Uncertainty Principle and Compatible Operators

Chapter 6

Uncertainty principle and

Compatible operators

c© B. Zwiebach

The uncertainty of a Hermitian operator on a state of a quantum system vanishes ifand only if the state is an eigenstate of the operator. We derive the Heisenberg uncertaintyprinciple, which gives a lower bound for the product of uncertainties of two hermitian oper-ators. When one of the operators is the Hamiltonian we are led to energy-time uncertaintyrelations. The uncertainty inequality is used to derive rigorous lower bounds for the ener-gies of certain ground states. We discuss diagonalization of operators and prove the SpectralTheorem, which states that one can construct an orthonormal basis of eigenvectors for anyHermitian or, more generally, any normal operator. We prove that commuting Hermitianoperators can be simultaneously diagonalized, discussing in detail the issue of degeneracy.Finally, we explain what is meant by a complete set of commuting observables.

6.1 Uncertainty defined

Observables are associated to Hermitian operators. Given one such operator A we can useit to measure some property of the physical system, as represented by a state Ψ. If the stateΨ is an eigenstate of the operator A, we have no uncertainty in the value of the observable,which coincides with the eigenvalue of A. We only have uncertainty in the value of theobservable A if the physical state is not an eigenstate of A, but rather a superposition ofvarious eigenstates with different eigenvalues.

We want to define the uncertainty ∆A(Ψ) of the Hermitian operator A on the state Ψ.This uncertainty should vanish if and only if the state is an eigenstate of A. The uncertainty,moreover, should be a real number. In order to define such uncertainty we first recall thatthe expectation value of A on the state Ψ, assumed to be normalized, is given by

〈A〉 = 〈Ψ|A|Ψ〉 = 〈Ψ, AΨ〉 . (6.1.1)

121

122 CHAPTER 6. UNCERTAINTY PRINCIPLE AND COMPATIBLE OPERATORS

The expectation 〈A〉 is guaranteed to be a real number since A is Hermitian. We then definethe uncertainty ∆A(Ψ) as the norm of the vector obtained by acting with (A−〈A〉I), whereI is the identity operator, on the physical state Ψ:

∆A(Ψ) ≡∣∣∣

(A− 〈A〉I

)Ψ∣∣∣ . (6.1.2)

The uncertainty, so defined, is manifestly non-negative. We can quickly see that zerouncertainty means the state is an eigenstate of A. Indeed, a state of zero norm must be thezero state and therefore

∆A(Ψ) = 0 →(A− 〈A〉I

)Ψ = 0 → AΨ = 〈A〉Ψ . (6.1.3)

Since 〈A〉 is a number, the last equation shows that Ψ is an eigenstate of A. You shouldalso note that 〈A〉 is, on general grounds, the eigenvalue. Taking the eigenvalue equationAΨ = λΨ and forming the inner product with another Ψ we get

〈Ψ, AΨ〉 = λ〈Ψ,Ψ〉 = λ → λ = 〈A〉. (6.1.4)

Alternatively, if the state Ψ is an A eigenstate, we now know that the eigenvalue is 〈A〉 andtherefore the state (A − 〈A〉I)Ψ vanishes and its norm is zero. We have therefore shownthat

The uncertainty ∆A(Ψ) vanishes if and only if Ψ is an eigenstate of A . (6.1.5)

To compute the uncertainty one usually squares the expression in (6.1.2) so that

(∆A(Ψ))2 =⟨ (A− 〈A〉I

)Ψ ,(A− 〈A〉I

)Ψ 〉 (6.1.6)

Since the operator A is assumed to be Hermitian and consequently 〈A〉 is real, we have(A − 〈A〉I)† = A − 〈A〉I, and therefore we can move the operator on the first entry ontothe second one to find

(∆A(Ψ))2 =⟨Ψ ,(A− 〈A〉I

)2Ψ 〉 . (6.1.7)

While this is a reasonable form, we can simplify it further by expansion

(∆A(Ψ))2 =⟨Ψ ,(A2 − 2〈A〉A + 〈A〉2I

)Ψ 〉 = 〈Ψ, A2Ψ〉 − 2〈A〉2 + 〈A〉2 . (6.1.8)

The last two terms combine and we find

(∆A(Ψ))2 = 〈A2〉 − 〈A〉2 . (6.1.9)

6.1. UNCERTAINTY DEFINED 123

Figure 6.1: A state Ψ and the one-dimensional subspace UΨ generated by it. The projection of AΨto UΨ is 〈A〉Ψ. The norm of the orthogonal component (AΨ)⊥ is the uncertainty ∆A(Ψ).

