Prob (solution by Michael Fisher) 1

Transcription

1 Prob 975 (solution by Michael Fisher) We begin by expressing the initial state in a basis of the spherical harmonics, which will allow us to apply the operators ˆL and ˆL z θ, φ φ() = 4π sin θ sin φ = (Y, Y, ) = i Now we apply the Hamiltonian to θ, φ φ() i (Y, Y, ) ˆL θ, φ Ĥ φ() = ( I + ω ˆL o z ) θ, φ φ() = Now we can use these results to find θ, φ φ(t) i (( I + ω o)y ( I ω o)y, ) θ, φ φ(t) = θ, φ e iĥt/ φ() = e i( I +ωo)t Y e i( I ωo)t Y, Finally, we compute ˆL x ˆL x = φ(t) (ˆL + + ˆL ) φ(t) = φ(t) (()Y + ()Y ) =

2 Maggie Regan Physics Boccio 977 In this problem we are given the three operators on a -dimensional Hilbert space, L x = L y = i i i i,, L z = In part (a) we want to find the possible values that we can obtain if L z is measured Since L z is already diagonalized, the possible values are the diagonal entries of,, and In part (b), a state of L z = is taken Then, we are asked to find L x, L x, and L x = L x L x So, L z = = Also, L x = Then, L x =, L x =, and thus, L x =, For part (c), we want to find the normalized eigenstates and eigenvectors of L x in the L z basis The eigenvalues of the L x matrix are,, and - The corresponding normalized eigenvectors are, v = v =

3 , and v = These vectors and eigenvalues are already in the L z basis, so no change of basis needs to be done For part (d) we are given that the particle is in the state with L z = Then, L x is measured We want to find the possible outcomes and their probabilities To do this we find the projection operators for the eigenvalues of the L x matrix So, =, =, and 4 = 4 In part (e) we have the state ψ =, which is in the L z basis We have that L z is measured and a result of is obtained So, for L z, is a degenerate eigenvalue, so the state will be a linear combination of two vectors To find this linear combination we have that for L z ψ, prob() =, prob() =, and prob(-) = We get these values by renormalizing the vector This renormalization factor is because from the original state, we have that prob() = + = Thus, the state is v + v, where v = and Thus, v = ψ = For part (f), we have that a particle is in the state where prob() = prob(-) = and 4 prob() = This means that the particle is in the state with the basis of L z Then, we know that the generalized state for this particle is ψ = L z = + L z = + L z = We know this because the probabilities listed are simply the squares of the coefficients of the basis states However, we need to include phase factors for each term, because these phase factors will cancel when L z is operatingo n the state So, it is easy to see that the most general, normalized state with this property is ψ = eiδ L z = + e iδ L z = + e iδ L z =

4 Then, we want to determine whether the phase factors are necessary To see this, let s calculate P (L x = ) So, P (L x = ψ) = L x = ψ = e iδ e iδ = ( cos(δ 4 δ )) Therefore, the relative phase differences are not irrelevant The arbitrary phase factors must be included when the state doesn t commute with ˆL z

5 Maggie Regan Physics Boccio 977 In this problem the z-component of the spin of an electron is measured and found to be For part (a), we want to know that if a measurement is then made of the x-component of the spin, what the possible values could be If one component of the spin is known, then the other components cannot be known Also, there are only ever two values of spin that are possible Therefore, we have the possible values of and Then, for part (b) we are asked to find the probabilities of finding these various results So, we have that there are equal possibilities for the two values, and therefore the probabilities are both However, for a more mathematical method, we have that +x = ( ) and x = ( ) Thus, the probability of measuring the up spin of the x-component is + z +x = + z ( ) = Then, the probability of measuring the down spin of the x-component is + z x = + z ( ) = In part (c), the axis defining the measured spin direction an angle θ with respect to the z-axis We then want to find the probabilities of the various possible results So, we have that ( ) cos(θ/) + ˆn = e iφ sin(θ/) and ( ˆn = sin(θ/) e iφ cos(θ/) Then, we have + z = cos(θ/) ˆn + sin(θ/) ˆn Using this, we have that P ( z) = +n z = cos (θ/) and P ( z) = n z = sin (θ/) For part (d) we want to find the expectation value of the spin measurement of part (c) We can find this by multiplying the eigenvalues by their respective probabilities So, measurement = cos (θ/) + ( ) sin (θ/) = (cos (θ/) sin (θ/)) = cos(θ) )

