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  #1  
Old 01-20-2006, 04:14 AM
thefisherman thefisherman is offline
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Default Quantum Problem -- Different Electrons

I'm new to this forum, but I just took a QM final today and this was on it. Curious if any science people out there can wrap their heads around this and give the right answer.

Electrons are spin 1/2 particles. If electrons were instead spin 1 particles, how much wider would the rows on the periodic table be? [Hint: The period table is separated so that, for the chemists out there, each shell makes a row, or for physicists, each row has different angular momentum numbers 'l' and 'm'.]

Just wondering if anyone knows the answer...
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  #2  
Old 01-20-2006, 04:41 AM
bobman0330 bobman0330 is offline
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Default Re: Quantum Problem -- Different Electrons

Haven't taken a chem class in a few years, but:

Pauli exclusion principle says there can only be as many electrons as there are distinct quantum numbers. N, L, and ml are unaffected, I think. However, if the particle spin is increased, maybe more values of ms are permissible? Possibly 4 (-1, -1/2, 1/2, and 1). In that case, each shell could take twice as many electrons, and the periodic table would be twice as wide.
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  #3  
Old 01-20-2006, 06:05 AM
Charon Charon is offline
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Default Re: Quantum Problem -- Different Electrons

Im too lazy to figure it out, but couldnt you use the eigenvalues of the spinmatrices, i.e. in real life it's

S_z[s m>=h_bar*m*[s m>

where S_z is spin in z-direction, m the magnetic quantum number and [s m> the vector representation. Now normally:

s=0, 1/2, 1, 3/2 ...

and m=-s, -s+1, ..., s-1, s.

So if you now only allow integer values of s, m will differ as well, and since

m=-l, -l+1,...,l-1, l

l will differ also. So there you have a different set of quantum numbers. This is just my first guess...could be totally wrong.
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  #4  
Old 01-20-2006, 11:58 AM
kevyk kevyk is offline
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Default Re: Quantum Problem -- Different Electrons

The Pauli exclusion doesn't apply to integer-spin particles.

Link

I'm sure I'm missing something, but if you can cram all the "electrons" into the lowest-energy quantum state, wouldn't the periodic table just be one long column?
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  #5  
Old 01-20-2006, 08:28 PM
thefisherman thefisherman is offline
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Default Re: Quantum Problem -- Different Electrons

That's the winner.
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  #6  
Old 01-21-2006, 10:18 AM
Charon Charon is offline
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Default Re: Quantum Problem -- Different Electrons

Nice!

I vaguely remember that you need quantum field theory to derive Pauli's exclusion principle, right? Can anybody give some insight in this or post a link or something?

Thanks
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  #7  
Old 01-21-2006, 01:19 PM
Metric Metric is offline
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Default Re: Quantum Problem -- Different Electrons

[ QUOTE ]
Nice!

I vaguely remember that you need quantum field theory to derive Pauli's exclusion principle, right? Can anybody give some insight in this or post a link or something?

[/ QUOTE ]
The actual exclusion principle is fairly easy to see (I'll show it below), but the spin-statistics theorem from which it derives is very non-trivial -- it's actually rather peculiar in this way: the rule is very easy to remember, but the theorem itsself requires full-tilt QFT to derive, and although I'm a theorist myself I'm not even close to reproducing it on demand (I haven't looked at this particular stuff for some time).

But as for the exclusion principle, recall that in quantum mechanics, the general rule for composing two systems is to take the tensor product of their states. I.e. if one system is described by state |a>, and another system is described by state |b>, their combined state is given by: |a>x|b>

If I want to do something to system 1, I apply operators to the first of the two states (|a&gt, if I want to do something to system 2, I apply operators to the second of the two states (|b&gt.

This seems like a perfectly satisfactory state of affairs, until the two systems happen to be identical particles -- then the following ambiguity arises: If I measure one of the two particles, I can't in general tell which one I interacted with, since they are identical particles! E.g. suppose I start out with one electron in state |a> and another in state |b>, and wait around a while for their states to interact and evolve, and then I perform a measurement on the combined system and observe that "an electron" is now in the state |c>. Does that mean that I've interacted with "system 1" or with "system 2"? This ambiguity is dealt with if instead of representing the initial state by |a>x|b>, we represent it by a superposition of |a>x|b> and |b>x|a> -- effectively making the mathematical statement that there is no distinction between "system 1" and "system 2" -- all we know is that we have one particle in state |a> and one in state |b>. Now, the way in which we take the superposition is crucial. It turns out that integer spin particles are represented as (|a>x|b> + |b>x|a&gt (symmetric), while half-odd integer particles are represented as (|a>x|b> - |b>x|a&gt (antisymmetric) -- this is what the spin-statistics theorem tells us.

Now, since electrons have spin 1/2, they get the "-" sign -- multi-electron states are antisymmetric. Note, then, that if I want to put two electrons in the same state |a>, their full state would be described by (|a>x|a> - |a>x|a&gt, otherwise known as zero. Using the probability interpretation of the state function, we could say "the probability to find two electrons in the same state is zero." This is a statement of the Pauli exculsion principle.
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  #8  
Old 01-21-2006, 01:35 PM
Borodog Borodog is offline
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Default Re: Quantum Problem -- Different Electrons

Very nice.
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  #9  
Old 01-21-2006, 03:27 PM
Charon Charon is offline
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Default Re: Quantum Problem -- Different Electrons

Thank you, very good explanation.

I guess I have to study QFT if I want a thorough derivation, but this gives a nice intuitive explanation. Interesting subject!
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