Since the left-hand side is greater than or equal to zero, this shows that the expectationvalue of A2 is larger than the square of the expectation value of A:

〈A2〉 ≥ 〈A〉2 . (6.1.10)

Geometric interpretation. An interesting geometrical interpretation of the uncertaintyis obtained by thinking about the relation between the states Ψ and AΨ. Let UΨ be theone-dimensional vector subspace generated by Ψ. Consider now the state AΨ. Since Ψ isnot assumed to be an eigenstate of A, the state AΨ state does not generally lie on UΨ. Wenow claim that

1. The orthogonal projection of AΨ to UΨ is in fact 〈A〉Ψ.

2. The component of AΨ in the orthogonal subspace U⊥Ψ has norm is equal to ∆A.

To prove these claims we consider the orthogonal projector PUΨto UΨ:

PUΨ= |Ψ〉〈Ψ| . (6.1.11)

Claim number one is quickly verified by calculating the projection of AΨ to UΨ:

PUΨA|Ψ〉 = |Ψ〉〈Ψ|A|Ψ〉 = |Ψ〉〈A〉 . (6.1.12)

Moreover, the vector A|Ψ〉 minus its projection to UΨ must be a vector |(AΨ)⊥〉 orthogonalto |Ψ〉

A|Ψ〉 − 〈A〉|Ψ〉 = |(AΨ)⊥〉 , (6.1.13)

as is easily confirmed by taking the overlap with the bra Ψ. Since the norm of the aboveleft-hand side is the uncertainty, we confirm that ∆A = |(AΨ)⊥|, as claimed. These resultsare illustrated in Figure 6.1. The uncertainty is the shortest distance from AΨ to thesubspace UΨ.


6.2 The Uncertainty Principle

The uncertainty principle is an inequality that is satisfied by the product of the uncertaintiesof two Hermitian operators that fail to commute. Since the uncertainty of an operator on anygiven physical state is a number greater than or equal to zero, the product of uncertaintiesis also a real number greater than or equal to zero. The uncertainty inequality gives us alower bound for the product of uncertainties. When the two operators in question commute,the uncertainty inequality gives no information: it states that the product of uncertaintiesmust be greater than or equal to zero, which we already knew.

Let us state the uncertainty inequality. Consider two Hermitian operators A and B anda physical state Ψ of the quantum system. Let ∆A and ∆B denote the uncertainties of Aand B, respectively, in the state Ψ. Then we have

(∆A)2(∆B)2 ≥(⟨

Ψ| 12i[A,B]

∣∣Ψ⟩)2

. (6.2.1)

The left hand side is a real, non-negative number. For this to be consistent inequality, theright-hand side must also be a real number that is not negative. Since the right-hand sideappears squared, the object inside the parenthesis must be real. This can only happen forall Ψ if the operator

1

2i[A,B] (6.2.2)

is Hermitian. For this first note that the commutator of two Hermitian operators is anti-Hermitian:

[A,B]† = (AB)† − (BA)† = B†A† −A†B† = BA−AB = −[A,B] . (6.2.3)

The presence of the i then makes the operator in (6.2.2) Hermitian. Note that the uncer-tainty inequality can also be written as

Uncertainty inequality: ∆A∆B ≥∣∣∣

⟨Ψ| 12i [A,B]

∣∣Ψ⟩∣∣∣ . (6.2.4)

The bars on the right-hand side denote absolute value. Here the i is just for convenience,and the right hand side is just the norm of the expectation value of 1

2i [A,B] on the state Ψ.

Before we prove the theorem, let’s do the canonical example!

Example. Taking A = x and B = p gives the position-momentum uncertainty relationyou have certainly worked with:

∆x∆p ≥∣∣〈Ψ| 12i [x, p]|Ψ〉

∣∣ . (6.2.5)

Since [x, p]/(2i) = ~/2 and Ψ is normalized, we get

∆x∆p ≥ ~

2 . (6.2.6)

6.2. THE UNCERTAINTY PRINCIPLE 125

We are interested in the proof of the uncertainty inequality for it gives the information thatis needed to find the conditions that lead to saturation.

Proof. We define the following two states:

|f〉 ≡ (A− 〈A〉I)|Ψ〉|g〉 ≡ (B − 〈B〉I)|Ψ〉 .

