Physics 3 for Electrical Engineering

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Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers Daniel Rohrlich, Ron Folman Teaching
Assistants Daniel Ariad, Barukh Dolgin
Week 10. Quantum mechanics Schrödingers
equation for the hydrogen atom eigenvalues and
eigenstates atomic quantum numbers
Stern-Gerlach and spin Pauli matrices
spin-orbit coupling Sources Feynman Lectures
III, Chap. 19 Sects. 1-5 Merzbacher (2nd
edition) Chap. 9 Merzbacher (3rd edition) Chap.
12 ????? ??????? ???????, ????? 8, ?????
1-2 Tipler and Llewellyn, Chap. 7 Sects. 2-5.
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Schrödingers equation for the hydrogen atom The
most common isotope of hydrogen contains just one
proton and one electron. Their potential energy
is so Schrödingers equation depends on rp
and re , i.e. on six coordinates But a
clever change of variables makes the equation
simpler
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Schrödingers equation for the hydrogen atom
Let r re rp and
then Schrödingers equation
becomes where is
called the reduced mass. For a proton and an
electron, the reduced mass is essentially the
electron mass, since mp 2000 me. But for a
positronium atom (which has a positron in the
place of the proton), µ me /2.
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Schrödingers equation for the hydrogen atom
Let r re rp and
then Schrödingers equation
becomes where is
called the reduced mass. There are solutions
of the form ? T(t)?(R)?(r), where T(t) is the
usual time dependence and ?(R) is a solution to
the free Schrödinger equation in R (the
center-of-mass coordinate), which has no
potential term.
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Schrödingers equation for the hydrogen atom
The equation for ?(r) is then with the
Coulomb energy as a central potential V(r). We
have seen that, for a central potential,
Schrödingers equation reduces to
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We replace m by µ and V(r) by and obtain
the eigenvalues of they are
. We solved this equation by expressing ?(r,?,f)
as a product of two functions ?(r,?,f)
R(r)Ylm(?,f) , where
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Eigenvalues and eigenstates Hence the equation
for R(r) is The solutions Rnl(r) of this
equation have the form with Fnl (r/a0) a
polynomial and n l 1. The constant a0 is
called the Bohr radius and equals
Å
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Eigenvalues and eigenstates The energies depend
only on n where 1 eV 1.602176 
10-19 J. Question What is the normalization
condition for ?nlm(r,?,f)?
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Eigenvalues and eigenstates The energies depend
only on n where 1 eV 1.602176 
10-19 J. Question What is the normalization
condition for ?nlm(r,?,f)? Answer
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Eigenvalues and eigenstates Question What is
the normalization condition for Rn0(r)?
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Eigenvalues and eigenstates Question What is
the normalization condition for Rn0(r)? Answer
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Eigenvalues and eigenstates Here are the lowest
normalized solutions ?nlm Rnl(r)Ylm(?,f) of the
hydrogen atom Schrödinger equation (see also
here)
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Radial eigenfunctions Rnl(r) and probability
distributions Pnl(r) for the lowest eigenstates
of the hydrogen atom. From here. Question How
can Pn0(r) vanish at r 0 if Rn0(r) does not?
Pnl(r)
Rnl(r)
Radius (a0)
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Radial eigenfunctions Rnl(r) and probability
distributions Pnl(r) for the lowest eigenstates
of the hydrogen atom. From here. Question How
can Pn0(r) vanish at r 0 if Rn0(r) does
not? Answer On a sphere of radius r, we have
Pn0(r) r2 Rn0(r)2.
Pnl(r)
Rnl(r)
Radius (a0)
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For a single electron bound by a nucleus
containing Z protons, the solutions of
Schrödingers equation are almost unchanged the
reduced mass µ is even closer to me, while the
potential term is Thus by replacing e2 with
Ze2 in the eigenstates and eigenvalues we
obtained for Z 1, we obtain the Z gt 1
eigenstates and eigenvalues. For example, we
obtain the energy eigenvalues
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Exercise Show that the energy of the ground
state is half the expectation value of the
potential energy in the ground state.
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Exercise Show that the energy of the ground
state is half the expectation value of the
potential energy in the ground state.
