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Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Eternal Student on 13/04/2023 16:33:42

Title: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Eternal Student on 13/04/2023 16:33:42
Hi.

     The scientists Stern and Gerlach developed some equipment that passed Silver atoms (Ag) through a region with a magnetic field.   The important point is the magnetic field, B, has a gradient (let's say in the z-direction).   So a beam of silver atoms moving through it will be deflected according to any magnetic moment it has.  Specifically, μz the magnetic component of the z-direction will be important (if its ambiguous later in this post, we are considering z components only).

    This might be familiar to people who have studied some QM.   Here's a Wiki article:  https://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment      although it's not the best reference.   (I'll provide more if required - but your own favourite textbook will probably cover it).

    Anyway, the usual result is to obtain just two spots, corresponding to an up or down deflection, on the far side of the apparatus.   From which the usual conclusion is that the electron in the Ag atom had an intrinsic property called spin.   The z-component of this spin can be in only two states (which we'll call "spin up" and "spin down"   or   +  and  -     or  quantitatively the spin z component takes the values +ħ/2   or  -ħ/2 only).    Obviously the Ag atom didn't have just one electron but it's the outermost lone electron that is usually in the 5s orbital which is important.   All the other electrons are assumed to form a reasonably complete cloud with enough symmetry that they contribute a net 0 magnetic moment.   Just to explain why they bothered with Ag atoms instead of using just a simple electron beam - they wanted an overall Neutral charged particle.  A charged beam would have shown deflection just due to the charge moving across a magnetic field which would have dominated over the subtle difference due to spin.

    Now, just to check I've understood this experiment.   If, for some reason, the outer electron was not in the 5s orbital (corresponding to quantum numbers n=5, l=0) but instead it was in the 5p orbital  (quantum numbers n=5,  l=1),  would we get a different set of spots?    Specifically with l=1  there are now  2l+1 = 3  different magnetic quantum numbers m = -1, 0, +1 that it can have.    These should contribute a Z component to the magnetic dipole.  So overall, there would now seem to be a larger set of magnetic dipole moments the Ag atom can have.

  The angular momentum Lz component due to the magnetic quantum number m should contribute a dipole moment of mμB.   Where μB  =  eħ/2me = a constant called the Bohr magneton.    Meanwhile the spin contributes a magnetic dipole moment of approximately  ±μB   (It's approximate because I've taken the Lande factor g ≈ 2   in the relationship between the value of spin and the magnetic dipole moment it creates - details shouldn't be too important).

     Overall we have a magnetic dipole moment in the z-direction given by:
[Contribution from magnetic quantum number, m]   +  [contribution from elctron spin]  =   mμB  ±  μB     with  m = -1, 0, +1.
    This gives a set of four values:     -1½ ,    - ½ ,    +½,  +1½     (in units of μB ).   

So, I would have thought, we get 4 different spots on the far side of the apparatus.    That's the first bit I wouldn't mind getting checked.   Is that right so far?

    Now since the Ag atoms were made into a beam by being heated in a furnace,  I would have thought there was plenty of energy around and quite a few of the outer electrons in the Ag atoms could have been in the more energetic 5p orbital rather than the 5s orbital.   So what I'm puzzled with is why did the experiment actually work so well?    There should be multiple spots on the far side of the apparatus and not just two.   It gets worse the more energetic the outer electron might have been, we could have had something that did look like a continuous smear of atoms running the range from maximum up and down deflections instead of just two clear spots where the atoms pile up on the screen / detector.   To say it another way - we could have obtained the classically predicted result and not the quantised splitting into two beams for which the experiment is famous.

    I don't know.   Just seems that there must have been some reason why the outer electron of the Ag atoms did tend to stay in the 5s orbital.   If anyone can remember their QM or possibly has an idea why a beam of Ag atoms produced from a furnace would tend to keep the outer electron in the 5s orbital, then I'd appreciate some discussion.

Best Wishes.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Eternal Student on 13/04/2023 16:40:30
Hi again.

   I appreciate the above post was long and boring, sorry.   Here's a short video that illustrates the (idealised) experiment that was done.

Turn the sound down, unless you like a bit of...?... I'm going to call it drum-and-bass.

https://en.wikipedia.org/wiki/File:Quantum_spin_and_the_Stern-Gerlach_experiment.ogv
       
[Video from Wikipedia]

Best Wishes.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Halc on 13/04/2023 17:51:37
Well this is uninformed me commenting on just your post without reading anything from a website. So hardly a position of authority I'm saying.