(6.2.7)

Note that by the definition (6.1.2) of uncertainty,

〈f |f〉 = (∆A)2 ,

〈g|g〉 = (∆B)2 .(6.2.8)

The Schwarz inequality immediately furnishes us an inequality involving precisely the un-certainties

〈f |f〉〈g|g〉 ≥ |〈f |g〉|2 , (6.2.9)

and therefore we have

(∆A)2(∆B)2 ≥ |〈f |g〉|2 = (Re〈f |g〉)2 + (Im〈f |g〉)2 . (6.2.10)

Our task is now to compute each of the two terms in the above right-hand side. Forconvenience we introduce

A ≡ (A− 〈A〉I) ,B ≡ (B − 〈B〉I) .

(6.2.11)

We now compute:

〈f |g〉 = 〈Ψ|AB|Ψ〉 = 〈Ψ|(A− 〈A〉I)(B − 〈B〉I)|Ψ〉= 〈Ψ|AB|Ψ〉 − 2〈A〉〈B〉 + 〈A〉〈B〉 ,

(6.2.12)

so that simplifying we have

〈f |g〉 = 〈Ψ|AB|Ψ〉 = 〈Ψ|AB|Ψ〉 − 〈A〉〈B〉,〈g|f〉 = 〈Ψ|BA|Ψ〉 = 〈Ψ|BA|Ψ〉 − 〈B〉〈A〉.

(6.2.13)

where the second equation follows because |f〉 and |g〉 go into each other as we exchange Aand B. We now use this to find a nice expression for the imaginary part of 〈f |g〉:

Im〈f |g〉 = 12i (〈f |g〉 − 〈g|f〉) = 1

2i〈Ψ|[A,B]|Ψ〉 . (6.2.14)

For the real part the expression is not that simple because the product of expectationvalues do not cancel. It is best to write the real part as the anticommutator of the checkedoperators:

Re〈f |g〉 = 12(〈f |g〉+ 〈g|f〉) = 1

2〈Ψ|{A, B}|Ψ〉 . (6.2.15)


Back in (6.2.10) we get

(∆A)2(∆B)2 ≥(

〈Ψ| 12i [A,B]|Ψ〉)2

+(

〈Ψ|12{A, B}|Ψ〉)2. (6.2.16)

This can be viewed as the most complete form of the uncertainty inequality. It turns out,however, that the second term on the right hand side is seldom simple enough to be of use,and many times it can be made equal to zero for certain states. At any rate, the term ispositive or zero so it can be dropped while preserving the inequality. This is often done,thus giving the celebrated forms (6.2.1) and (6.2.4) that we have now established. �

Now that we have a proven the uncertainty inequality, we can ask: What are the conditionsfor this inequality to be saturated? If the goal is to minimize uncertainties, under whatconditions can we achieve the minimum possible product of uncertainties? As the proofshows, saturation is achieved under two conditions:

1. The Schwarz inequality is saturated. For this we need |g〉 = β|f〉 where β ∈ C.

2. Re(〈f |g〉) = 0, so that the last term in (6.2.16) vanishes. This means that 〈f |g〉 +〈g|f〉 = 0.

Using |g〉 = β|f〉 in Condition 2, we get

〈f |g〉+ 〈g|f〉 = β〈f |f〉+ β∗〈f |f〉 = (β + β∗)〈f |f〉 = 0 , (6.2.17)

which requires β + β∗ = 0 or that the real part of β vanish. It follows that β must bepurely imaginary. So, β = iλ, with λ real, and therefore the uncertainty inequality will besaturated if and only if

|g〉 = iλ|f〉, λ ∈ R . (6.2.18)

More explicitly this requires

Saturation Condition: (B − 〈B〉 I)|Ψ〉 = iλ (A − 〈A〉I)|Ψ〉 . (6.2.19)

This must be viewed as an equation for Ψ, given any two operators A and B. Moreover,note that 〈A〉 and 〈B〉 are constants that happen to be Ψ dependent. What is λ, physically?Well, the norm of λ is actually fixed by the equation. Taking the norm of both sides we get

∆B = |λ|∆A → |λ| =∆B

∆A. (6.2.20)

The saturation condition can be written as an eigenvalue equation. By moving the Aoperator to the left-hand side and the B expectation value to the right-hand side we have

(B − iλA)|Ψ〉 = (〈B〉 − iλ〈A〉)|Ψ〉 . (6.2.21)