Solution The ground state energy is The
expectation value of the potential energy is
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Atomic quantum numbers According to what we have
seen so far, every eigenstate of the hydrogen
atom can be associated with three quantum numbers
n, l and m, where n 1, 2, 3,
principle quantum number l 0, , n 1 m
l, l 1, , l 1, l . The degeneracy of
the energy eigenvalue En is therefore 1 3 5
2(n 1) 1 n2 . In atomic
physics, the l quantum numbers have special
names s for l 0, p for l 1, d for l 2,
f for l 3, etc. Then 2s means n 2 and l
0 3p means n 3 and l 1 and so on.
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Since the energy of an electron in a stationary
hydrogen atom can only be one of the En , where
a stationary atom can absorb and emit photons
only if the photon energy equals Ephoton En
En . And since the energy of a photon is
related to its frequency ? by Ephoton h?, the
frequencies of electromagnetic radiation emitted
or absorbed by a hydrogen atom must obey the
rule for some n and n. This formula,
derived by Bohr 13 years before Schrödinger, was
soon verified via spectroscopy.
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Energy levels for hydrogen, and transitions among
the levels
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Energy levels for hydrogen, and transitions among
the levels
Corresponding spectral lines
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Setup for emission spectroscopy
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The magnetic quantum number m When an atom is
immersed in a uniform magnetic field, the
energies En split! Let B be the strength of the
field and let point up the z-axis. A state with
quantum numbers n, l, m has energy where µB
eh/2me is the Bohr magneton. How can we explain
this effect? An electron moving in a circular
orbit of radius r, at speed v, produces a current
I ev/2pr and a magnetic moment µz I(pr2)
evr/2 eLz/2me. Since Lz mh, we have µz
µBm. The corresponding extra potential term for
the hydrogen atom is
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The magnetic quantum number m But this is not
the only magnetic effect. In a uniform magnetic
field, there is a torque on the atom (to make it
anti-parallel to the magnetic field). But in a
non-uniform magnetic field, there is also a force
on the atom. Assume again that the field points
up the z-axis, so that VB µz B. Then if dB/dz
? 0, the force is So, what happens if a beam
of neutral atoms with lh Lz lh crosses a
non-uniform magnetic field?
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The magnetic quantum number m But this is not
the only magnetic effect. In a uniform magnetic
field, there is a torque on the atom (to make it
anti-parallel to the magnetic field). But in a
non-uniform magnetic field, there is also a force
on the atom. Assume again that the field points
up the z-axis, so that VB µz B. Then if dB/dz
? 0, the force is So, what happens if a beam
of neutral atoms with lh Lz lh crosses a
non-uniform magnetic field? We expect the beam
to split into 2l1 beams, one for each value of
µz. In some cases, the beam indeed splits into
2l1 beams.
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Stern-Gerlach and spin But O. Stern and W.
Gerlach saw a beam of silver atoms split into two
beams!
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Stern-Gerlach and spin But O. Stern and W.
Gerlach saw a beam of silver atoms split into two
beams! How can have an even number of
eigenvalues? G. Uhlenbeck and S. Goudsmit
suggested that each electron has its own
intrinsic angular momentum spin with only
two eigenvalues. But electron spin has odd
features. For example, its magnitude never
changes, just its direction and it has only two
directions.
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Stern-Gerlach and spin Lets try to understand
spin better by reviewing the algebra of
Consider l 1 and m
1, 0, 1. The matrix representation of in
a basis of eigenstates of is since the
eigenvalues are 0 and h.
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Stern-Gerlach and spin What is ? We
know it must equal but what is a? We
have
Hence , up to an
overall phase. Similarly, we can show that
, hence
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Stern-Gerlach and spin Similarly,
Now, since
, we can write
Since we can write
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Pauli matrices It is straightforward to check
that these matrix representations have the
correct commutation relations But Pauli
discovered 2 2 matrices with the same
commutation relations (The Pauli matrices
are these matrices without the h/2
factors.) These are the operators for the
components of electron spin!
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Pauli matrices We can write the eigenstates of
as
for Sz h/2 and as for
Sz h/2. Since
for s ½,
we refer to electron spin as spin-½.
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Stern-Gerlach and spin More odd features of
electron spin The eigenvalues of
are h/2. We can write an eigenstate
of with eigenvalue mh as
but an analogous eigenstate of , namely
, would not be single-valued. Yet
experiments show that these electron spin
eigenstates are not invariant under rotation by
2p, but they are invariant under rotation by
4p! This is reminiscent of a trick with a
twisted ribbon.
