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Handout 5 from PHYS 560
Northern Illinois University
Fall 2007
(Dated: version printed September 12, 2007)
Brief (well fairly expanded) notes on concepts and topics from Chapter 4, section 1 and 2 and 3.
Shankar’s Chapter Four: The postulates - a general discussion
4.1. THE POSTULATES
- The postulates (be aware of comparison to the classical equivalents given in book). Note that some texts have
more than 4 postulates, but it is merely separating them differently.
- State of particle represented by |ψ(x)〉 (in physical Hilbert space)
- Variables x and p are Hermitian operators
X and
P
- matrix elements expressed in eigenbasis of
X are 〈x
′ |
X|x〉 = xδ(x − x
′ ) and 〈x
′ |
P |x〉 = −iℏδ
′ (x − x
′ )
- matrix elements expressed in eigenbasis of
X are 〈p
′ |
X|p〉 = iℏδ
′ (p − p
′ ) and 〈p
′ |
P |p〉 = pδ(p − p
′ )
- Operators corresponding functions ω(x, p) are replaced with Hermitian operators
X,
P ) ⇔ ω(x →
X, p →
P )
- Observables are associated with Hermitian operators.
- If particle in state |ψ〉, then observation (measurement) of
Ω will yield ONE of its eigenvalues ω. If another
particle in state |ψ〉 is measured, also measurement will yield ONE of the eigenvalues of
Ω but not necessarily
the same one. The probability of measuring any particular eigenvalue ωi is P (ωi) ∝ |〈ωi|ψ〉|
2 .
And, upon the measurment, the state vector (i.e., wave function) describing the particle will no longer be
|ψ〉, but will be the |ω i
〉 that is
Ω’s eigenket associated with the eigenvalue ω i
- State vectors |Ψ〉 obey Schrodinger equation.
H is the quantized Hamiltonian operator that is related to H for
classical problem (with the x and p changed to corresponding operators
X and
P ).
iℏ
d
dt
|Ψ(t)〉 =
H|Ψ(t)〉
4.2. DISCUSSION OF POSTULATES I, II, AND III
- Ponder the implicaton from the postulates above: Why was it so important from the first chapter to become
fluent in expanding a ket in arbitrary basis sets? Also, how does it apply to the whole endeavour that the eigen-
kets of an operator (assuming non-degenerate eigenvalues) form a complete basis set, (and that of a Hermitian
operators, a complete orthonormal basis set)?
i
|ω i
〉〈ω i
|ψ〉
2 = 〈ψ|ω〉〈ω|ψ〉
- Ideal measurement (designed to not disturb the system) still by virtue of measuring quantity described by
means that final state vector of particle will be the eigenket of
Ω. The state changes! (Unless, of course, |ψ〉 is
already an eigenket of
- Given |ψ〉 prepared in particular state. Probability that particular ωi will be measured is given above.
- Implication that logically follows is that AFTER first measurement is done, system is in |ω i
〉 state. All
subsequent measurements of
Ω will give ω i
with 100% probability!
(until we alter the state vector by measuring
- Ponder the above statements to see that it provides way to ‘prepare’ system in a state that is a mixed state.
- Assume that observables associated with
Λ and
Ω and that these operators do not commute. Therefore,
they do NOT have the same eigenkets and at least two terms will be non-zero in following expansion
|λ i
j
|ω j
〉〈ω j
|λ i
〉. And vice versa.
- Start with state |ψ〉. Perhaps we do not know the probabilities, however, upon measuring parameter
associated with
Ω we measure a particular ω i
- Now, if we were to measure parameters associated with
Λ, the system is in a mixed state with respect to
the eigenkets |λ〉.
- Ponder the above statements for measurements that correspond to observables with operators that commute.
Let [
Γ] = 0. Then, the same eigenkets describe both operators.
- For reminder, I will denote the eigenkets by |ω, γ〉 merely to emphasize that the same eigenket is in the
|ωi〉 and |γi〉 eigenket equations below. Different values for the eigenvalues but same eigenkets.