6.3. THE ENERGY-TIME UNCERTAINTY 127

Since the factor multiplying |Ψ〉 on the right-hand side is a number this is indeed an eigen-value equation. It is a curious one, however, since the operator B − iλA is not hermitian.The eigenvalues are not real and they are not expected to be quantized either. Coherentstates, are in fact eigenstates of non-hermitian operators. The presence of expectation val-ues on the right-hand side may obscure the fact that this is a standard eigenvalue problem.Indeed, consider the related equation

(B − iλA)|Ψ〉 = (b− iλa〉)|Ψ〉 . (6.2.22)

with b, a, real constants to be determined. This is the familiar eigenvalue problem. Takingexpectation value by applying 〈Ψ| to both sides of the equation we get

〈B〉 − iλ〈A〉 = b− iλa . (6.2.23)

Since a, b, 〈A〉, 〈B〉, and λ are real this equation implies b = 〈B〉 and a = 〈A〉. Therefore,when solving (6.2.21) you can simply take 〈A〉, 〈B〉 to be numbers and if you get a solutionthose numbers will indeed be the expectation values of the operators in your solution!

The classic illustration of this saturation condition is worked out for the x, p uncertaintyinequality ∆x∆p ≥ ~/2. You will find that gaussian wavefunctions satisfy the saturationcondition.

6.3 The Energy-Time uncertainty

A more subtle form of the uncertainty relation deals with energy and time. The inequality issometimes stated vaguely in the form ∆E∆t & ~. In here there is no problem in defining ∆Eprecisely, after all we have the Hamiltonian operator, and its uncertainty ∆H is a perfectcandidate for the ‘energy uncertainty’. The problem is time. Time is not an operator inquantum mechanics, it is a parameter, a real number used to describe the way systemschange. Unless we define ∆t uncertainty in a precise way we cannot hope for a well-defineduncertainty relation.

We describe a familiar setting in order to illustrate the spirit of the inequality. Considera photon that is detected at some point in space, as a passing oscillatory wave of exactduration T . Without any quantum mechanical considerations, we can ask the observer forthe value of the angular frequency ω of the pulse. In order to answer our question theobserver will attempt to count the number N of complete oscillations of the waveform thatwent through. Of course, this number N is given by T divided by the period 2π/ω of thewave:

N =ω T

2π. (6.3.1)

The observer, however, will typically fail to count full waves, because as the pulse getsstarted from zero and later on dies off completely, the waveform mayl cease to follow thesinusoidal pattern. Thus we expect an uncertainty ∆N & 1. Given the above relation, andwith T known exactly, this implies an uncertainty ∆ω in the value of the angular frequency

∆ω T & 2π . (6.3.2)


This is all still classical, the above identity is something electrical engineers are well awareof. It represents a limit on the ability to ascertain accurately the frequency of a wave thatis observed for a limited amount of time. This becomes quantum mechanical if we speak ofa single photon, whose energy is E = ~ω. Then ∆E = ~∆ω, so that multiplying the aboveinequality by ~ we get

∆E T & h . (6.3.3)

In this uncertainty inequality T is the duration of the pulse. It is a reasonable relation butthe presence of & betrays its lack of precision.

A way out was found by Russian physicists L. Mandelstam and Tamm shortly after theformulation of the uncertainty principle. Given a Hermitian operator Q that measures Q-ness(!) we can find a precise energy/Q-ness uncertainty inequality. We do this by applyingthe uncertainty inequality to the Hamiltonian H and Q, finding

∆H∆Q ≥∣∣∣

⟨Ψ| 1

2i[H,Q]

∣∣Ψ⟩∣∣∣ . (6.3.4)

This starting point is interesting because the commutator [H,Q] encodes something veryphysical about Q. Indeed, let us consider henceforth the case in which the operator Q hasno time dependence. It could be, for example some function of x and p, or for a spin-1/2particle, the operator |+〉〈−|. Such operator Q can easily have time-dependent expectationvalues, but the time dependence originates from the time dependence of the states, not fromthe operator Q itself.

