Exercise 18.21 If the perturbation H = 0, show that the rotating frame state ) will be constant over time. The perturbation H will produce changes in ). A rotating frame version of the Schrödinger picture governs such changes: m) = -Ho l&m) + Uon' (Ho + H) Ug(?) |& c1) . (18.18)

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18.21
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Nuclear spins
Exercise 18.20 Verify Eq. 18.15, either by working out the power series for exp(ial) or
by using matrix representations.
The Hamiltonian of an actual nuclear spin, however, is not completely determined by
the external field. The local molecular environment and interactions with other nuclei
contribute to the total Hamiltonian, which we will write Ho + H. The spin state |v(1)} is
governed by the Schrödinger equation
lý)) = (Ho + H) |V(1)).
(18.16)
dt
For the external magnetic fields used in NMR experiments, the Hamiltonian H can be
regarded as a small perturbation of Họ. That is, the spin precession at the Larmor frequency
2 is much more rapid than state changes produced by H. In this case, it is convenient to
adopt the rotating frame picture, which compensates for the simple precession of the spin
state due to Ho. We define the rotating frame state
(18.17)
%3D
Exercise 18.21 If the perturbation H = 0, show that the rotating frame state v) will be
constant over time.
The perturbation H will produce changes in v). A rotating frame version of the
Schrödinger picture governs such changes:
0) = -Ho |&) + U6(n* (Ho + H) Uo() |& c).
(18.18)
Exercise 18.22 First show that iU,(1) = H,Uo(1). Then verify Eq. 18.18 .
dt
We now define the rotating frame Hamiltonian Ĥ(t) = Uo(t)*HUo(t), so that our evolution
equation for (1)) takes the familiar form
(18.19)
The rotating frame Hamiltonian H is a function of time; but notice that the original H might
also be time-dependent.
How is the rotating frame Hamiltonian H(t) different from the original H? Let us write
H in terms of the Pauli operators, which form a basis for the operators on the qubit Hilbert
space Q.
H = €gl + €,X+€,Y+e;Z.
(18.20)
The Pauli operators are affected by the rotation operators as follows:
UŽIU, = 1,
U XU, = cos 2t X + sin 2t Y,
UŽZU, = Z. uYU, = cos 2tY – sin 2t X.
(18.21)
Transcribed Image Text:No SIM ? 9:56 PM O 90% Done quantum-processes-systems... 377 Nuclear spins Exercise 18.20 Verify Eq. 18.15, either by working out the power series for exp(ial) or by using matrix representations. The Hamiltonian of an actual nuclear spin, however, is not completely determined by the external field. The local molecular environment and interactions with other nuclei contribute to the total Hamiltonian, which we will write Ho + H. The spin state |v(1)} is governed by the Schrödinger equation lý)) = (Ho + H) |V(1)). (18.16) dt For the external magnetic fields used in NMR experiments, the Hamiltonian H can be regarded as a small perturbation of Họ. That is, the spin precession at the Larmor frequency 2 is much more rapid than state changes produced by H. In this case, it is convenient to adopt the rotating frame picture, which compensates for the simple precession of the spin state due to Ho. We define the rotating frame state (18.17) %3D Exercise 18.21 If the perturbation H = 0, show that the rotating frame state v) will be constant over time. The perturbation H will produce changes in v). A rotating frame version of the Schrödinger picture governs such changes: 0) = -Ho |&) + U6(n* (Ho + H) Uo() |& c). (18.18) Exercise 18.22 First show that iU,(1) = H,Uo(1). Then verify Eq. 18.18 . dt We now define the rotating frame Hamiltonian Ĥ(t) = Uo(t)*HUo(t), so that our evolution equation for (1)) takes the familiar form (18.19) The rotating frame Hamiltonian H is a function of time; but notice that the original H might also be time-dependent. How is the rotating frame Hamiltonian H(t) different from the original H? Let us write H in terms of the Pauli operators, which form a basis for the operators on the qubit Hilbert space Q. H = €gl + €,X+€,Y+e;Z. (18.20) The Pauli operators are affected by the rotation operators as follows: UŽIU, = 1, U XU, = cos 2t X + sin 2t Y, UŽZU, = Z. uYU, = cos 2tY – sin 2t X. (18.21)
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