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Feynman's hydrogen molecule

  1. Jun 26, 2014 #1
    In his lectures on quantum mechanics, Feynman treats the hydrogen molecule as a two-state system to give a general understanding of the covalent bond. He starts with base state |1> as electron a in ground state of left proton and electron b in ground state of right proton and base state |2> as electron b in ground state of left proton and electron a in ground state of right proton. Then, by supposing that there is an amplitude -A that the electrons switch places (which they cannot do classically), he arrives at two stationary states (let's ignore the time dependence): state |I> = 0,707|1> - 0,707|2> corresponding to an energy E0 + A and state |II> = 0,707|1> + 0,707|2> corresponding to an energy E0 - A. Since A increases as the atoms get closer and the protons are repelled when the distance is too small, there is a minimum in energy for state |II>, so it corresponds to the bonding state.

    Then, Feynman says that if the electrons have identical spins, state|II> is not allowed. He says that because electrons are fermions, state |II> must become -|II> when electrons a and b are exchanged, which it does not. I have a hard time justifying this assertion. Here's the justification I would give:

    Suppose the system is in state |ψ> and that we want the amplitude for it to be in state |II>. Let's say the spins of the electrons are the same. Since we have to subtract the amplitude corresponding to the exchange of indistinguishable electrons a and b, we have

    <ψ|II> = (0,707<ψ|1> + 0,707<ψ|2>) - (0,707<ψ|2> + 0,707<ψ|1>) = 0 = <II|ψ>

    Since the amplitude is 0 and state |ψ> is general, state |II> is impossible and so is bonding with parallel spins.

    Now, if the spins are opposite, we can distinguish electrons a and b and we have no amplitude to substract, so the total amplitude is nonzero and state |II> is possible. By using the same derivation, I get that the amplitude for state |ψ> to be in state |I> is the same whenever the spins are parallel or antiparallel. Am I right?

    I have to say I am a newbie for quantum mechanics and I am not too familiar with approaches different from Feynman's. Thank you!
     
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  3. Jun 26, 2014 #2

    PeterDonis

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    Which assertion? The assertion that state |II> does not become - |II> when electrons a and b are interchanged? That's obvious: just interchange them. You get the same state |II> back again, since the signs of both terms are positive. State |I>, OTOH, goes to - |I> when the electrons are interchanged, since one term has a + sign and the other has a - sign.

    Or are you having a hard time justifying the assertion that the state (whatever it is) must flip its sign when the electrons are interchanged? That's just the Pauli exclusion principle. You simply state the same thing a different way when you say:

    For bosons, you would add the amplitude instead of subtract.
     
  4. Jun 26, 2014 #3
    The assertion that |II> not becoming -|II> when a and b are interchanged means that state |II> is impossible if the spins are parallel.
     
  5. Jun 27, 2014 #4

    PeterDonis

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    [Edited to correct references to OP.]

    Ok. Do you have a specific reference for which lecture (I assume you are looking at Feynman's Lectures on Physics) you are talking about? I see a couple of issues in your OP, but it's possible that they're issues in translation, so to speak, between Feynman's lecture and your post.

    First, the difference between the electron spins being parallel and antiparallel is not that the electrons are distinguishable in the latter case but not the former. Electrons are never distinguishable, in the sense that they don't have little labels on them that let us keep track of which electron is which. So, for example, if we have a hydrogen molecule with two electrons, we can't actually describe its state as, say, "electron a is with atom #1 and electron b is with atom #2". We can only describe it as "there is an electron with atom #1 and an electron with atom #2". That is true whether the spins are parallel or antiparallel.

    Second, the fact that the state has to switch sign when the electrons are exchanged is, as I noted before, just the Pauli exclusion principle; and it is also true of any two-fermion state, i.e., it's true whether the spins are parallel or antiparallel. But note that this applies to the *entire* state, i.e., the state with both position (which atom each electron is with) and spin taken into account. So, for example, if the spins are parallel, the spin part of the state does not switch sign under electron interchange, so the *position* part must; but if the spins are antiparallel, the spin part of the state switches sign under electron interchange, so the position part must *not*.

    It looks to me like the states you are calling |I> and |II> in the OP are states in which the position part is antisymmetric and symmetric, respectively. So for the entire state to be antisymmetric (i.e., to switch sign on electron interchange), the spins of the electrons would have to be parallel and antiparallel, respectively; i.e., if we use |P> and |A> to designate the parallel and antiparallel spin states, the allowed two-electron states (including both position and spin parts) would be |I>|P> and |II>|A>; |I>|A> and |II>|P> would not be allowed by the Pauli exclusion principle (because they would not change sign on electron interchange). Of the two allowed states, |II>|A> is the one that has the lower energy, so it will be the hydrogen molecule ground state.
     