Ω|ωi〉 = ωi|ωi〉 ≡ ωi|ωi, γi〉 and
Γ|γi〉 = γi|γi〉 ≡ γi|ωi, γi〉
- When a measurement of a system gives a particular ω i
, then the system is in the ket |ω i
〉 ≡ |ω i
, γ i
〉. Just as
any subsequent measurements of
Ω are performed, 100% of the time that particular ωi is measured. And,
if
Γ is measured, 100% of the time the associated, γ i
is measured.
- Until things get mixed up again by measuring quantity associated with
Λ which does not commute with
or
- The points above are related to the idea of Incompatible and Compatible Variables, that is, if the asso-
ciated operators commute, the observables/variables are compatible, and if the operators do not commute, the
variables are incompatible.
Occasionally there may be some common eigenkets between two operators, even though the operators do
not commute. (If all eigenkets were the same, then obviously the two operators would commute). However,
doesn’t change the way we would use the postulates.
- The points above are also related to the idea that by measuring, we Collapse the State vector. More on
this later.
- Normalization, typically we will want ‘normalized’ states. Then
i
P (ω i
) = 1, and other things will be auto-
matically a bit easier.
- Example, a Normalized state |ψ〉 composed of two eigenkets |ω 1
〉 and |ω 2
|ψ〉 =
α|ω 1
〉 + β|ω 2
(|α|
2
2 )
1 / 2
Complications
- Operator is degenerate.
- The projection operator includes those parts of the eigenspace with the degenerate eigenvalues ω.
Eigenkets are distinct and different, even though the eigenvalues are the same (or at least someone
should have gone to the trouble to complete the eigenket basis set)
P
ω
over i degeneracies
|ω, i〉〈ω, i|
P (ω) =
over i degeneracies
|〈ω, i|ψ〉|
2
But if the initial state were unknown (i.e., the coefficients of the expansion are not known), and a measure-
ment gave a particular results ω that was part of a degenerate subspace in
Ω, we only know that the final state
is mixture. For example, the α’s are unknown below, and are just denoting a mixture of the degenerate states.
|ψ〉 =
over i degen
α i
|ω i
over i degen
α
2
i
1 / 2
However, then the fun begins as theorists and experimentalists can apply various models of the system and
its interactions, and whoever correctly predicts the measurements wins.
The Uncertainty
- Remember that the Expectation Value (that is, the ‘average’ value of a measurement of an observable
when measured on an ensemble of systems prepared in the same initial state |ψ〉) is 〈
Ω〉 = 〈ψ|
Ω|ψ〉.
- What is root-mean-square deviation or standard deviation of a set of measurements?
σy =
[
N ∑
n
(yn − y)
2
]
1 / 2
where y is the average of the measurements of y. Note that y is in our notation, 〈y〉.
- If we rewrite this a little, noting that for all N measurements of y, I may have only K different possible
values of y, each distinct value denoted by y i
. Let N i
be the the number of times a particular y i
value is
measured and N the total number of measurements, then Ni/N is the weighting (the probability, P (yi))
of a particular yi in the set. Using this notation, I can rewrite the exact same equation above as
σy =
[
K ∑
i
P (yi)(yi − y)
2
]
1 / 2
- This tranlates directly over to our dealings with measurements, although typically it wll be called the
root-mean-square deviation or y, or the Fluctuations in y, or (important sound effects please go
here) the Uncertainty in y, ∆y.
- Putting this in our notation for our observables, (for discrete eigenvalue spectrums or for continuous eigenvalue
spectrum)
[
i
P (ω i
)(ω i
2
]
1 / 2
or
[∫
P (ω)(ω − 〈
2
]
1 / 2
- This equals the definition, which is ∆
[
〈ψ|(
2 |ψ〉
]
1 / 2
- Note that this notation, ∆
Ω can be a bit confusing, as it is not an operator, (nor is it two operators ∆ and Ω
acting is succession), but it denotes the scalar quantity as calculated by the above equations. Sometimes it will
be noted by ∆ω.