Quote from: Eternal Student on 13/04/2023 16:33:42
Obviously the Ag atom didn't have just one electron but it's the outermost lone electron that is usually in the 5s orbital which is important.
'Usually'. BC can probably comment better on this, but I presume by that wording that it is not always the case. The higher orbitals can only get 'full' if there are electrons in higher orbitals. The outer one can only hold 8. So I guess Ag is where you get this new lone guy in the outermost orbital, but I can also imagine an atom in an excited state that has more than one out there.

Quote
All the other electrons are assumed to form a reasonably complete cloud with enough symmetry that they contribute a net 0 magnetic moment.
Today I learned ...
But if in that excited state, some inner orbital suddenly loses its symmetry and the defection might be due to that instead of the outer electron(s).

Quote
Just to explain why they bothered with Ag atoms instead of using just a simple electron beam - they wanted an overall Neutral charged particle.
Any atom fits that bill. Surely there's something other than Ag that fits the bill of having a lone electron in the higher orbital. Lithium seems the obvious choice.

Quote
Now, just to check I've understood this experiment.   If, for some reason, the outer electron was not in the 5s orbital (corresponding to quantum numbers n=5, l=0) but instead it was in the 5p orbital  (quantum numbers n=5,  l=1),  would we get a different set of spots?
OK, that seems to be the excited state I mentioned. Not a lower electron joining in the 5s, but the lone one jumping to a higher energy state. The 5p orbitals are not spherically symmetric, so given three possible 5p orbitals and 2 spin states, perhaps that's 6 dots not including the two 5s dots. BC perhaps can correct me on this. It was a very naive answer.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Eternal Student on 13/04/2023 19:01:37
Hi.

Thanks for your input @Halc ,  you do seem to have the right idea for how electronic orbitals are filled.  The details aren't too important (but don't tell that to chemist).
 
Quote from: Halc on 13/04/2023 17:51:37
BC can probably comment better on this
   I used the chemists preference for  s, p, d  orbitals instead of the  physicist's preference for l quantum numbers just because I know there are a few chemists in the audience.  Double the audience, double the replies (was my thinking).
Quote from: Halc on 13/04/2023 17:51:37
OK, that seems to be the excited state I mentioned.
    Electrons being excited from their usual ground state is exactly the problem, or issue that concerns me.   It could be the outer 5s electron going elsewhere (as described in my post)  OR   some of the inner electrons going outwards.   Either of these should change the total magnetic dipole moment of the atom.
LATE EDITING:   And possibly dipole moments from the nucleons doing something.

Quote from: Halc on 13/04/2023 17:51:37
so given three possible 5p orbitals and 2 spin states, perhaps that's 6 dots not including the two 5s dots
    Recall that the z-component of the dipole moment is going to be just a scalar not a vector quantity.   The choice of p orbital and the choice of spin all contribute to just one scalar quantity.  There are 3 different p orbitals exactly as you stated and depending on which you choose, the contribution is +1, 0 or -1  (in Bohr magneton units).     The contribution from spin will be +½  or  -½.   The contributions from each thing are of very similar sizes and not well spread, so you do not get 6 different total dipole moments.
      Notice that  1 - ½   =   0 + ½ .   There are some combinations that give identical final dipole moments.   Assuming I've done the arithmetic correctly we have the result as stated:
 
Quote from: Eternal Student on 13/04/2023 16:33:42
This gives a set of four values:     -1½ ,    - ½ ,    +½,  +1½     (in units of μB ).

     This is only a rough approximation,  depending on how accurately you take the Lande factor to be,  the  dipole contribution from spin is not exactly ½ of a Bohr magneton unit.   As such there could be 6 different dipole moments you could obtain but I doubt the equipment would reliably show the difference between some atoms deposited  at  0.499 units up and   0.501 units up,   they'd all look like they had been deposited at 0.5 units of length up the screen.

