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These notes are for the two-semester graduate level quantum mechanics class taught at ... where δ(x−x ) is a Dirac delta function, as opposed to the usual ...
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These notes are for the two-semester graduate level quantum mechanics class taught at Michi- gan State University. Although they are more terse than a typical text book, they do cover all the material used in PHY 851/852. The notes presume a familiarity with basic undergraduate concepts in quantum mechanics.
Anybody is welcome to use the notes to their heart’s content, though the text should be treated with the usual academic respect when it comes to copying material. If anyone is interested in the LATEX source files, they should contact me ([email protected]). Solutions to the end-of-chapter problems will also be provided on the course web site, https://people.nscl.msu.edu/~pratt /phy851. Please beware that this is a web manuscript, and is thus a living document and subject to change at any time.
Quantum mechanics consists of states and operators. For any finite system, states are discrete and can be assigned labels. The discrete nature of states is what is what is meant by the word “quantized”. For some systems the number of such discrete states is finite, e.g. a particle in either spin-up or spin-down state, whereas in other systems the number of states are infinite, e.g. levels of a harmonic oscillator. In this chapter we concentrate on those systems where the number of states are finite, and proceed to the latter case in the next chapter.
Any physical state can be assigned a label, in this case ψ. This state can always be expressed as a linear combination of basis states, |i〉, in an orthonormal basis. These states can be expressed as vectors in a vector space of size n,
, · · · |n〉 = ˆn =
By inspection, the basis is orthonormal,
〈i|j〉 ≡ ˆi∗^ · ˆj = δij. (1.2)
In quantum mechanics the vector algebra is complex, and the adjoint vector 〈ψ| is represented by the complex transpose. Any orthonormal set of vectors can be expressed in this basis. The notational choice of using 〈ψ| to refer to the adjoint vector and |ψ〉 to denote the vector is known as bras and kets respectively (a take on the word “bracket”), and is known as Dirac notation, though it derives from Hermann Grassmann’s work, https://en.wikipedia.org/wiki/Herman n_Grassmann, one hundred years earlier.
Whenever the notation has a bra followed by a ket, it implies the dot product between the two vectors, i.e.
〈v|u〉 =
i
v∗ i ui, (1.3)
and is sometimes referred to as an inner product. A sum over vector indices is not implied if the ket precedes the bra, and
|u〉〈v| = uiv∗ j (1.4)
is known as an outer product. However, if one sees the notation,
|α〉〈α|, (1.5)
a sum over the vectors α is implied (unless otherwise stated), as the summation
α is implicit. Ultimately, for any physical observable, all bras and kets will be closed with a bra being the first
This is easy to see with the simple basis states in Eq. (1.1). In that case
The completeness relation will also work for any set of basis states, because the basis can always be transformed to the simple basis, and because
i |i〉〈i|^ =^ I^ and the unit matrix is unchanged by a transformation of basis, a.k.a. a unitary transformation. This will be demonstrated ahead.
As mentioned previously, all states can be expressed as vectors. For instance, if a basis has two states, the two states might be defined in terms of vectors as,
The choice of these two vectors is arbitrary. As long as the two vectors are orthonormal, any state can be represented as a linear sum of the two states. Any set of orthonormal states that spans the space is known as a “basis”, with each orthonormal state being a basis state.
Example 1.1: Probabilities and Overlaps Consider the state
|ψ〉 =
i
a) Find Z so that the state is normalized. b) What is the probability that |ψ〉 would be measured in the | ↓〉 state?
Solution : a) Squaring |ψ〉,
〈ψ|ψ〉 =
1 −i
i
b) The probability is
P (↓) = |〈↓ |ψ〉|^2
=
i
2
(−i)(i) =
Operators operate on vectors and return another vector. All operators may be described in terms of bras and kets as a linear combination of outer products,
A =
ij
aij |i〉〈j|. (1.13)
Just as any state can be equivalently expressed as a vector, any operator can be represented by a matrix, in this case aij. Just as a state’s expression in terms of a vector depends on the basis, so does the expression of an operator in terms of a matrix. Knowing the coeficients aij is sufficient to define the matrix. If the basis is defined by
the matix and the coefficients are synonymous Aij = aij. Otherwise, one can express the states |i〉 and |j〉 as vectors, v(i)m and v(j)n. In that case
Amn =
ij
aij v(i)mv∗(j)n, (1.15)
where v(i)m is the mth^ component of the basis vector v(i).
Consider a matrix element, 〈φ|A|ψ〉. Expressing the operators as matrices and the states as vectors, then taking the complex conjugate one sees that,
〈φ|A|ψ〉 = φ∗ i aij ψj (1.16) 〈φ|A|ψ〉∗^ =
φ∗ i aij ψj
= ψ∗ j a∗ ij φj.
Thus, if one wishes to define an operator A†^ such that
〈ψ|A†|φ〉 = 〈φ|A|ψ〉∗, (1.17)
for any states |φ〉 and |ψ〉, the operator A†^ must be represented by the complex-conjugate of the matrix that represents A, then transposed.