To explore the meaning of [H,Q] we begin by computing the time-derivative of theexpectation value of Q:

d

dt〈Q〉 =

d

dt

⟨Ψ , QΨ

⟩=⟨∂Ψ

∂t,QΨ

⟩

+⟨

Ψ , Q∂Ψ

∂t

⟩

(6.3.5)

where we did not have to differentiate Q as it is time-independent. At this point we canuse the Schrodinger equation to find

d

dt〈Q〉 =

⟨ 1

i~HΨ , QΨ

⟩

+⟨

Ψ , Q1

i~HΨ

⟩

=i

~

(⟨HΨ , QΨ

⟩−⟨Ψ , QHΨ

⟩)

=i

~

⟨Ψ , (HQ−QH)Ψ

⟩=

i

~

⟨Ψ , [H,Q]Ψ

⟩

(6.3.6)

where we used the Hermiticity of the Hamiltonian. We have thus arrived at

d

dt〈Q〉 =

i

~

⟨[H,Q]

⟩for time-independent Q . (6.3.7)

This is a very important result. Each time you see [H,Q] you should think ‘time derivativeof 〈Q〉’. In classical mechanics one usually looks for conserved quantities, that is, functions

6.3. THE ENERGY-TIME UNCERTAINTY 129

of the dynamical variables that are time independent. In quantum mechanics a conservedoperator is one whose expectation value is time independent. An operator Q is conservedif it commutes with the Hamiltonian!

With this result, the inequality (6.3.4) can be simplified. Indeed, using (6.3.7) we have

∣∣∣

⟨ 1

2i[H,Q]

⟩∣∣∣ =

∣∣∣1

2i

~

i

d〈Q〉dt

∣∣∣ =

~

2

∣∣∣d〈Q〉dt

∣∣∣ (6.3.8)

and therefore

∆H∆Q ≥ ~

2

∣∣∣d〈Q〉dt

∣∣∣ , for time-independent Q . (6.3.9)

This is a perfectly precise uncertainty inequality. The terms in it suggest a definition of atime ∆tQ

∆tQ ≡∆Q∣∣∣d〈Q〉dt

∣∣∣

. (6.3.10)

This quantity has units of time. It is the time it would take 〈Q〉 to change by ∆Q if both

∆Q and the velocity d〈Q〉dt were time-independent. Since they are not necessarily so, we can

view ∆tQ as the time for “appreciable” change in 〈Q〉. This is certainly so when 〈Q〉 and∆Q are roughly of the same size. In terms of ∆tQ the uncertainty inequality reads

∆H∆tQ ≥~

2. (6.3.11)

This is still a precise inequality, given that ∆tQ has a precise definition.

As you will consider in the homework, (6.3.9) can be used to derive an inequality fortime ∆t⊥ that it takes for a state to evolve into a state orthogonal to itself. If we call theinitial state Ψ(0), we call ∆t⊥ the smallest time for which 〈Ψ(0),Ψ(∆t⊥)〉 = 0. You will beable to show that

∆H∆t⊥ ≥h

4. (6.3.12)

The speed in which a state can turn orthogonal depends on the energy uncertainty, and inquantum computation it plays a role in limiting the maximum possible speed of a computerfor a fixed finite energy.

The uncertainty relation involves ∆H. It is natural to ask if this quantity is timedependent. As we show now, ∆H is time independent if the Hamiltonian is also time-independent. Indeed, if H is time independent, we can use H and H2 for Q in (6.3.7) sothat

d

dt〈H〉 =

i

~

⟨[H,H]

⟩= 0 ,

d

dt〈H2〉 =

i

~

⟨[H,H2]

⟩= 0 .

(6.3.13)


It then follows thatd

dt(∆H)2 =

d

dt

(〈H2〉 − 〈H〉2

)= 0 . (6.3.14)

showing that ∆H is a constant. So we have shown that

If H is time independent, the uncertainty ∆H is constant in time. (6.3.15)

The concept of conservation of energy uncertainty can be used to understand someaspects of atomic decays. Consider, for illustration the hyperfine transition in the hydrogenatom. Due to the existence of proton spin and the electron spin, the ground state of hydrogenis fourfold degenerate, corresponding to the four possible combinations of spins (up-up, up-down, down-up, down-down). The magnetic interaction between the spins actually breaksthis degeneracy and produces the so-called “hyperfine” splitting. This is a very tiny split:5.88 × 10−6ev (compare with about 13.6 ev for the ground state energy). For a hyperfineatomic transition, the emitted photon carries the energy difference: Eγ = 5.88 × 10−6evresulting in a wavelength of 21.1cm and a frequency ν = 1420.405751786(30)MHz. Theeleven significant digits of this frequency attest to the sharpness of the emission line.