Quote from: Halc on 13/04/2023 17:51:37
Any atom fits that bill  (being Neutral). Surely there's something other than Ag that fits the bill of having a lone electron in the higher orbital. Lithium seems the obvious choice.
     I'm not sure why Ag was choosen.   Some reasons are known to me and were declared.   There are probably practical reasons for not choosing a simpler atom.  For example,  Lithium would react violently and you just won't have pure Lithum in your experiment for long.  Silver is considerably less reactive for something that does usually have a unpaired electron in an outer s orbital.    I don't know.... my best guess.

Best Wishes.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Bored chemist on 13/04/2023 20:01:54
Fundamentally, a silver atom has an odd number of electrons.
So, when they try to pair up, there will always be one left over.

The magnetic effects of the paired electrons will cancel out.
That's why you can ignore them and also why you can pretty much ignore what orbital the leftover one is in.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Halc on 13/04/2023 20:08:19
Quote from: Eternal Student on 13/04/2023 19:01:37
Recall that the z-component of the dipole moment is going to be just a scalar not a vector quantity.   The choice of p orbital and the choice of spin all contribute to just one scalar quantity.  There are 3 different p orbitals exactly as you stated and depending on which you choose, the contribution is +1, 0 or -1  (in Bohr magneton units).     The contribution from spin will be +½  or  -½.   The contributions from each thing are of very similar sizes and not well spread, so you do not get 6 different total dipole moments.
OK, that makes sense

Quote
There are probably practical reasons for not choosing a simpler atom.  For example,  Lithium would react violently and you just won't have pure Lithum in your experiment for long.
Ah yes, spitting acid (or in this case, a strong base) at a screen might not be the best option.

Quote from: Bored chemist on 13/04/2023 20:01:54
Fundamentally, a silver atom has an odd number of electrons.
So, when they try to pair up, there will always be one left over.
Well that is satisfied by half the elements then.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Bored chemist on 13/04/2023 22:57:02
The experiment was done in a high vacuum.
Exactly what is the lithium or whatever expected to react with?

Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Eternal Student on 13/04/2023 23:27:02
Hi.

Quote from: Bored chemist on 13/04/2023 20:01:54
The magnetic effects of the paired electrons will cancel out.
That's why you can ignore them and also why you can pretty much ignore what orbital the leftover one is in.
    I'll agree that the dipole moment due to spin cancels.   For paired up electrons you'll obviously have one electron with spin up and one spin down.

     However, the contribution from the magnetic or m quantum number that the electron(s) have, should still be important.

     Putting this in chemistry terms is tricky because although Pz orbitals correspond nicely to a physicist's magnetic quantum number m=0,   the  Px and Py orbitals just don't correspond to m=+1 and -1  respectively.   Instead the Px and Py are actually superpositions of the two states.   For example, the Px state is  (1/√2) . (   |m = -1 state >    -   |m = +1 state> )  giving an expected dipole moment which is still  0  ( 50% probability of +1,  50% of -1).    While  Py is found as the addition of the two states followed by multiplication with i/√2 (complex number i) and has the same expected dipole moment = 0 for the same reasons.

   I think there are two reasons for chemists choosing to consider the states as being Pz, Px and Py  instead of m=+1, 0 and -1.   The first  is simply that you can draw them nicely and they give probability clouds that look like dumb-bells lined up on the appropriate axis.   The second is that this particular combination of states keeps most the wave functions real valued.   There is no loss or harm done by considering this set {Px, Py, Pz}  to be an eigenbasis for the states,  any linear combination of the  m= -1,  m=0 and m=+1  states is a solution to the Schrodinger Equation, so the Px, Py, Pz orbitals are valid solutions.   Furthermore, the particular combination chemists have taken is clearly seen to remain linearly indpendant such that all possible solutions can still be written as a combination of the basis  Px, Py, Pz  instead of the states m= +1 , m=0 and m= -1.
    Overall there is a certain amount of sense to the choice that Chemist's have made - but the Physicist's choice of using an m quantum number isn't without some sense of it's own.   The dipole moment is determined by that m or magnetic quantum number and that is the sort of thing the Physicist's often care about.
     The Px and Py orbitals are a superposition of the m=+1 and m =-1 states and so they are going to be made to collapse to the physicist's  m=+1  and m=-1 states when the magnetic dipole moment is actually measured.   In this situation, simply that there is a magnetic field with a gradient in the z direction is effectively making a measurement of that m quantum number.