- Finally, there is much easier form to remember instead, ∆
[
2 〉 − 〈
2
]
1 / 2
. Derive it as practice in playing
with the notation. (See page 136 Shankar).
Finding Uncertainty as applied to get the famous result on ∆
ˆ X∆
ˆ P , and coincidentally, working with continuous function
The ∆
ˆ X construction
- A gaussian function with width ∆, normalized so that
|ψ(x)|dx = 1, and positioned at x = a is given by
ψ(x) =
(π∆
2 )
1 / 4
exp
−(x − a)
2 /2∆
2
and it is valid in range from x = −∞ to x = +∞.
- As a ket |ψ〉, this is the x representation, 〈x|ψ〉. By the way, this is good function to know along with its
normalization. Gaussians pop up everywhere in life.
- Note, the eigenkets of the
X operator are |x〉. The eigenvalues are x. Thus, the eigenvalue spectrum for
X is continuous and P (x) = |〈x|ψ〉|
2 must be interpreted and used as a probability density.
P (x)dx = |〈x|ψ〉|
2 dx is the probability of measuring the position of a particle between x and x + dx
(when the initial state vector of the particle is described by ket |ψ〉).
X for the particle described by the |ψ〉 noted above. It should be equal to (〈
X
2 〉 − 〈
X〉
2 )
1 / 2 , also as
noted even earlier.
X〉 =
∞
−∞
dx〈ψ|x〉〈x|
X|ψ〉
∞
−∞
∞
−∞
dxdx
′
〈ψ|x〉〈x|
X|x
′
〉〈x
′
|ψ〉
∞
−∞
∞
−∞
dxdx
′
〈ψ|x〉(xδ(x − x
′
))〈x
′
|ψ〉 =
∞
−∞
dx〈ψ|x〉x〈x|ψ〉 =
∞
−∞
dx ψ(x)
∗
x ψ(x) =
∞
−∞
dx |ψ(x)|
2
x
Do the same for 〈
X
2 〉. If one wanted to go through each insertion of identity operators without taking
shortcuts that experienced quantum operators eventually recognize, it would eventually give something that
looked like this,
〈ψ|
X
2 |ψ〉 =
∞
−∞
∞
−∞
∞
−∞
dxdx
′ dx
′′ 〈ψ|x〉〈x|
X|x
′′ 〉〈x
′′ |
X|x
′ 〉〈x
′ |ψ〉 =
∞
−∞
dx|ψ(x)|x
2
The ∆
ˆ P construction
- Perform it in momentum space. Note that 〈p|ψ〉 =
ψ(p). We only have a representation in x, ψ(x). But,
〈p|ψ〉 =
dx〈p|x〉〈x|ψ〉. Note that p = ℏk and thus we already know (Class Notes set 4 regarding 〈k|x〉) that
〈p|x〉 =
e
−ipx/ℏ
√
2 πℏ
and 〈x|p〉 =
e
+ipx/ℏ
√
2 πℏ
. So
ψ(p) and its complex conjugate, are related to the Fourier transforms of
ψ(x) and ψ
∗ (x). Example, 〈p|ψ〉 =
dx
e
−ipx/ℏ
√
2 πℏ
ψ(x) =
ψ(p). Assuming that we can do those, continue.
〈ψ|
P |ψ〉 =
dpdp
′ 〈ψ|p〉〈p|
P |p
′ 〉〈p
′ |ψ〉
dpdp
′ ˜ ψ
∗ (p) pδ(p − p
′
)
ψ(p
′
) =
dp p|
ψ(p)|
2
- Perform it in position space. Then, it would be
〈ψ|
P |ψ〉 =
dxdx
′ 〈ψ|x〉〈x|
P |x
′ 〉〈x
′ |ψ〉
dxdx
′
〈ψ|x〉 (−iℏδ
′
(x − x
′
)) 〈x
′
|ψ〉
= −iℏ
dx〈ψ|x〉
[
d
dx
〈x|ψ〉
]
= −iℏ
dxψ
∗
(x)
[
dψ(x)
dx
]
Note, the order stays quite important here. It is not equal to −iℏ
dx d(|ψ(x)|
2 )/dx.