    Last edited: Jun 27, 2014
  6. Jun 28, 2014 #5
    I was refering to section 10-3 of Volume III of the Feynman lectures.

    What led me to think that electrons were distinguishable if they had opposite spins is section 3-4 in which Feynman talks about scattering of electrons on electrons. There he says that if the incident electron has spin up and the target electron has spin down, we could tell by measuring the spin after scattering which electron is in which detector (he assumes no spin change due to interaction), so there would be no interference, just addition of the two probabilities.

    Your explanations helped a lot, thanks!
     
    Last edited: Jun 28, 2014
  7. Jun 28, 2014 #6
    It is easy to show algebraically that the space part |I> = 0,707|1> - 0,707|2> of the state does switch sign under electron interchange: |I'> = 0,707|2> - 0,707|1> = -|I>.

    Is it possible to show in similar way that the spin part does switch sign under electron interchange if the spins are antiparallel?
     
  8. Jun 28, 2014 #7

    PeterDonis

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    Sure, it's basically the same thing. The parallel and antiparallel spin states |P> and |A> that I referred to look like this (leaving out normalization factors, and using the subscripts ##a## and ##b## to refer to the spin states of electrons ##a## and ##b##):

    $$
    \vert P \rangle = \vert \uparrow_a \rangle \vert \uparrow_b \rangle + \vert \downarrow_a \rangle \vert \downarrow_b \rangle
    $$

    $$
    \vert A \rangle = \vert \uparrow_a \rangle \vert \downarrow_b \rangle - \vert \downarrow_a \rangle \vert \uparrow_b \rangle
    $$

    Exchanging the electrons just means swapping the ##a## and ##b## subscripts; obviously this leaves |P> unchanged but takes |A> to -|A>.
     
  9. Jun 28, 2014 #8

    PeterDonis

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    Exactly: he's ruling out the possibility of the spins changing, so of course you can use the spins to distinguish the electrons. :wink: But this is only an approximation, which works OK for the scattering case, but for the hydrogen molecule it doesn't work. If you rule out the spins of the electrons changing, you can't even write down an antiparallel spin state for the hydrogen molecule, like the |A> I wrote down, at all; if you look at the |A> state, you will see that it requires both electrons to have an amplitude for having both spins (up and down).
     
  10. Jun 29, 2014 #9
    It is getting clearer and clearer. Now, do the following states exist, and if they don't, why not?

    $$
    \vert P' \rangle = \vert \uparrow_a \rangle \vert \uparrow_b \rangle - \vert \downarrow_a \rangle \vert \downarrow_b \rangle
    $$

    $$
    \vert A' \rangle = \vert \uparrow_a \rangle \vert \downarrow_b \rangle + \vert \downarrow_a \rangle \vert \uparrow_b \rangle
    $$
     
  11. Jun 29, 2014 #10

    PeterDonis

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    Yes, they exist; more precisely, they exist with the appropriate normalization factors included, which I've been leaving out--I'll include them in this post. In fact, |P'>, since it also is unchanged by electron exchange, is actually another possible parallel spin state that could be used in the simple model Feynman is using. The most general such parallel spin state would be

    $$
    \vert P_{general} \rangle = C_1 \vert \uparrow_a \rangle \vert \uparrow_b \rangle + C_2 \vert \downarrow_a \rangle \vert \downarrow_b \rangle
    $$

    where ##C_1## and ##C_2## are any pair of complex constants whose squared moduli add up to 1. Any such state is unchanged by electron exchange.

    The antiparallel states are different; the one I wrote down, |A>, is the *only* antiparallel spin state that goes to minus itself under electron exchange, so it's the only one that can be combined with state |II> to form an allowed state for the hydrogen molecule in Feynman's model. Any other linear combination of ##\vert \uparrow_a \rangle \vert \downarrow_b \rangle## and ##\vert \downarrow_a \rangle \vert \uparrow_b \rangle## will not go to minus itself under electron exchange.

    Most of those other states will have something more complicated happen to them under electron exchange, so they don't fit into the model Feynman is using (which requires spin states that are either symmetric or antisymmetric). But the state you wrote down, |A'> (more precisely, |A'> with appropriate normalization factors included), is actually *unchanged* by electron exchange, since the coefficients of both terms are equal and electron exchange takes each term into the other. So this state could actually fit into Feynman's model combined with state |I>! In other words, the most general state in Feynman's model in which the position part is antisymmetric is ##\vert I \rangle \left( K_1 \vert P_{general} \rangle + K_2 \vert A' \rangle \right)##, where ##K_1## and ##K_2## are again any pair of complex constants whose squared moduli add up to 1. (Note that this means I told a little white lie in an earlier post :redface:, when I said the electron spins had to be parallel if the position part of the state was antisymmetric; the precisely correct statement is that the spin part of the state has to be symmetric, which, as we've just seen, allows one particular antiparallel spin state as well.) But, as Feynman shows, any such state will have higher energy than the states with a symmetric position part, so they aren't possible ground states for the hydrogen molecule.