6 Prob 97 (solution by Michael Fisher) a We know that the probability of, is and the probability of, is So, we can 4 4 write this state as φ =, + eiθ, where θ is an arbitary phase angle b In order to determine the effect of a measurement of ˆL x, we must switch to the basis of the eigenvectors of L x From problem 977, we know that in an L = state, the matrix L x is given by which has eigenvalues L x = λ = λ = λ = and corresponding eigenvectors v = v =

7 Prob 97 (solution by Michael Fisher) in the L z basis Let v = w = This is the eigenvector associated with the eigenvalue of L z of, which is the state we are in So, we can write the state, in the L x basis as, = L z = = w v v + w v v + w v v = v + v + v So, prob( ) = prob(λ ) = prob( v ) = 4 prob() = prob(λ ) = prob( v ) = 4 prob() = prob(λ ) = prob( v ) = c We can write, in the L x basis as spatial wavefunctions by simply converting to the spherical harmonics So, θ, φ ( v + v + v ) = Y, + Y, + Y,

8 Prob Z5-6 (solution by Alexandra Werth) We are given Ψ(θ, φ) = Y 8 (θ, φ) + Y 8 (θ, φ) + AY (θ, φ) We can quickly see that Y (θ, φ) is the Hilbert basis of Ψ(θ, φ) therefore we can rewrite Ψ in terms of its ket vectors Ψ(θ, φ) =, +, + A, 8 8 (a) To normalize Ψ the square of the coefficients infront of the kets must sum to one A = A = (b) Next, we want to find ˆL + Ψ(θ, ψ) First, lets define the operater ˆL + ˆL + Y l,m (θ, φ) = l(l + ) m(m + )Y l,m+ (θ, φ) We can now apply this operator to Ψ(θ, φ) defined above ˆL + Ψ(θ, φ) =, +, 8 =, +, (c) Now, we want to solve for ˆL x Ψ(θ, φ) The operator ˆL x is defined as ˆL x = (ˆL + + ˆL The operator ˆL is, ˆL Y l,m (θ, φ) = l(l + ) m(m )Y l,m (θ, φ) Therefore, ˆL Ψ(θ, φ) =, +, 8 8 =, +, So, ˆL x = (ˆL + + ˆL = (, +, +, +, ) =, + +, +, 4 4 4

9 Prob Z5-6 (solution by Alexandra Werth) The probability of finding the angular momentum in the x direction while in state Ψ is Ψ ˆL x Ψ Ψ ˆL x Ψ = = + 4 We can also solve for Ψ ˆL Ψ ˆL Y l,m = l(l + )Y l,m (θ, φ) ˆL Ψ(θ, φ) = Y 8 + Y 8 + Ψ ˆL Ψ = Y (d) We can solve for the probability of having angular momentum in the z direction Angular momentum in the z direction, L z cooresponds to Y Therefore, Y Ψ(θ, φ) = 8 (e) Now, if we are given Φ(θ, φ) = operator ˆL z Y l,m = m Y l,m 8 Y Y 5 + Y 5 We also know that the So, We can also solve, Φ ˆL z Ψ = 5 Φ ˆL Ψ = 5

10 Maggie Regan Physics Zettili 5 In this problem there is a spin particle We are given the Hamiltonian of Ĥ = ɛ (Ŝ x Ŝ y Ŝ z) From the text, we have that Ŝ z = Ŝ = Ŝ + = Then, we can calculate Ŝx, Ŝy, Ŝ x, Ŝ y, and Ŝ z Ŝ x = Ŝ y = Ŝ x = 4 7 7

11 Ŝ y = Ŝ z = 9 9 Thus, Ĥ = ɛ Using mathematica, we are then able to find the eigenvalues, which correspond to the possible energy values of this Hamiltonian These are ɛ 4, ɛ 4, ɛ 4, and ɛ 4 Therefore, the eigenvectors are v =, v =,, and v = v 4 =

12 Prob Z5- (solution by Alexandra Werth) Given the hamiltonian Ĥ = ɛ (Ŝ x + Ŝ y) + ɛ Ŝz A spin 5 particle has possible magnetic numbers of m = [ 5,,,,, 5 ] If we use the identity that Ŝ x + Ŝ y = Ŝ Ŝ z, we can write E m = 5, m ĥ 5, m = ɛ ( ( 5 ( 5 + ) m ) + ɛ m = ɛ ( 5 4 m + m) The possible energy eigenvaules can be solve for by plugging in for the various m values m 5 5 E 5ɛ 8ɛ 9ɛ 8ɛ 5ɛ