value signs is in case we were dealing with complex valued functions. And I like to be general and just include what goes into a formula in its most general form. But we don't really need them because we're dealing with real valued functions. So when you substitute everything in for these particular wave functions, this is what you obtain and it looks like this. Now notice we finally haven't even function over ah symmetric interval, and you could evaluate this using particular methods. You could research galaxy and in a girls, for example, and see how they're evaluated. You can just look in a table of any girls, which is what a lot of people dio and because we don't have a lot of time. We're just going to state what this integral is equal to. So when you look it up in a table, for example, or if you do like to just see how it's done, you could look it up. You can use polar coordinates to evaluate this type of integral Um, this is what you get. And so there's some cancellation that will go on when you when you have all these new miracle or basically all these factors multiplying each other and you finally get one divided by the square root.

So first we start with the plot of the wave function and the probability density. And we note that the wave function and the probability density both approach zero as X approaches infinity or minus infinity. And both are also symmetric about the vertical axis, the Y axes that will be important in calculating in a girls so no also, the way function is continuous. Ah, even though the derivative is not the first, riveted may not be continuous, but way function is so Another property that we already noted is that side the wave function goes to zero as X goes to positive or minus infinity. Um, this means that the particle is in a bound state, so the particle is bound to X equals zero, which is the origin Also. This means that we have a finite area under the curve of the wave function, so the particle can or the way function can be normalized. So we go ahead normalized and we'll obtain a value for the normalization constant. So we start with the usual integral but no, using the symmetry of the way function, we if we look at this curve, for example, we can say that the area on one side of the vertical axis. Ah, if we double, that's the area on that one side. It is twice the area is equal to the area of the under the entire curve for the entire way function. So using that property, then this integral can be written as twice the integral under one side of the curve. And that's what we've done here so that we can just integrate from zero to infinity rather than integrating for a minus infinity to positive infinity. And that makes the job of integration a lot easier. The whole process of integration simplifies. So carrying out the integral, um, we'll get the following. There should not be a square on the lower case A there, so we'll just continue and we get that Well, yeah, here we go. So you carry out the integral on you evaluated the in 0.0, and X approaches positive infinity, and you finally get that capital a squared over little a must equal one. Which means that capital a equals the square root of little A. So that's the normalization constant. Now we can calculate a probability over an interval, and this probability calculation is pretty much like the calculation for normalization. And in that sense we go ahead and just calculate for 1 1/2 of the ah area under the curve of the entire way function. And then we multiply by two. So we use that same trick and we get following result for the probability this ends up being one minus either the minus one, which equals 0.632

When we are asked to show that it is normalized. So to do this we can calculate the probability density by taking the integral from zero to infinity of this way functions where times four pi r squared. So, of course, if you square this way function the square root is gonna go out of the denominator up here. And that pie is going to cancel with this pie here. So you're left with four r squared over a not cubed times e to the minus two. You get this minus two from the square are over a Not taking the derivative perspective d r This could be actually be a pretty ugly integral. You could go ahead and try to do integration by parts. And this kind of turns into account to problem more than a physics problem. So toe work efficiently. I just used a integration table. I just looked up in the integration of r squared E to the minus. Two are over. A Not I found a table that looked like that and found the answer that way again. You could do integration by parts, but that's more of a calculus problem than a physics problem. So why not let integration table or an integration system do that integral for you? That's what we get here because we have four over a not cube that gets pulled out. Since those are just constants and left inside the integral will box this in we have it gives us minus r squared times a not over to e to the minus two or over a night minus a not squared times are over too. Time to get to the minus two are ovary not ca minus a not cubed over four time t to the monies to our ovary. Not missing girl was taken from minus and or from, uh, zero to infinity. So the first thing we do is plug in infinity into these values. So when R is equal to infinity, we have negative infinity minus infinity, right? So we said equals four over a not cubed outfront. So if you plug in infinity for our you have negative infinity here and negative infinity here. But if you take e to the minus two infinity over a not this is equal to zero so e to the minus. Infinity is equal to zero. So maybe we should put that off to the side. So we know what we're doing here. E to the minus. Infinity is equal to zero, therefore from plugging infinity. And we have zero minus zero minus zero. Now we're gonna plug in zero for our. So if you plug in zero for are, of course, the first value is gonna be zero. You have zero squared. So anything times zero is zero. The second value also has a zero in it. When you replace our was zero here. So this is gonna be minus zero, and then the third value we have ain't out cubed over four, so there's no are there, So this doesn't get multiplied by a zero, and then it's e to the minus zero. Well, e to the zero is equal to one. So let's also put that up here so he can say he 20 equals one. So we're gonna be utilizing that information. So what this end up with is minus a not cute over four times. One box. All that remained close off that print to see there. Okay, So what we're left with carrying this out? This is equal to four over a not cubed well supplied by you have minus and negative. So this becomes a positive. This is a not Cuba over four. So carrying out this operation, we climb that this is equal to one. I can box that in as a solution for a because since it's equal to one, it's normalized. So I'm thinking maybe type off to the side rehs formalized for part B. It wants us to find in a numerical value, um, for the probability between a not over two and three a not over to, but we're integrating with respect to the same probability density. So we're gonna essentially get just by using simple logic. We're gonna get to this point here when you use a red arrow to draw it. We're gonna get to this point here, but our integration parameters are gonna be from zero to infinity. They're gonna be from a not over 2 to 3 a. Not over to. So, using that logic, we'll call this part B. We're gonna write that same equation here because it's the same integral. So we have, uh, the probability being equal to We have four over a not cubed multiplied by negative R squared times a not over to e to the minus two, but e to the minus two are over A Not if we go back up here and look you to the minus. Two are over a Not as in every single one. So I'm just gonna pull that out. So we have negative r squared over a not divided by two minus a squared times are over too minus in our to the third over four. And then all of these get multiplied by E to the minus to our over A Not in an hour. Integration parameters go from a not over to 23 a. Not over to. So now we complying in those integration parameters for our into these expressions, and we find that this is equal to we have for over a not cute outfront Still times. So first plugging in, uh, club three A. Not over two. So this gives us negatives. Line over eight CA minus three. Force minus 1/4 multiplied by. So we hav e to the minus. Two are over. A not were re plugging three a. Not over to for our. So this is e to the minus three We're gonna subtract from that. Playing in a knot over to this gives us negative 1/8 minus 1/4 minus 1/4. Oh, I'm actually sorry about that. I did make a mistake here. So this is just for outfront The A Knock it, uh, not Cube gets canceled there from plugging in our A not values and out over two values. We lose that a not cute terms. So we just have four written out front. Okay, then this gets multiplied by plugging e to the minus two or over. A not. And using r equals a not over to. This becomes e to the minus one. So all of these air just a numerical value. So we just simply have to carry out this new miracle operation, and we find that this probability is equal to 0.497 We can box set in as the solution to our question.


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