The issue of uncertainty arises because the excited state of the hyperfine splitting hasa lifetime τH for decay to the ground state and emission of a photon. This lifetime isextremely long, in fact τH ∼ 11 million years (= 3.4 × 1014 sec, recalling that a year isabout π × 107sec, accurate to better than 1% ). This lifetime can be viewed as the timethat takes some observable of the electron-proton system to change significantly (its totalspin angular momentum, perhaps) so by the uncertainty principle it must be related to someenergy uncertainty ∆E ∼ ~/τH ≃ 2× 10−30ev. of the original excited state of the hydrogenatom. Once the decay takes place the atom goes to the fully stable ground state, withoutany possible energy uncertainty. By the conservation of energy uncertainty, the photonmust carry the uncertainty ∆E. But ∆E/Eγ ∼ 3× 10−25, an absolutely infinitesimal effecton the photon. There is no broadening of the 21 cm line! That’s one reason it is so useful inastronomy. For decays with much shorter lifetimes there can be an observable broadeningof an emission line due to the energy-time uncertainty principle.

6.4 Lower bounds for ground state energies

You may recall that the variational principle could be used to find upper bounds onground state energies. The uncertainty principle can be used to find lower bounds for theground state energy of certain systems. We use below the uncertainty principle in the form∆x∆p ≥ ~/2 to find rigorous lower bounds for the ground state energy of one-dimensionalHamiltonians. This is best illustrated by an example.

Consider a particle in a one-dimensional quartic potential considered earlier

H =p2

2m+ αx4 , (6.4.1)

6.4. LOWER BOUNDS FOR GROUND STATE ENERGIES 131

where α > 0 is a constant with units of energy over length to the fourth power. Ourgoal is to find a lower bound for the ground state energy 〈H〉gs. Taking the ground stateexpectation value of the Hamiltonian we have

〈H〉gs =〈p2〉gs2m

+ α 〈x4〉gs , (6.4.2)

Recalling that(∆p)2 = 〈p2〉 − 〈p〉2 , (6.4.3)

we see that〈p2〉 ≥ (∆p)2 , (6.4.4)

for any state of the system. We should note however, that for the ground state (or anybound state) 〈p〉 = 0 so that in fact

〈p2〉gs = (∆p)2gs , (6.4.5)

From the inequality 〈A2〉 ≥ 〈A〉2 we have

〈x4〉 ≥ 〈x2〉2 . (6.4.6)

Moreover, just like for momentum above, (∆x)2 = 〈x2〉 − 〈x〉2 leads to

〈x2〉 ≥ (∆x)2 , (6.4.7)

so that〈x4〉 ≥ (∆x)4 , (6.4.8)

for the expectation value on arbitrary states. Therefore

〈H〉gs =〈p2〉gs2m

+ α 〈x4〉gs ≥(∆pgs)

2

2m+ α (∆xgs)

4 (6.4.9)

From the uncertainty principle

∆xgs∆pgs ≥~

2→ ∆pgs ≥

~

2∆xgs. (6.4.10)

Back to the value of 〈H〉gs we get

〈H〉gs ≥~2

8m(∆xgs)2+ α (∆xgs)

4 . (6.4.11)

The quantity to the right of the inequality is a function of ∆xgs. This function has beenplotted in Figure 6.2.

If we knew the value of ∆xgs we would immediately know that 〈H〉gs is bigger than thevalue taken by the right-hand side. This would be quite nice, since we want the highestpossible lower bound. Since we don’t know the value of ∆xgs, however, the only thing


Figure 6.2: We have that 〈Hgs〉 ≥ f(∆xgs) but we don’t know the value of ∆xgs. As a result, wecan only be certain that 〈Hgs〉 is greater than or equal to the lowest value the function f(∆xgs) cantake.

we can be sure of is that 〈H〉gs is bigger than the lowest value that can be taken by theexpression to the right of the inequality as we vary ∆xgs:

〈H〉gs ≥ Min∆x

(~2

8m(∆x)2+ α (∆x)4

)

. (6.4.12)

The minimization problem is straightforward. In fact you can check that

f(x) =A

x2+Bx4 is minimized for x2 =

1

21

3

(A

B

) 1

3

yielding f = 21

332 (A

2B)1

3 . (6.4.13)

Applied to (6.4.12) we obtain

〈H〉gs ≥ 21

33

8

(~2√αm

) 2

3 ≃ 0.4724(~2√αm

) 2

3

. (6.4.14)

This is the final lower bound for the ground state energy. It is actually not too bad, for theexact ground state the prefactor would be 0.668 instead of 0.4724.