- That final evaluation of 〈
P 〉 = 0 for the particular gaussian wavefunction ψ(x) is left as exercise. (Exercise 4.2.
It is zero for any real wavefunction, i.e., when ψ(x) = ψ
∗ (x)).
P
2 〉, similar shenanigans can be used inserting identity operators until all the terms are terms
that we know.
- This is denoted by 〈Ω〉. (Sort of a combination of the (〈 〉) notation denoting expectation value calculation,
and the ( ) notation of taking the mean (classical averaging).
i
ρi〈i|
Ω|i〉
- Ponder the derivation (Equation 4.2.22) for the Expectation Value of the distribution. (and convince yourself
that you can repeat the steps) proving
〈Ω〉 = Tr(
Ωˆρ)
This is the expected value that an experimenter would report for the observable
Ω if she performed the
measurement on the ensemble.
- Ponder derivation (equations following 4.2.22) and convince yourself that you can repeat it. What is the
probability with a particular ensemble for obtaining a particular measurement ωα, (assuming that the experiment
could pull out one system at random and measure it independently.)
P (ωα) = Tr(Pω α
ρˆ)
caution: distinguish between notation for probability P and projection operator P.
- In the derivations, is it necessary that the |i〉 be eigenkets of
- In any measurement of any particular one system, would the measurement leave that system in an eigenket
of
- When calculating the P (ω α
), is this the probability of measuring a particular eigenvalue of
Ω or the
probability of measuring a particular eigenvalue of
Λ for which the kets were an orthonormal basis?
- Very useful properties to memorize and derive.
ρˆ
†
= ˆρ , Trˆρ = 1 , Trˆρ
2
≤ 1
ρˆ =
k
I : for uniform distribution over k states, (i.e., same number in each state)
ρˆ
2
= ˆρ , Trˆρ
2
= 1 : for pure ensembles
Generalization to more degrees of freedom
- We come back to this and related questions in Chapter 10 Systems of N Degrees of Freedom.
- Also some interesting related discussion on issues arising from ‘Quantizing’ a system (converting H(x, p) →
H(
X,
P ) when trying to work in spherical coordinates. This is in Chapter 7 The Harmonic Oscillator, for
the last few pages of section 4. (on page 214 (in second edition), and right after exercise 7.4.10 in 1st and 2nd
editions). No wonder we usually do the changes when it is ‘easy’ and in cartesian coordinates.
- More degrees of freedom can mean
- More than one particles but confined in one dimension
- One particle but not confined to one dimension, (e.g., 2D or 3D... ).
- Or of course, both.
- Postulate II changes as (note the cartesian coordinate formulation)
Corresponding to N Cartesian coordinates x 1 ,... , xN in classical, exist N mutually commuting operators
X
1
X
N
. Eigenbasis is coordinate basis |x 1
, x 2
, · · · , x N
- Normalization 〈x 1 , x 2 , · · · , xN |x
′
1
, x
′
2
, · · · , x
′
N
〉 = δ(x 1 − x
′
1
)δ(x 2 − x
′
2
) · · · δ(xN − x
′
N
- Similarly 〈x 1 , x 2 , · · · , xN |ψ〉 = ψ(x 1 , x 2 , · · · , xN )
- 〈x 1 , x 2 , · · · , xN |
Xi|ψ〉 = xiψ(x 1 , x 2 , · · · , xN )
, x 2
, · · · , x N
P
i
|ψ〉 = −iℏ
∂
∂xi
ψ(x 1
, x 2
, · · · , x N
- Dependent dynamical variable ω(xi, pi) →
Xi,
Pi).