    [Edit: It's also worth noting that the antiparallel spin state has a different total spin than the parallel one, so measuring the total spin of the hydrogen molecule can distinguish them. I assumed that the total spin of the molecule was not known in the above; but the total spin also affects the total energy, so a more precise analysis would treat ##\vert I \rangle \vert P_{general} \rangle## and ##\vert I \rangle \vert A' \rangle## as separate possibilities with different energies, rather than treating them together as I did above. Both will still have higher energy than ##\vert II \rangle \vert A \rangle##, though.]
     
  12. Jun 30, 2014 #11
    I can't thank you enough for your help. But there are still a few things I don't get, probably because Feynman's model is a peculiar one.

    Why is $$
    \vert A \rangle = 0,707 \vert \uparrow_a \rangle \vert \downarrow_b \rangle - 0,707 \vert \downarrow_a \rangle \vert \uparrow_b \rangle
    $$ the only spin state that can be combined with |II> ? I mean, why does it absolutely have to go to minus itself under electron exchange? I thought it was only important that it wasn't unchanged by electron exchange, so that the amplitude that a random state would be found in this spin state would not be zero.

    For example, why couldn't $$ \vert A \rangle = 0,8 \vert \uparrow_a \rangle \vert \downarrow_b \rangle - 0,6 \vert \downarrow_a \rangle \vert \uparrow_b \rangle $$ combine with |II>?
     
    Last edited: Jun 30, 2014
  13. Jun 30, 2014 #12

    PeterDonis

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    Because the state |II> is unchanged by electron exchange, and the total state (including both the position and spin parts) must go to minus itself under electron exchange (because that has to be true for *any* fermion state, by the Pauli exclusion principle).

    In other words, Feynman is using a simplified model where he only considers position and spin parts of the state that are either symmetric (unchanged) or antisymmetric (change sign only) under electron exchange. Since the total state must be antisymmetric under electron exchange by the Pauli exclusion principle, that means one part (position or spin) must be symmetric and one part (spin or position) must be antisymmetric. There are no other options in this simplified model.

    We could, of course, try a much more complicated model that looked at all kinds of other possibilities for building a total state that was antisymmetric under electron exchange, without assuming that either the position or the spin parts by themselves had to be symmetric or antisymmetric. But that would be a lot more effort, and it would end up with the same answer.

    No, that by itself is not enough to satisfy the Pauli exclusion principle. See above.
     
  14. Jun 30, 2014 #13
    OK, so if Pauli exclusion principle is that the total state (including both the position and spin parts) must go to minus itself under electron exchange, it all makes sense. But the only thing that Feynman says that resembles this in Vol. III of his Lectures is the second sentence of the following quote from section 10-3:

    "...if there are two ways something can happen by exchanging the electrons, the two amplitudes will interfere with a negative sign. This means that if we switch which electron is which, the sign of the amplitude must reverse." He then adds that this is only true if both electrons have the same spin, since he is talking about the position part of the state.

    I just don't see how the first sentence implies the second one!

    Also, I don't understand why Feynman says a couple paragraphs later that the electrons a and b are in the same position state because we get back exactly |II> when we exchange them.
     
    Last edited: Jun 30, 2014
  15. Jun 30, 2014 #14

    PeterDonis

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    When he says "these arguments are correct if both electrons have the same spin", he's not talking about his statement about the amplitudes and how they interfere (the one you quoted). He's talking about the argument he just made in the previous paragraph, that says that the position state |II> is not allowed. He's saying that argument is only valid if the electron spins are parallel.

    In other words, he's saying that if the electron spins are parallel, the spin part of the state is symmetric, so the position part must be antisymmetric; but state |II> is symmetric. That's why, if the spins are parallel, the only allowed position state is |I>, as he goes on to say in the next sentence. But, as he then points out, the position state |I> does not lead to a bound state; it has no minimum of energy. So the position part of the state *has* to be |II> if the state is going to be a bound state, since |II> does have a minimum of energy. And *that* then requires that the spin part of the state is antisymmetric.

    I think Feynman was leaving out a piece of the logic, at least in this particular section. (He goes into it a little more in section 4.7, which is referenced in this section.)

    It's easier to think about this if we take a simpler example; the mixing of position and spin parts of the state in the hydrogen molecule makes things more complicated. Suppose we consider a helium atom in its ground state, with two electrons in the 1s orbital. This means the electrons are both at the same "position", so there is no position part of the state to consider; the only nontrivial part of the two-electron state is the spin part.