6.5 Diagonalization of Operators

When we have operators we wish to understand, it can be useful to find a basis on thevector space for which the operators are represented by matrices that take a simple form.Diagonal matrices are matrices where all non diagonal entries vanish. If we can find a setof basis vectors for which the matrix representing an operator is diagonal we say that theoperator is diagonalizable.

6.5. DIAGONALIZATION OF OPERATORS 133

If an operator T is diagonal in some basis (u1, . . . un) of the vector space V , its matrixtakes the form diag (λ1, . . . λn), with constants λi, and we have

Tu1 = λ1u1 , . . . , Tun = λnun . (6.5.1)

The basis vectors are recognized as eigenvectors with eigenvalues given by the diagonalelements. It follows that a matrix is diagonalizable if and only if it possesses a set

of eigenvectors that span the vector space.Recall that all operators T on finite-dimensional complex vector spaces have at least

one eigenvalue and thus at least a one eigenvector. But not even in complex vector spacesall operators have enough eigenvectors to span the space. Those operators cannot be diag-onalized. The simplest example of such operator is provided by the two-by-two matrix

(0 10 0

)

. (6.5.2)

The only eigenvalue of this matrix is λ = 0 and the associated eigenvectors are all in thespan of (1, 0). This is just a one-dimensional subspace, so the eigenvectors fail to span thetwo-dimensional vector space. This matrix cannot be diagonalized.

Suppose we have a vector space V and we have chosen a basis (v1, . . . , vn) such that alinear operator has a matrix representation Tij({v}) that is not diagonal. As we learnedbefore, if we change basis to a new one (u1, . . . , un) using a linear operator A such that

uk = Avk , (6.5.3)

the matrix representation Tij({u}) of the operator in the new basis takes the form

T ({u}) = A−1T ({v})A or Tij({u}) = (A−1)ikTkp({v})Apj , (6.5.4)

where the matrix Aij is the representation of A in the either the original v-basis or the newu-basis. T is diagonalizable if there is an operator A such that Tij({u}) is diagonal.

There are two pictures of the diagonalization of T :

1. We can say that the matrix representation of T is diagonal when using the u basisobtained by acting with A on the original v basis.

2. Alternatively, the operator A−1TA that is diagonal in the original v basis.

The second viewpoint requires justification. Since T is diagonal in the u-basis, Tui = λiui (inot summed). This implies that TAvi = λiAvi and acting with A−1 that (A−1TA) vi = λivi,which confirms that A−1TA is represented by a diagonal matrix in the original v basis. Bothviewpoints are valuable.

Using the second viewpoint we write the following matrix equation in the original basis:

A−1TA = DT . (6.5.5)


HereDT is a diagonal matrix, and we say that A is a matrix that diagonalizes T by similarity.It is useful to note that

The columns of the matrix A are the eigenvectors of T. (6.5.6)

We see this as follows. Recall that the eigenvectors of T are the uk and therefore

uk = Avk =∑

i

Aikvi =

A1k...

Ank

. (6.5.7)

In the last step we noted that the basis vector vi is represented by a column vector of zeroeswith a single unit entry at the i-th position. This confirms that the k-th column of A is thek-th eigenvector of T .

While not all operators on complex vector spaces can be diagonalized, the situation ismuch improved for Hermitian operators. Hermitian operators can be diagonalized, and socan unitary operators. But even more is true: these operators take the diagonal form in anorthonormal basis!

An operator M is said to be unitarily diagonalizable if there is an orthonormal basis inwhich its matrix representation is a diagonal matrix. That basis, therefore, is an orthonor-

mal basis of eigenvectors. This uses the first viewpoint on diagonalization.Alternatively, start with an arbitrary orthonormal basis (e1, . . . , en) where the matrix

representation of M is written simply as the matrix M . It follows by the discussion insection 5.5 that there is a unitary operator U that maps the {ei} basis vectors to theorthonormal basis of eigenvectors. With U−1 = U † and using the {ei} basis we have thefollowing matrix equation, with DM a diagonal matrix.

U †M U = DM . (6.5.8)

We say that the matrixM has been diagonalized by a unitary transformation. This equationholds for the same reason that equation (6.5.5) holds.