, x 2
, · · · x N
2 dx 1
dx 2
· · · dx N
is the probability that particle coordinates lie between x 1
, x 2
, · · · , x N
and
x 1 + dx 1 , x 2 + dx 2 , · · · , xN + dxN
- Note the point that the postulate is stated in terms of Cartesian coordinates, since those have the simple operator
assignments xi →
Xi and pi →
Pi → iℏ
∂
∂xi
that is, what if we were already in spherical coordinates? What is 〈r|
Pr |ψ〉? Is it −iℏ
∂
∂r
? More discussion
on this topic and some subtleties in Chapter 7 section 4 after exercise 7.4.10.
4.3. THE SCHRODINGER EQUATION (DOTTING YOUR i’S AND CROSSING YOUR ℏ’S
d
dt
|Ψ(t)〉 =
H|Ψ(t)〉
Often, one will see Ψ as statevector with time dependence, and ψ for time independent solution. However,
it will usually be clear from the context, even if the notation is not consistent.
- Examples of converting H(x, p) to
H(
X,
P ),
- Eventually we will have problems with spin,
S, (like spin on electrons) which have no classical counterpart,
(i.e., what is
S in
X
i
and
P
j
). See Chapter 14.
- Simple examples such as H = p
2 / 2 m + mω
2 x
2 /2 convert to
H =
P
2 / 2 m + mω
X
2 / 2
- Example of more complicated Hamiltonian that requires symmetrization because the term − 2
P ·
A is
ambiguous as the two operators do not commute, however, the classical equation it was quantized from
|p − (q/c)A|
2 does not specify order from any fundamental physics involved. Answer becomes to keep the
same functional form
P ˆA but to convert it into form that does not depend on order.
H =
|p − (q/c)A(r, t)|
2
2 m
H =
P ·
P − (q/c)
P ·
A − (q/c)
A ·
P +
A ·
A
2 m
φ (2)
H has No explicit time dependence then |E(t)〉 = |E〉e
−iEt/ℏ The |E〉 are the stationary states. (In the
harmonic oscillator problem, also called the normal modes. In this chapter, Shankar is calling |E〉 normal
modes, mostly because he motivated some equations in this sub section with referring to derivations in Chapter
1 using a simple harmonic oscillator, however, I think it is a bit non-standard to call them normal modes unless
we are specifying
H that have a bit of harmonic oscillator logic. The results are valid anywhere.)
We have seen (review discussion Chapter one) that for NO explicit time dependence of an operator, then
the time dependent eigenkets are merely,
|Ψ(t)〉 = U (t)|Ψ(0)〉
H|E〉 = E|E〉 is time-independent Schrodinger equation
It is just an eigenvalue equation. Find the eigenvalues and eigenkets.
H is hermitian (or better be) and those eigenkets from basis set that spans the relevant space.
- Be able to follow and repeat derivation that leads to equation 4.3.13,
U (t) =
E
|E〉〈E|e
−iEt/ℏ
- If E is continuous, then the sum becomes
U (t) =
dE|E〉〈E|e
−iEt/ℏ
Choosing a basis set for solving Schrodinger’s equation
- The choices typically are either position or momentum representations, however, we will come across a few cases
where can solve it without referring to either of these bases but staying in the energy representation.
- Obviously, choose the basis that is ‘simpler’ to solve for particular Hamiltonian. Because a solution is a solution,
one can convert back to the other basis if necessary. The solution is STILL a solution.
- We often work in x basis set, because we think in x, and because many problems have a classical V (ˆx that is
rather in
X already (and the only operator that depends on
P is the same for many Hamiltonians, as it enters
only in the kinetic energy term
P / 2 m. However, it is possible that V can depend on momentum or velocity
(friction?), although I don’t remember any examples that we solve in this class.
- Harmonic oscillator will be solved three ways. X, P , and E.
- However, typically, constant force potentials are a bit easier to solve in momentum basis, of the type where
V (x) = f x.