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Stern-Gerlach and spin This is
reminiscent of a trick with a twisted ribbonone
twist cannot be undone, but two twists are
equivalent to no twist.
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Stern-Gerlach and spin One more odd feature of
electron spin For orbital angular momentum, we
found that µz eLz/2me. For spin angular
momentum, experiment shows that µz eSz/me.
That is, electronic spin produces an anomalous
double magnetic moment. Therefore, the total
magnetic moment of an electron with orbital
angular momentum mh and spin angular momentum
h/2 is
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Atomic quantum numbers (again) We associated
every eigenstate of the hydrogen atom with three
quantum numbers n, l and m. But now we have to
introduce a fourth quantum number, the spin ms
½ . The degeneracy of the energy eigenvalue
En is therefore not n2 but 2n2, since there are
two spin states for every set of quantum numbers
n, l and m. The nucleus, too, has spin angular
momentum. But its magnetic moment is relatively
tiny because the mass of a proton is about 2000
times the electron mass. In this course we
neglect the spin and magnetic moment of the
nucleus.
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Exercise Show that the superposition of wave
functions is normalized if each wave function
is, and calculate
and ?Lz.
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Exercise Show that the superposition of wave
functions is normalized if each wave function
is, and calculate
and ?Lz. Solution Since the components have
different eigenvalues, they are orthonormal, and
the normalization is obtained from the absolute
value of the squares of the coefficients
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Exercise Show that the superposition of wave
functions is normalized if each wave function
is, and calculate
and ?Lz. Solution
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Exercise Show that the superposition of wave
functions is normalized if each wave function
is, and calculate
and ?Lz. Solution
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Exercise What happens in a Stern-Gerlach
experiment, if each electron in an incident beam
of hydrogen atoms has l 1?
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Exercise What happens in a Stern-Gerlach
experiment, if each electron in an incident beam
of hydrogen atoms has l 1? Solution The
magnetic moment of the electron depends on Lz and
Sz according to Since m 1, 0, 1 and,
independently, ms ½, we get five possible
values of m 2ms 2, 1, 0, 1, 2. We therefore
expect to see 5 separate spots on the screen.
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Spin-orbit coupling We discussed atomic magnetic
moments in a magnetic field that is uniform or
non-uniform. But even without any external
magnetic field, an electron feels an effective
field. Why?
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Spin-orbit coupling We discussed atomic magnetic
moments in a magnetic field that is uniform or
non-uniform. But even without any external
magnetic field, an electron feels an effective
field. Why? The electron moves relative to the
nucleus. Transforming the Coulomb field to the
electrons rest frame yields a magnetic field B'
v E/c2. Since E is radial, v E/c2 is
dV(r)/dr times r p /emec2r L/emec2r. Since
the electrons magnetic moment e /me interacts
with B', the spin-orbit interaction contains also
. It enters the Hamiltonian as where
V(r) is the Coulomb potential.
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Spin-orbit coupling To compute the eigenvalues
of , we must know how to add angular
momenta. Defining , we find that
and so on, i.e. the components of follow
exactly the same algebra as the components of
and . We immediately infer that the
eigenvalues of are j(j1)h2 and that the
eigenvalues of are jh, (j1)h,, (j1)h,
jh.
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Spin-orbit coupling To compute the eigenvalues
of , we must know how to add angular
momenta. Defining , we find that
and so on, i.e. the components of follow
exactly the same algebra as the components of
and . We immediately infer that the
eigenvalues of are j(j1)h2 and that the
eigenvalues of are jh, (j1)h,, (j1)h,
jh. Now from
we derive
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Exercise The spin-orbit coupling splits the
degeneracy between the hydrogen states ?2,1,1, ½
and ?2,1,1,½ by ?E 4.5 10-5 eV. Estimate the
magnetic field B' felt by the electron.
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Exercise The spin-orbit coupling splits the
degeneracy between the hydrogen states ?2,1,1, ½
and ?2,1,1,½ by ?E 4.5 10-5 eV. Estimate the
magnetic field B' felt by the electron.
Solution The energy splitting is due to the
interaction of the electrons magnetic moment
with the effective magnetic field B'. In the rest
frame of the electron only the spin magnetic
moment contributes µzB' (eSz/me)B'
eB'h/2me, hence ?E eB'h/me and
B' me?E/eh ?E/2µB (4.5 10-5
eV) / 2 (5.79 10-5 eV/T) 0.39 T .