Spin Algebra, Spin Eigenvalues, Pauli Matrices

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1 C/CS/Phys 9 9/5/3 Spin Algebra, Spin Eigenvalues, Pauli Matrices Fall 3 Lecture Spin Algebra Spin is the intrinsic angular momentum associated with fundamental particles To understand spin, we must understand the quantum mechanical properties of angular momentum The spin is denoted by S In the last lecture, we established that: S S S x ˆx+S y ŷ+s z ẑ S x + S y + S z [S x,s y ] i S z [S y,s z ] i S x [S z,s x ] i S y [S,S i ] for i x,y,z Because S commutes with S z, there must exist an orthonormal basis consisting entirely of simultaneous eigenstates of S and S z (We proved that rule in a previous lecture Since each of these basis states is an eigenvector of both S and S z, they can be written with the notation a,b, where a denotes the eigenvalue of S and b denotes the eigenvalue of S z Now, it will turn out that a and b can t be just any numbers The word quantum in quantum mechanics refers to the fact that many operators have quantized eigenvalues eigenvalues that can only take on a limited, discrete set of values (In the example of the position and momentum, from previous lectures, the position and momentum eigenvalues were not discrete or quantized in this sense; they were continuous However, the energy of the particle on a ring was quantized Question: What values a and b can have? We ll give away the answer first, and most of the lecture will be spent proving this answer: Answer: a can equal n(n+, where n is an integer or half of an integer Given that a n(n+,b can equal ( n, ( n+, (n, (n, n Now, let s prove it First, define the raising and lowering operators S + and S : S + S x + is y, S S x is y Let s find the commutators of these operators: [S z,s + ] [S z,s x ]+i[s z,s y ] i S y + i( i S x (S x + is y S + Therefore [S z,s + ] S + Similarly, [S z,s ] S C/CS/Phys 9, Fall 3, Lecture

2 Now act S + on a,b Is the resulting state still an eigenvector of S? If so, does it have the same eigenvalues a and b, or does it have new ones? First, consider S : What is S (S + a,b? Since [S,S + ], the S eigenvalue is unchanged: S (S + a,b S + (S S + (a s,m a(s+ The new state is also an eigenstate of S with eigenvalue a Now, consider S z : What is S z (S +? Here, [Sz,S + ] S + ( That is, S z S + S + S z S + So S z S + S + S z + S +, and: S z (S + (S+ S z + S + a,b (S + b+ S + S z (S + (b+ S+ Therefore S + is an eigenstate of Sz But S + raises the S z eigenvalue of a,b by! S+ changes the state a,b to + but S + raises the S z eigenvalue of s,m by! Similarly, S z (S s,m (b (S (Homework So S lowers the eigenvalue of S z by Now, remember that S is like an angular momentum S represents the square of the magnitude of the angular momentum; and S z represents the z-component But suppose you keep hitting s,m with S + The eigenvalue of S will not change, but the eigenvalue of S z keeps increasing If we keep doing this enough, the eigenvalue of S z will grow larger than the square root of the eigenvalue of S That is, the z-component of the angular momentum vector will in some sense be larger than the magnitude of the angular momentum vector That doesn t make a lot of sense perhaps we made a mistake somewhere? Or a fault assumption? What unwarranted assumption did we make? Here s our mistake: we forgot about the ket, which acts like an eigenvector of any operator, with any eigenvalue I don t mean the ket ; I mean the ket For instance, if we were dealing with qubits, any ket could be represented as the α +β What ket do you get if you set both α and β to? You get the ket Which is not the same as Remember in our proof above when we concluded that S z (S + a,b (b+ S + a,b? Well, if S + a,b, then this would be true in a trivial way That is, S z (b+ But that doesn t mean that we have succesfully used S + to increase the eigenvalue of S z by All we ve done is annihilate our ket So the resolution to our dilemma must be that if you keep hitting a,b with S+, you must eventually get Let a,btop (a be the last ket we get before we reach (b top (a is the top value of b that we can reach, for this value of a We expect that b top (a is no bigger than the square root of a Then S z top (a b top (a a,btop (a Similarly, there must exist a bottom state a,bbot (a, such that S bot (a And S z bot (a b bot (a a,bbot (a Now consider the operator S + S (S x + is y (S x is y Multiplying out the terms and using the commutation relations, we get C/CS/Phys 9, Fall 3, Lecture

