UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet

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1 UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet Solution for take home exam: FYS3, Oct. 4, 3. Problem. Ĥ ɛ K K + ɛ K K + β K K + α K K For Ĥ Ĥ : ɛ ɛ, β α. The operator ˆT can be written ˆT K K K K. Therefore ˆT K K K K ˆT ˆT is Hermitian. Unitarity can be shown by forming the product ˆT ˆT K K K K K K K K K K + K K Î ˆT ˆT. Note that if you use a specific state to prove operator properties such as hermiticity and unitarity, you also need to show that these properties holds regardless of your choice of state. Since ˆT is unitary and hermitian, ˆT ˆT. This implies that the eigenvalues of ˆT are ±. Labeling the eigenstates ± we have ˆT ˆT Adding these we see that ˆT Thus we can identify + + c K and + c K, where c is a constant. Adding and subtracting these we get + c normalizing these we find c thus c K K, K + K ± K K.3 Using the fact that ˆT is unitary, ˆT ˆT, we have ˆT Ĥ ˆT Ĥ Ĥ ˆT ˆT Ĥ

2 [Ĥ, ˆT ]. Then using Ehrenfests theorem which was proven in class and in Griffiths for an operator which does not explicitly depend on time, we have d dt ψt ˆT ψt ī ] [Ĥ, h ψt ˆT ψt ψt ˆT ψt constant not dependent on t }{{} Another way of showing this is to consider the unitary time-evolution operator Ût i Ht/ h e [ and reckognize that when ˆT, Ĥ], ˆT will also commute with any powers of Ĥ, thus [ ] ˆT, Û t, and so ψt ˆT ψt ψ Û t ˆT Ût ψ ψ Û t Ût ˆT ψ ψ ˆT ψ.4 ˆT ] is a symmetry if [Ĥ, ˆT. To evaluate the commutator we need Ĥ ˆT ɛ K K + ɛ K K + α K K + β K K K K K K α K K β K K ɛ K K ɛ K K ˆT Ĥ K K K K ɛ K K + ɛ K K + α K K + β K K β K K α K K ɛ K K ɛ K K therefore ] [Ĥ, ˆT β α K K K K α β Hermiticity requires β α so for ˆT to be a symmetry, α must be real..5 When ˆT is a symmetry, α β R, there exists a common complete set of states which are simultaneously eigenstates of both ˆT and Ĥ. The eigenstates of ˆT were found in above. So we will evaluate the action of Ĥ on these Ĥ + ɛ K K + ɛ K K + α K K + α K K K K ɛ α K K ɛ α + Ĥ ɛ K K + ɛ K K + α K K + α K K K + K ɛ + α K + K ɛ + α

3 So the common complete set of eigenfunctions are + K K, +, ɛ α K + K,, ɛ + α.6 The time dependent state can be written in terms of the energy eigenkets as K + +. Thus the time-dependent state starting from K at t is ψt e iĥt/ h K e iĥt/ h + + e iĥt/ h e iɛ αt/ h + + e iɛ+αt/ h e iɛt/ h cosαt/ h + i sinαt/ h + + cosαt/ h i sinαt/ h e iɛt/ h cosαt/ h i sinαt/ h + e iɛt/ h cosαt/ h K i sinαt/ h K Therefore the probabilities are P K t K ψt e iɛt/ h cosαt/ h cos αt/ h P Kt K ψt ie iɛt/ h sinαt/ h sin αt/ h alternatively, since there are only two possibilities: P Kt P K t. One can also solve problem entirely using the matrix representation. Representing the states K and K. The operators become Ĥ ɛ α β ɛ ˆT Problem. A state is equal to the zero ket if and only if its norm is zero. The norm of Ŝ is [Ŝx ] Ŝ+ Ŝ Ŝx Ŝ x + Ŝy Ŝ y i, Ŝy 3 h 4 h h h 3 Ŝ Ŝ z + hŝz Ŝ