A†^ =
ij
(a†)ij |i〉〈j|, (1.18)
(a†)ij = a∗ ji
The operator A†^ is known as the Hermitian conjugate of A. A Hermitian operator is one that obeys the relation, A = A†. (1.19)
operator that satisfies this condition is called unitary. The word unitary follows from the fact that a state U |ψ〉 has the same norm as |ψ〉, implying that the net probability is unchanged by the unitary transformation U.
Hermitian operators are often used to generate unitary transformations,
U = e−iKθ. (1.23)
It is easy to see that such an operator is unitary if K = K†,
U U †^ = e−iKθeiK
†θ (1.24) = e−iKθeiKθ^ = e−i(K−K)θ = I.
If an operator U is unitary, it represents unitary transformations, under which a vector |ψ〉 transforms as
|ψ′〉 = U |ψ〉, (1.25)
and an operator A transform as
A′^ = U AU −^1. (1.26)
The latter definition allows the vector |φ〉 = A|ψ〉 to transform as
|φ′〉 = (U AU −^1 )U |ψ〉 (1.27) = U (A|ψ〉) = U |φ〉.
Summarizing, for a unitary transformation U ,
U †^ = U −^1 , (1.28) |ψ′〉 = U |ψ〉, 〈ψ′| = 〈ψ|U †, A′^ = U AU †, < ψ′|A′|ψ′〉 = 〈ψ|A|ψ〉.
This last line emphasizes the fact that if you transform BOTH the operators and states, the matrix element is unchanged. However, in some cases you transform the states, leaving the operator unchanged, or you might transformed the operator and leave the state unchanged. This leads to new matrix elements. It can be painful to keep track of which objects are being transformed. This is akin to performing a rotation, where rotating an object by φ or rotating the coordinate system by −φ has the same effect. Also, if you rotate both the coordinate system and the object by the same angle, nothing changes.
First, we stop and consider what forms of matrix elements might be considered as an observable. Observables must be real and independent of the basis, i.e. all unitary transformations that act
both on the bras and kets and on the operators should leave the observable unchanged. All observables can be expressed as either the expectation of a Hermitian operator,
〈ψ|K|ψ〉, (1.29)
or as the squared overlap of two states,
|〈φ|U |ψ〉|^2. (1.30)
Instead of describing a state ψ by a vector, one could describe it by a density matrix,
ρψ = |ψ〉〈ψ|, (1.31)
or as a matrix
(ρψ)ij = ψiψ j∗. (1.32)
By inspection, one can see that the density matrix is Hermitian. From the definition of the density matrix, one can write the state |ψ〉 as a vector and the operator A as a matrix, and that
〈ψ|A|ψ〉 =
i,j
ψ∗ i Aij ψj (1.33)
= Tr ρψA,
and that
|〈φ|A|ψ〉|^2 =< ψ|A†|φ〉〈φ|A|ψ〉 (1.34) = Tr ρψA†ρφA.
Thus, density matrices are sufficient to generate all observables. The trace of any product of op- erators, or matrices, is invariant to unitary transformations so the answer should be independent of basis, as long as the density matrices and the operator A are all transformed.
Example 1.2: Density Matrix for a Two-Component System Consider the state
|ψ〉 =
cos θ eiφ^ sin θ
ρψ =
cos θ eiφ^ sin θ
cos θ e−iφ^ sin θ
cos^2 θ e−iφ^ sin θ cos θ eiφ^ sin θ cos θ sin^2 θ
However, density matrices can be more general. If one has a density matrix which is an incoher- ent sum over several states, the resulting diagonalized density matrix could have more than one non-zero element, though the trace would have to remain equal to unity. Thus, density matrices can be used to express non-pure states such as unpolarized beams. For instance, one can define the following density matrix,
ρ =
As with the pure state, the trace remains at unity, but after being diagonalized the strength is spread along the diagonal. This density matrix describes being in state | 1 〉 50% of the time and in state | 2 〉 50% of the time. A way to write a state that is described by this density matrix, would be 1 √ 2
| 1 〉 + eiφ^
with φ being treated as a random phase to justify ignoring the off-diagonal terms in the density matrix. When performing calculations with such a state, one would ignore all terms in the density matrix with a leftover phase, eiφ, to account for the randomness of the phase. Note that a density matrix ρ for an impure state is not a projection operator, i.e. ρ^2 6 = ρ.
Density matrices play an essential part in thermodynamics. In that case, one considers incoher- ent sums over many states weighted by the energy. If one is in a basis where the Hamiltonian is diagonalized, the density matrix takes the form
ρthermal =
e−βE^1 0 · · · 0 0 e−βE^2 · · · 0 .. .
0 · · · 0 e−βEN
i
e−βEi^ ,
where β is the inverse temperature.
One might ask the question whether it makes sense to do quantum mechanics in terms of den- sity matrices rather than with wave functions. Density matrices are Hermitian, and given the constraint that the trace is unity, and N × N density matrix can be represented by N 2 − 1 real numbers. A wave function has N complex components, and given the normalization, have 2 N − 1 components. Of course, there is also a meaningless phase, so a wave function can be physically represented by 2 N − 2 numbers. Clearly, the wave function is more efficient. How- ever, the density matrix can also account for incoherent mixtures of states. This explains the need for additional information.