3 S + S S x + S y i(s x S y S y S x S S z + S z Hence S S + S + S z S z ( Similarly S S S + + S z + S z ( Now act S on top (a and bot (a S a,btop (a (S S + + S z + S z a,btop (a by ( (+b top (a + b top (a a,btop (a S top (a b top (a(b top (a+ top (a Similarly, S a,bbot (a (S + S + S z S z a,bbot (a by ( (+b bot (a b bot (a a,bbot (a S a,bbot (a b bot (a(b bot (a a,bbot (a So the first ket has S eigenvalue a b top (a(b top (a +, and the second ket has S eigenvalue a b bot (a(b bot (a But we know that the action of S + and S on leaves the eigenvalue of S unchanged An we got from a,b top (a to bot (a by applying the lowering operator many times So the value of a is the same for the two kets Therefore b top (a(b top (a+ b bot (a(b bot (a This equation has two solutions: b bot (a b top (a+, and b bot (a b top (a But b bot (a must be smaller than b top (a, so only the second solution works Therefore b bot (a b top (a Hence b, which is the eigenvalue of S z, ranges from b top (a to b top (a Furthermore, since S lowers this value by each time it is applied, these two values must differ by an integer multiple of Therefore b top (a ( b top (a N for some N So b top (a N Hence b top (a is an integer or half integer multiple of Now we ll define two variables called s and m, which will be very important in our notation later on Let s define s b top(a Then s N, so s can be any integer or half integer And let s define m b Then m ranges from s to s For instance, if b top(a f rac3, then s 3 and m can equal 3,,, or 3 C/CS/Phys 9, Fall 3, Lecture 3

4 Then: a s(s+b m Since a is completely determined by s, and b is completely determined by m, we can label our kets as s,m (instead of without any ambiguity For instance, the ket s,m, is the same as the ket a,b 6, In fact, all physicists label spin kets with s and m, not with a and b (The letters s and m are standard notation, but a and b are not We will use the standard s,m notation from now on For each value of s, there is a family of allowed values of m, as we proved Here they are: (table omitted for now Fact of Nature: Every fundamental particle has its own special value of s and can have no other m can change, but s does not If s is an integer, than the particle is a boson (Like photons; s If s is a half-integer, then the particle is a fermion (like electrons, s So, which spin s is best for qubits? Spin sounds good, because it allows for two states: m and m The rest of this lecture will only concern spin- particles (That is, particles for which s The two possible spin states s,m are then, and, Since the s quantum number doesn t change, we only care about m ± Possible labels for the two states (m ± :,, + All of these labels are frequently used, but let s stick with,, since that s the convention in this class Remember: state representing ang mom w/ z-comp up state representing ang mom w/ z-comp down So we have derived the eigenvectors and eigenvalues of the spin for a spin- proton: and are simultaneous eigenvectors of S and S z system, like an electron or C/CS/Phys 9, Fall 3, Lecture 4

5 S s(s+ ( S s(s+ 3 4 S z S z m Results of measurements: m S 3 4,S z +, Since S z is a Hamiltonian operator, and from an orthonormal basis that spans the spin- space, which is isomorphic to C So the most general spin state is Ψ α + β ( α β Question: How do we represent the spin operators (S,S x,s y,s z in the -d basis of the S z eigenstates and? Answer: They are matrices Since they act on a two-dimensional vectors space, they must be -d matrices We must calculate their matrix elements: S s s s s,s z s z s z s z s z,s x s x s x s x s x, etc (S y Calculate S matrix: We must sandwich S between all possible combinations of basis vector (This is the usual way to construct a matrix! s S s S 3 4 s S 3 4 s S So S 3 4 ( C/CS/Phys 9, Fall 3, Lecture 5

6 Find the S z matrix: So S z s z S z + s z Sz s z Sz + s z Sz ( Find S x matrix: This is more difficult What is S x Sx? is not an eigenstate of Sz, so it s not trivial Use raising and lowering operators: S ± S x ± is y S x (S + + S,S y i (S + S S x (S + + S S +, since is the highest S z state But what is S? Since S is the lowering operator, we know that S That is S A for some complex number A which we have yet to determine Similarly, S + A+ Question: What is A? (This is a homework problem Answer: A + s(s+ m(m+ S + s,m A+ s,m+ A s(s+ m(m S s,m A s,m So S + S + ( + ( ( + S ( + ( ( S S x (S+ + S [S+ + S ] C/CS/Phys 9, Fall 3, Lecture 6

7 So S x S x [+ ] S x (S + + S [ + ] S x (S + + S [+ ] S x (S + + S [ + ] ( Find S y matrix: Use S y i (S + S Homework: find the S y,s y,s y,s y matrix elements ( i Answer: S y i Define σ ( ( σ ( i σ i ( σ 3 S 3 4 σ,s x σ,s y σ,s z σ 3 σ,σ,σ,σ 3 are called the Pauli Spin Matrices They are very important for understanding the behavior of two-level systems C/CS/Phys 9, Fall 3, Lecture 7

PHY4604 Introduction to Quantum Mechanics Fall 2004 Practice Test 3 November 22, 2004

PHY4604 Introduction to Quantum Mechanics Fall 2004 Practice Test 3 November 22, 2004 PHY464 Introduction to Quantum Mechanics Fall 4 Practice Test 3 November, 4 These problems are similar but not identical to the actual test. One or two parts will actually show up.. Short answer. (a) Recall