    So how do we write down this two-electron spin state? If we call the two electrons ##a## and ##b##, then this state looks exactly like the state |A> that we wrote down. Why? Let's look at the two sentences you quoted to see.

    First sentence: "if there are two ways something can happen by exchanging the electrons, the two amplitudes will interfere with a negative sign". Here the "two ways" are just the two terms in the state |A>: we can have either electron a up and b down, or b up and a down. And these two ways are indeed combined with a negative sign in state |A>.

    Second sentence: "This means that if we switch which electron is which, the sign of the amplitude must reverse". Now obviously this is the case for state |A>, as we've already seen. And the reason it's the case, as we've seen, is that the coefficients in front of both terms in state |A> have the same magnitude, but opposite sign. The opposite sign part is what the first sentence asserts; so the missing piece in the logic, so to speak, is, what forces the two terms in |A> to have the same magnitude?

    Remember what the Pauli exclusion principle says: it says that no two fermions can ever be in exactly the same state. So in the case of the helium atom, the two electrons can never be in the same spin state (since every other aspect of their state *is* the same). In other words, if one electron is up, the other *must* be down; there must be *zero* amplitude for both electrons to be up, or for both to be down.

    Now consider a state of similar form to |A>, but where we allow the two coefficients to have different magnitudes:

    $$
    |C> = C_1 \vert \uparrow_a \rangle \vert \downarrow_b \rangle - C_2 \vert \downarrow_a \rangle \vert \uparrow_b \rangle
    $$

    where for simplicity we assume ##C_1## and ##C_2## are both real and positive (we could run a similar argument with arbitrary ##C_1## and ##C_2##, it would just look more complicated while still giving the same answer), and ##C_1^2 + C_2^2 = 1##. So if we measure the sum of the two spins (i.e., we measure both spins and then add the results), we get the result:

    $$
    R = C_1 \langle \uparrow_a \vert \uparrow_a \rangle \langle \downarrow_b \vert \downarrow_b \rangle - C_2 \langle \downarrow_a \vert \downarrow_a \rangle \langle \uparrow_b \vert \uparrow_b \rangle = C_1 - C_2
    $$

    We must have ##R = 0## in order for there to be zero amplitude for the two spins to be the same. But this requires ##C_1 = C_2##, i.e., it requires the magnitudes of the coefficients to be the same.
     
  16. Jun 30, 2014 #15

    PeterDonis

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    I don't see the words "are in the same position state"; can you give an exact quote?
     
  17. Jul 1, 2014 #16
    Still in section 10-3:

    "On the other hand, our state |II> is perfectly symmetric for the two electrons. In fact, if we interchange which electron we call a and which we call b we get back exactly the same state. We saw in Section 4-7 that if two Fermi particles are in the same state, they must have opposite spins."

    He just says "in the same state" in last sentence, but he could have said "in the same position state", right?
     
  18. Jul 1, 2014 #17

    PeterDonis

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    Ah, I see. What he's saying is that if the position part of the state is |II>, then it's symmetric under electron exchange; but he's saying it in a way that emphasizes the indistinguishability of the electrons. Bear in mind that, as I said before, electrons (like all quantum particles) aren't really distinguishable; labeling the electrons "a" and "b" is a mathematical convenience so we can write expressions for states and doesn't actually represent anything physical.

    So the position state |II> is not really best thought of as a superposition of "electron a at atom 1 and electron b at atom 2" and "electron b at atom 1 and electron a at atom 2". It's better to think of it as something like an orbital of a single atom, like the 1s orbital of the helium atom that I talked about in an earlier post, which is occupied by both electrons; so both electrons have "the same position state" in much the same way as the two electrons in the helium atom's ground state. That's why Feynman is using the language he's using.

    Of course the hydrogen molecule is not exactly the same as a helium atom in terms of electron configurations; for a single atom, there is no analogue of the antisymmetric position state |I> that Feynman describes for the hydrogen molecule. But since the position state |I> can't lead to a bound state anyway, as we saw before, its existence doesn't really affect the question of what the ground state of a bound hydrogen molecule is.
     
  19. Jul 3, 2014 #18
    Is this the result of "amplitude to measure up-up + amplitude to measure up-down + amplitude to measure down-up + amplitude to measure down-down" (which should give 1)? In that case I don't see what the relation is between this being zero and the amplitude for parallel spins being zero.

    I just thought that the fact that state |C> didn't have any up-up or down-down term implied zero amplitude for same spin while C1 and C2 just had to have equal magnitude by symmetry.
     
    Last edited: Jul 3, 2014
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