Quantum theory of spin: algebraic approach in analogy with classical physics, we believe that a spinning mass carries intrinsic angular momentum, along.

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Presentation transcript:

Quantum theory of spin: algebraic approach in analogy with classical physics, we believe that a spinning mass carries intrinsic angular momentum, along with the angular momentum of its center-of-mass, which is L we call this electron spin angular momentum S Stern and Gerlach experimentally demonstrated, by putting a beam of electrons in a non-uniform magnetic field, that electrons seemed to carry two possible values of one spin component, so the suggestion is that electrons have spin quantum number s = ½ and a z-component of spin quantum number m = ± ½  ‘spin-up’ or ‘spin-down’ we will work almost exclusively with operator algebra, and push the analogy with what we have seen already to the hilt the explicit representation for spin will be 2 x 2 matrix algebra we have a new notation now for our humble electron’s spin | s m > is the ‘spin state’ for an electron with spin angular momentum (√3ħ /2) and spin z-component ±ħ /2

What do our spin operators look like? what does this tell us? Same operator behavior: start with commutation relations: assume verysame structure eigenstates (simultaneously measurable) of those two operators while we’re on the subject, we have raising and lowering operators otherwise results are zero: can’t raise m = ½ or lower m =  ½ that √ prefactor is exactly zero as a guarantee in fact, s may be an integer (like l was) or a ‘half-integer’ (as here)

What do our spin states look like, and are they eigenstates? the basis spinors are eigenfunctions of S 2 and S z operators the eigenvalues are s(s+1) ħ 2 and mħ, respectively in this notation, the operators are 2 x 2 matrices inner (scalar) products are row vectors times column vectors to get of S 2 as a matrix, look at its effect on ‘up’ and ‘down’ spin what are our spin states? The general state is a spinor   we have an alternate, more explicit, column vector notation

What do more of our spin operators look like? to get S z as a matrix, look at its effect on ‘up’ and ‘down’ spin we do it in class! Result: now get the raising and lowering operators! Results:

to get S x and S y as a matrix recall that the raising and lowering operators were expressed in terms of them we also define the 3 Pauli spin matrices {  } with the identity matrix I the four are said to span spin space, since any complex 2 x 2 matrix can be expressed as a linear combo Now to get the last couple of them by working backwards, and define the Pauli spin matrices

Effect of S x and S y on eigenvectors of S z what is the effect of (say) S x on an eigenspinor of S z ? so that’s interesting… it ‘flips’ the basis spinor what is the effect of (say) S y on an eigenspinor of S z ? so that’s even more interesting… it ‘flips’ the basis spinor and it does a strange sign/imaginary operation on it, depending note that we can normalize any spinor by taking its inner product with its adjoint (in this representation, column  row and c.c.) example of how this works:

Eigenvectors for S x what are the eigenspinors and eigenvalues of S x ? but any multiple of an eigenvector is also an eigenvector of the same eigenvalue, so let’s rescale by dividing by  (bad news if  = 0)

Eigenvectors for S y all operators have real eigenvalues and are observable the normalized eigenspinors for S z are obvious building blocks for an arbitrary spinor however, the normalized eigenspinors for S x and for S y are equally good: we need a ‘Fourier’s trick’ in spin space!!

Decomposing an arbitrary spinor into a basis set [notation modified for consistency with x and y] assume  is normalized: how to express  as a linear combination of (say) eigenspinors of S x we can also express  as a linear combination of eigenspinors of S y

Do these matrices obey the requisite commutation relations? for example, does S x commute with S z ? for example, does S 2 commute with S z ?

How does one combine angular momentum? this might be a situation where two electrons need to have the total spin added or it might be a situation where you have one electron and you want to add its spin to its orbital angular momentum or you might want to combine an electron’s angular momentum with the angular momentum of the nucleus… classically, it is merely the vector sum of each part – changes with time, if torques act, etc. quantally, it is a more probabilistic cloud of somewhat quantized possibilities the classical picture of adding two vectors is helpful, but one must realize that in the end the sum is quantized both in length and in z- component, and that there will usually be more than one way to achieve a given length/z-component combo we have two angular momenta: S (1) and S (2) represented by operators of the usual form (operators, matrices..) to start, assume two s = ½ ‘spins’

What will our notation be? What can we cook up? if this is really a s = 1 situation, then where is s =1, m = 0 ? problem lies in the two middle states, which appear twice NOT eigenstates of the operator S 2 = (S (1) + S (2) )· ( S (1) + S (2) ) realize that the ‘1’ operators only operate on the first argument of the ket, and the ‘2’ operators only on the second argument a naïve attempt to create a couple of angular momenta together

Check out whether one of those middle states works obviously, the other ‘naïve’ state will fail too Are either of them eigenstates of S z = S z (1) + S z (2) ? NO are the first and last states eigenstates of either operator?

What to do with the middle states, and recover the missing combination to boot what can we cook up? Linear combinations of those two states!! strategy: try lowering that ‘top’ state

Another, singlet state: m = s = 0 all four are orthornormal to one another all are eigenstates of S z = S z (1) + S z (2) as well as of S 2 the Clebsch-Gordan coefficients tell how to assemble more general linear combinations of two angular momenta the new state had a + sign in it.. Try a .. the triplet states have s = 1, m =  1, 0, 1 the singlet state has s = 0, m = 0