As an example of a unitary transformation we consider rotations in a two-component system. For spin-1/2 systems the spin operator is
~σ. (1.42)
For two-component systems, all operators can be written as a linear combination of the Pauli σ matrices and the unit matrix.
σz =
, σx =
, σy =
0 −i i 0
As we will see later, under rotations the three Pauli matrices transform like components of a vector. For now, we only notice that each matrix is Hermitian and that when each sigma matrix is squared it gives the unit matrix. Note that the Pauli matrices have the properties,
σ i^2 = 1, (1.44) {σi, σj } = 2δij , [σi, σj ] = 2iijkσk, σiσj = δij + iijkσk.
where the anti-commutator is noted by {A, B} ≡ AB + BA for operators A and B, and the regular commutator is noted by [A, B] ≡ AB − BA. From these expressions one can readily show that for a unit vector nˆ,
(~σ · nˆ)^2 = I. (1.45)
For i 6 = j one can see
σiσj 6 =i = iijkσk, (1.46) σxσy = iσz , σyσz = iσx, σz σx = iσy.
For a state with spin-up or spin-down along the z axis, we choose the basis
Rotations by an angle θ are given by:
R(~θ) = e−i ~θ·~σ/ 2 , (1.48)
where the direction of θ~ is along the axis of rotation nˆ, or equivalently, ~θ = θ nˆ.
From performing a Taylor expansion, and using Eq. (1.44),
e−iθ~σ·n/ˆ^2 = cos(θ/2) − i sin(θ/2)~σ · n.ˆ (1.49)
This trick comes in handy for a large number of physics examples, not just rotations.
Example 1.3: Rotating Spin-1/2 Systems Consider a fermion’s spin to originally be in the spin-up state, where “up” is defined by the z axis.
|ψ〉 = | ↑〉 =
additional factor of − 1. The fact that both kinds of transformations represent the same rotations is something that will be discussed during the brief discussion of group theory in 4.2.
Electromagnetic waves are comprised of quantized photons. For an electromagnetic wave trav- eling in some direction, the light can be polarized, with the electric field oscillating being in some direction perpendicular to the direction of propagation. If the light is moving in the z direction, the field might be oscillating in the x direction. For a photon the spin is quantized into two pos- sible polarization states. For a photon propagating in the z direction, those states might be |x〉 or |y〉, which would correspond to light polarized in the x or the y directions. The two states might be represented by the vectors
|x〉 =
, |y〉 =
Note that one switches from the |x〉 to the y〉 if rotated by 90 ◦. This is in contrast to spin- half particles, where one rotates by 180 ◦^ to switch to the orthonormal state. Of course, the other difference is that for photons we are considering only rotations about the direction of propagation, whereas for spin-half particles, one must consider three dimensions of rotations.
The wave function for photons moving in the z direction has the form
φ(x, t) = e−iωt+ikz
a b
where the vector is known as the polarization vector, which we consider to be normalized a^2 + b^2 = 1. The following polarization vectors describe linearly polarized photons,
( 1 0
, linear polarization along the x axis ( 0 1
, linear polarization along the y axis
√^1 2
, linear polarization along a direction 45◦^ from x axis.
Another linear combination of |x〉 and y〉 states are right- and left-circularly polarized photons. Those states are
(|x〉 + i|y〉) =
i
(|x〉 − |y〉) =
−i
To see why these are called linear polarized photons, one uses the fact that the electric fields are proportional to the corresponding components of the wave function. Looking at those points
with z = 0, for the polarization with upper and lower components a and b, the electric fields behave as
Ex = ReE 0 ae−iωt, (1.55) Ey = ReE 0 be−iωt.
For a = 1/
2 and b = i
2 , this becomes
Ex =
cos ωt, Ey =
sin ωt. (1.56)
At t = 0 the electric field is pointed along the positive z axis, but after one fourth of a pe- riod it is pointed along the positive y axis. At this fixed value of z, the direction of the field is then rotating about the z axis and is called right circularly polarized. If one had considered
the polarization vector a = 1/
2 , b = −i/
2 , the polarization vector would have rotated in the opposite direction, and be called left-circularly polarized. As long as one fixes the mo- mentum of the photon, there are two polarizations and the polarization states can be described in a two-component basis. If one wants the polarization vector to describe the direction of the polarization even when the photon is propagating away from the z axis, polarization vectors must then become three-component vectors. But because polarizations are perpendicular to the direction of propagation, pˆ, one must project away the contribution of those vectors in the pˆ direction.
Example 1.4: Photon Polarization Photons are traveling along the z axis and are polarized along the x axis. They pass through a polarization filter which only permits passage of photons whose polarization is at an angle 30 ◦^ to the x axis. What fraction of photons pass through the filter. Solution : Consider a single photon. The original polarization vector is
|φ = 0〉 =
and the state that passes through the filter is
|φ = 30◦〉 =
cos φ sin φ
(^21) 2
The probability of being in the new state is the squared overlap,
|〈φ = 30◦|φ = 0〉|^2 =