4 Note that points will be deducted if you use Ŝ to show that the norm is.. Expanding the exponential we have e µ h Ŝ + µ hŝ + µ Ŝ Since h!ŝ Ŝ Ŝ hŝ all higher powers than linear vanish. Using Ŝ h we get µ + µ hŝ + µ N N The corresponding bra is µ N + µ so that µ µ N + µ N + µ N + µ Using this normalization we get λ µ + λ + µ + λ + µ + λ µ + λ + µ.3 Using Ŝx µ Ŝ µ + µ + µ Ŝ + µ hµ + µ he iφ tanθ/ + tan θ/ he iφ sinθ/ cosθ/ h e iφ sinθ µ Ŝ+ µ µ Ŝ µ h eiφ sinθ Ŝ+ + Ŝ / we get θ, φ Ŝx θ, φ θ, φ Ŝ+ θ, φ + θ, φ Ŝ θ, φ h 4 sin θ e iφ + e iφ h sin θ cos φ Alternatively one can directly write down the state θ, φ first θ, φ + tanθ/e iφ cosθ/ + sinθ/e iφ + tan θ/.4 A very simple way to solve this problem is to use the result from problem.b and use the fact that since there are only two possible measurement outcomes the expectation value determines the probability. Denoting the probability for measuring + h/ by p x one has θ, φ Ŝx θ, φ h p x h p x p x + sin θ cos φ 4

5 One can also proceed in the more standard way: Note that for θ π/ and φ, Ω,,. Therefore the eigenstate of Ŝ x with eigenvalue h/ is θ π/, φ. The probability of measuring this value in the state θ, φ is thus p x π/, θ, φ θ, φ π/, + tanθ/e iφ tanπ/4e i + tan π/4 + tan θ/ + tanθ/e iφ + tan θ/ cosθ/ + sinθ/e iφ cosθ/ cosθ/ p x π/, θ, φ cosθ/ + sinθ/e iφ cosθ/ + sinθ/e iφ cos θ/ + sin θ/ + sinθ/ cosθ/ cos φ + sin θ cos φ.5 The eigenstates of the Hamiltonian are θ, φ and π θ, φ + π. The last fact follows from sinπ θ cosφ + π, sinπ θ sinφ + π, cosπ θ S π θ, φ + π h π θ, φ + π using sinπ θ sin θ, cosπ θ cos θ, cosφ + π cosφ and sinφ + π sin φ one sees that sin θ cos φ, sin θ sin φ, cos θ S π θ, φ + π h π θ, φ + π }{{} Ω in other words π θ, φ + π is an eigenstate of Ω S with eigenvalue h/. The explicit expression for the two energy eigenstates are θ, φ cosθ/ + sinθ/e iφ π θ, φ + π sinθ/ cosθ/e iφ At t immediately after the measurement of the spin x component that gave the value h/ the system is in the state π/,. This state can be decomposed into the Hamiltonian eigenstates as π/, α θ, φ + β π θ, φ + π 5

6 to find the coefficients α and β we multiply by the bras θ, φ and π θ, φ + π and use the orthonormality of these states. Using the results from problem.3 we find α α θ, φ π/, cosθ/ + sinθ/eiφ β π θ, φ + π π/, sinθ/ cosθ/eiφ β is found by letting θ π θ and φ φ + π in the expression for α. The time-dependent state is then ψt e iĥt/ h α θ, φ + β π θ, φ + π αe igt/ h θ, φ + βe igt/ h π θ, φ + π and the probability of measuring the spin-x component to be the value h/ is Now α P t π/, ψt β αe igt/ h π/, θ, φ + βe igt/ h π/, π θ, φ + π α e igt/ h + β e igt/ h cosθ/ + sinθ/eiφ + sinπ θ cosφ + π Therefore + sin θ cos φ P t e igt/ h + cosθ/ + sinθ/e iφ sin θ cos φ sin θ cos φ + sin θ cos φ e igt/ h cosgt/ h + i sin θ cos φ singt/ h cos gt/ h + sin θ cos φ sin gt/ h sin θ cos φ sin gt/ h + sin θ cos φ + sin θ cos φ cosgt/ h Note that you will not get credit on this subproblem if you changed the problem to a much easier one by choosing particular values for θ and φ, as for instance choosing ˆΩ,, or ˆΩ,,. The solutions to these simpler problems are of course contained in the solution for general θ and φ above..6 µ µ + µ + µ + µ + µ + µ + µ + µ 6

7 reparametrizing µ tanθ/e iφ one gets θ, φ θ, φ + tan θ/ + tanθ/e iφ + tanθ/e iφ + tan θ/ Doing the integral over φ from to π, the last two terms vanish thus π dφ π π dθ sin θ θ, φ θφ π π dθ sin θ + tan θ/ + tan θ/ cos θ/ dθ sin θ + tan cos θ/ + sin θ/ θ/ dcos }{{} θ cos θ/ + sin θ/ }{{}}{{} x +cos θ/ cos θ/ dx + x + + Î dx x The answer is the identity operator. 7