Quantum Engi Q and A_Part_21

Quantum Engi Q and A_Part_21 - 5.2. THE HYDROGEN MOLECULE...

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Unformatted text preview: 5.2. THE HYDROGEN MOLECULE 5.2.4.1 61 Solution hmold-a Question: Obviously, the visual difference between the various states is minor. It may even seem counter-intuitive that there is any difference at all: the states ψl ψr and ψr ψl are exactly the same physically, with one electron around each proton. So why would their combinations be any different? The quantum difference would be much more clear if you could see the full six-dimensional wave function, but visualizing six-dimensional space just does not work. However, if you restrict yourself to only looking on the z -axis through the nuclei, you get a drawable z1 , z2 plane describing near what axial combinations of positions you are most likely to find the two electrons. In other words: what would be the chances of finding electron 1 near some axial position z1 and electron 2 at the same time near some other axial position z2 ? Try to guess these probabilities in the z1 , z2 -plane as grey tones, (darker if more likely), and then compare with the answer. Answer: Here are the pictures, assuming the origin is halfway in between the protons: z2 z2 z1 z2 z1 z2 z1 z1 Figure 5.1: Wave functions on the z -axis through the nuclei. From left to right: ψl ψr , ψr ψl , the symmetric combination a(ψl ψr + ψr ψl ), and the antisymmetric one a(ψl ψr − ψr ψl ). These results can be explained as follows: For the state ψl ψr , electron 1 is around the left proton, so its likely z1 positions are clustered around the position zlp of that proton, indicated by a tick mark on the negative z1 -axis in figure 5.1. Similarly electron 2 is around the right proton, so its z2 positions are clustered around the positive value zrp indicated by the tick mark on the positive z2 axis. This means the wave function, Ψ(0, 0, z1 , 0, 0, z2 ), will look as shown in the left picture of figure 5.1. It will be mostly in the quadrant of negative z1 and positive z2 . Similarly ψr ψl will look as the second picture. Here the positions of electron 1 cluster around the positive position of the right proton and those of electron 2 around the negative position of the left proton. When you average the two states symmetrically, you get a two-blob picture like the third picture. Now it is electron 1 around the left proton and electron 2 around the right one or 62 CHAPTER 5. MULTIPLE-PARTICLE SYSTEMS vice-versa. But there is still almost no probability of finding both protons in the first quadrant, both near the right proton. Nor are you likely to find them in the third quadrant, both near the left proton. If you average the first two states antisymmetrically, you get the fourth picture. In the antisymmetric combination, the wave function is zero on the symmetry line between the blobs. You see that the states are really different when looked at in the full six-dimensional space. 5.2.4.2 Solution hmold-b Question: Based on the previous question, how would you think the probability density n(z ) would look on the axis through the nuclei, again ignoring the existence of positions beyond the axis? Answer: For any state, the probability of finding electron 1 near a given z , regardless of where electron 2 is, is found by setting z1 equal to z and integrating over all possible positions z2 for electron 2. For the two-dimensional state ψl ψr shown in the left column of figure 5.2, you are then integrating over vertical lines in the top picture; imagine moving all blank ink vertically towards the z1 axis and then setting z1 = z . The resulting curve is shown immediately below. As expected, electron 1 is in this state most likely to be found somewhere around zlp , the negative position of the left proton. Regardless of where electron 2 is. (Note that all possible positions of electron 2 should really be found by integrating over all possible positions in three dimensions, not just axial ones. The final row in the figure gives the total probabilities when corrected for that. But the idea is the same, just harder to visualize.) The probability density at a given value of z also needs to include the possibility of finding electron 2 there. That probability is found by setting z2 = z and then integrating over all possible values of z1 . You are now moving the blank ink horizontally towards the z2 axis, and then setting z2 = z . The resulting curve is shown in the second graph in the left column of figure 5.2. As expected, electron 2 is most likely to be found somewhere around zrp , the positive position of the right proton. To get the probability density, the chance of finding either proton near z , you need to add the two curves together. That is done in the third graph in the left column of figure 5.2. An electron is likely to be somewhere around each proton. This graph looks exactly like the correct three-dimensional curve shown in the bottom graph, but that is really just a coincidence. The states ψr ψl and a(ψl ψr ± ψr ψl ) can be integrated similarly; they are shown in the subsequent columns in figure 5.2. Note how the line of zero wave function in the antisymmetric case disappears during the integrations. Also note that really, the probability density functions 5.2. THE HYDROGEN MOLECULE z2 63 z2 z1 n1 z2 z1 n1 z2 z1 n1 z1 n1 n2 z n2 z n2 z n2 z n z n z n z n z n z n z n z n z z z z z Figure 5.2: Probability density functions on the z -axis through the nuclei. From left to right: ψl ψr , ψr ψl , the symmetric combination a(ψl ψr + ψr ψl ), and the antisymmetric one a(ψl ψr − ψr ψl ). From top to bottom, the top row of curves show the probability of finding electron 1 near z regardless where electron 2 is. The second row shows the probability of finding electron 2 near z regardless where electron 1 is. The third row shows the total probability of finding either electron near z , the sum of the previous two rows. The fourth row shows the same as the third, but assuming the true three-dimensional world rather than just the line through the nuclei. ...
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This note was uploaded on 11/13/2011 for the course PHY 4458 taught by Professor Garvin during the Fall '11 term at University of Florida.

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