   For a large number of atoms and where you are only taking an average, then it's not a noticeable issue.  As mentioned, the expected (or average) dipole moment due to being in the Px or Py orbital is 0.    For individual atoms, though, it matters a lot.   Under a magnetic field of the type in the apparatus,  electrons in Px (or Py) orbitals will be deflected as if they are contributing a magnetic dipole of +1 μB   or  -1μB (with equal likelihood), while those in Pz orbitals contribute 0 dipole moment -  and so you should see three piles of atoms being deposited on the other side.    It's actually 4 different piles when you also allow the spin of the electron to contribute and note that some combinations of the magnetic quantum number m and the spin give the same overall dipole moment  (discussed elsewhere, for example, with @Halc  just recently).   [Just to be clear - that's what I would have thought - but the experiment actually tends to show only two piles of Ag atoms on the far side.  This is something that I think is only likely if the outer electron stays in an s orbital].

    I hope that makes some sense.   It's just an attempt to keep the chemists and physicists both together and using compatible terminology.
    For example,  chemists may know that the emission spectrum of a Hydrogen atom will often exhibit a line splitting when a magnetic field is applied.  The typical "Zeeman effect" is that 1 line is split into  3 lines  (not just 2 lines).   Applying a magnetic field splits the physicist's  states   |n,l,m>   according to the magnetic quantum number m.  So there would be 3 possible values for m when  l=1 and we are talking about a p orbital.    For a chemist,  it's the Px and Py orbitals that could no longer stay intact, they had to split and a new pair of orbitals which I'll call Phigh and Plow appear,  with the slightly higher and lower energies respectively.   Meanwhile the Pz orbital could stay intact and it's energy level was unaffected by the field.   Just to be clear (assuming a general atom with mutiple electrons),  it's not that both electrons in the Px orbital went high but both in the Py went low,  i.e. it's not that something about the x-axis was favoured and the entire Px orbital was adjusted.   The Px and Py orbitals as you know them simply cannot be supported, that superposition of the magnetic m states is not a valid energy eigenstate of the Schrodinger equation any longer.   The electrons that end up in the Phigh energy level could easily have come one each from the Px and Py orbitals.   Only the Pz orbital can survive as you know it.
   So, if you wanted a different way to think about the situation:   When the atoms pass through the magnetic field, p orbitals are very likely to be split and then their true contribution to a magnetic dipole is exhibited (through the quantum number m that they will have).  Meanwhile s orbitals have quantum number m=0 always, they will not be split and do not contribute to the net dipole moment of the atom.  It seems important that the outer electron of the Ag atom stays in an s orbital because then the only thing that causes a deflection is the spin of the electron and not the magnetic quantum number of the orbit it is in - so we will only get two piles of Ag atoms on the far side. 

Best Wishes.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Eternal Student on 13/04/2023 23:32:58
Hi.

Quote from: Bored chemist on 13/04/2023 22:57:02
The experiment was done in a high vacuum.
Exactly what is the lithium or whatever expected to react with?
    I don't know.   I don't know why Silver was chosen other than the reasons outlined in the first post.   It does seem like an unusual choice.   I was hoping a chemist could tell me why it was chosen.
    I'll guess you just can't get many group I metals from a hole in the ground but silver is easy enough to dig up?   the experiment was done in approx. 1921~ 1922.   Maybe they just used silver because it's expensive and science can't be done properly without high costs?
   Why did they use silver?  i.d.k.   but they did and apparently it worked quite well.

Best Wishes.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Eternal Student on 13/04/2023 23:43:58
Hi again.

   Some more information I've found:
In 1927, T.E. Phipps and J.B. Taylor reproduced the effect using hydrogen atoms in their ground state, thereby eliminating any doubts that may have been caused by the use of silver atoms. .
from Wikipedia's history:   https://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment#History 

However, it still seems that Silver was used originally and it did work well, despite all the problems I would have thought should be there.  I do not know why it worked so well and there wasn't just something like a smear of atoms in a continuous line.

Best Wishes.
Title: Re: How many beams or spots emerge from Stern-Gerlach apparatus?
Post by: Bored chemist on 14/04/2023 08:53:24
You want a material which vapourises as single atoms (unlike nitrogen for example, which forms N2 molecules).
I also think you want a material with just 1 electron outside the "closed shells" in order to simplify the orbitals
And that leaves you with the alkali metals (which are a PITA to work with) and the coinage metals, of which silver is most volatile.