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!
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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