Work out the spin matrices for arbitrary spin , generalizing spin (Equations 4.145 and 4.147), spin 1 (Problem 4.31), and spin (Problem 4.52). Answer:

Sz=(s0000s-10000s-200000-s)Sx=2(0bs0000bs0bs-10000bs-10bs-20000bs-200000000b-s+10000b-s+10)Sy=2(0-ibs0000ibs0-ibs-10000-ibs-10-ibs-20000-ibs-200000000-ibs+10000-ibs+10)

where,bj(s+j)(s+1-j)

Short Answer

Expert verified

The spin matrices are, sy=2i0bs000-bs0-bs-100-bs-10-bs-200000-b-s-10000-b-s-10

Step by step solution

01

Definition of spin matrix

The spin related matrices are known as spin matrices. These are number of matrices. These matrices are complex that include involutory, unitary, and Hermitian.

02

Determination of spin matrices.

Write equation 4.135.

sz|sm=hm|sm

Write the matrix element of sz.

sznm=nszm=hmnm=hmδnm

Write the matrix (diagonal matrix) with values of m ranging from s to -s along the diagonal.

Sz=(s0000s-10000s-200000-s)

Determine the value of s+nm.

s+nm=ns+m=h(s-m)(s+m+1)nm+1=hbnδnm-1

Here,bm+1=(s-m)(s+m+1) .

Use the property of the δ function.

(s*)mw=bnδnm+1

Write the matrix.

s+=0bs00000bs-100000bs-20b-s+100000

Write the value of s_nm.

role="math" localid="1658146052379" s_nm=ns_m=h(s+m)(s-m+1)δnm-1=hbnδnm-1

Write the value of s_ .

s_=h000.....0bs000...00bs-1.......00bs-2.......0000sx=12s++s-

Write the value ofrole="math" localid="1658143674691" sx.

sx=20bs000bs0bs-1000bs-10bs-200bs-200b-s+1000b-s+10b-s+10

Write the value ofsy .

sy=12i[s+-s-]sy=2i0bs000-bs0-bs-100-bs-10-bs-200000-b-s-10000-b-s-10

Thus, the spin matrices are2i0bs000-bs0-bs-100-bs-10-bs-200000-b-s-10000-b-s-10

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Most popular questions from this chapter

Deduce the condition for minimum uncertainty inSx andSy(that is, equality in the expression role="math" localid="1658378301742" σSxσSy(ħ/2)|<Sz>|, for a particle of spin 1/2 in the generic state (Equation 4.139). Answer: With no loss of generality we can pick to be real; then the condition for minimum uncertainty is that bis either pure real or else pure imaginary.

[Attempt this problem only if you are familiar with vector calculus.] Define the (three-dimensional) probability current by generalization of Problem 1.14:

J=ih2m(ψψ*-ψ*ψ)

(a) Show that satisfies the continuity equation .J=-t|ψ|2which expresses local conservation of probability. It follows (from the divergence theorem) that sJ.da=-ddtv|ψ|2d3rwhere Vis a (fixed) volume and is its boundary surface. In words: The flow of probability out through the surface is equal to the decrease in probability of finding the particle in the volume.

(b) FindJfor hydrogen in the staten=2,l=1,m=1 . Answer:

h64ma5re-r/asinθϕ^

(c) If we interpretmJas the flow of mass, the angular momentum is

L=m(r×J)d3r

Use this to calculate Lzfor the stateψ211, and comment on the result.

(a) If you measured the component of spin angular momentum along the x direction, at time t, what is the probability that you would get +h/2?

(b) Same question, but for the ycomponent.

(c) Same, for the z component.

The (time-independent) momentum space wave function in three dimensions is defined by the natural generalization of Equation 3.54:

Φ(p,t)=12πhe-ipx/hψ(x,t)dx(3.54).ϕ(p)1(2πh)3/2e-i(p.r)Ihψ(r)d3r.(4.223).

(a)Find the momentum space wave function for the ground state of hydrogen (Equation 4.80). Hint: Use spherical coordinates, setting the polar axis along the direction of p. Do the θ integral first. Answer:

ψ100(r,θ,ϕ)=1πa3e-r/a(4.80).ϕ(p)=1π(2ah)3/21[1+ap/h2]2.(4.224).

(b) Check that Φ(p)is normalized.

(c) Use Φ(p)to calculate <p2>, in the ground state of hydrogen.

(d) What is the expectation value of the kinetic energy in this state? Express your answer as a multiple of E1, and check that it is consistent with the virial theorem (Equation 4.218).

<T>=-En;<V>=2En(4.218).

(a) Work out all of the canonical commutation relations for components of the operator r and p : [x,y],[x,py],[x,px],[py,pz],and so on.

(b) Confirm Ehrenfest’s theorem for 3 dimensions

ddt<r>=1m<p>andddt<p>=<-v>

(Each of these, of course, stand for three equations- one for each component.)

(c) Formulate Heisenberg’s uncertainty principle in three dimensions Answer:

σxσph2;σyσph2;σzσph2

But there is no restriction on, say, σxσpy.

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