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Unformatted text preview: 5.2. THE HYDROGEN MOLECULE
22.214.171.124 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
You see that the states are really diﬀerent when looked at in the full six-dimensional space. 126.96.36.199 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
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.
- Fall '11