6.6 The Spectral Theorem

While we could prove, as most textbooks do, that Hermitian operators are unitarily diago-nalizable, this result holds for the more general class of normal operators. The proof in thegeneral case is not harder than the one for hermitian operators. An operator M is said tobe normal if it commutes with its adjoint:

M is normal : [M †,M ] = 0 . (6.6.1)

Hermitian operators are clearly normal. So are anti-hermitian operators (M † = −M isantihermitian). Unitary operators U are normal because both U †U and UU † are equal to

6.6. THE SPECTRAL THEOREM 135

the identity matrix and thus U and U † commute. If an operator is normal, a similaritytransformation with a unitary operator gives another normal operator:

Exercise 1. If M is normal show that so is V †MV where V is a unitary operator.

It is a useful fact that a normal operator T and its adjoint T † share the same set ofeigenvectors:

Lemma: Let w be an eigenvector of the normal operator M : Mw = λw. Then w is alsoan eigenvector of M † with complex conjugate eigenvalue:

M †w = λ∗w . (6.6.2)

Proof: Define u = (M † − λ∗I)w. The result holds if u is the zero vector. To show this wecompute the norm-squared of u:

〈u, u〉 = 〈(M † − λ∗I)w , (M † − λ∗I)w〉 . (6.6.3)

Using the adjoint property to move the operator in the first entry to the second entry:

〈u, u〉 = 〈w , (M − λI)(M † − λ∗I)w〉 . (6.6.4)

Since M and M † commute, so do the two factors in parenthesis and therefore

〈u, u〉 = 〈w , (M † − λ∗I)(M − λI)w〉 = 0 , (6.6.5)

since (M − λI) kills w. It follows that u = 0 and therefore (6.6.2) holds. �

We can now state our main theorem, called the spectral theorem. It states that a matrix isunitarily diagonalizable if and only if it is normal:

Spectral Theorem: Let M be an operator in a complex vector space. The vector

space has an orthonormal basis of M eigenvectors if and only if M is normal.

(6.6.6)

Proof. It is easy to show that unitarily diagonalizable implies normality. Indeed, from(6.5.8) and dropping the reference to the e-basis,

M = UDMU† and therefore M † = UD†MU

† .

We then get

M †M = UD†MDMU† and MM † = UDMD

†MU

† .

so that

[M †,M ] = U(D†MDM −DMD†M )U † = 0 ,

because any two diagonal matrices commute.


Now let us prove by induction that for any normalM , viewed as a matrix on an arbitraryorthonormal basis, there is a unitary matrix U such that U †MU is diagonal. By our generaldiscussion this implies that the eigenvectors of M are an orthonormal basis.

The result is clearly true for dimV = 1. We assume that it holds for (n−1)-dimensionalvector spaces and consider the case of n-dimensional V . Let M be an n× n normal matrixreferred to the orthonormal basis (|1〉, . . . , |n〉) of V so that Mij = 〈i|M |j〉 . We know thereis at least one eigenvalue λ1 with a non-zero eigenvector |x1〉 of unit norm:

M |x1〉 = λ1|x1〉 and M †|x1〉 = λ∗1|x1〉 , (6.6.7)

in view of the Lemma. There is, we claim, a unitary matrix U1 such that

|x1〉 = U1|1〉 → U †1 |x1〉 = |1〉 . (6.6.8)

U1 is not unique and can be constructed as follows: extend |x1〉 to an orthonormal basis|x1〉, . . . , |xN 〉 using Gram-Schmidt. Then write U1 =

∑

i |xi〉〈i|. Define now

M1 ≡ U †1MU1 . (6.6.9)

M1 is also normal and M1|1〉 = U †1MU1|1〉 = U †1M |x1〉 = λ1U†1 |x1〉 = λ1|1〉 , so that

M1|1〉 = λ1|1〉 . (6.6.10)

Let us now examine the explicit form of the matrix M1:

〈j|M1|1〉 = λ1〈j|1〉 = λ1δ1,j , (6.6.11)

which says that the first column of M1 has zeroes in all entries except the first. Thenormality of M1 now helps us show that the first row of M1 is also zero except for the firstelement. Indeed,

〈1|M1|j〉 = (〈j|M †1 |1〉)∗ = (λ∗1〈j|1〉)∗ = λ1〈1|j〉 = λ1δ1,j , (6.6.12)

where we used M †1 |1〉 = λ∗1|1〉 which follows from the normality of M1. It follows from thetwo last equations that M1, in the original basis, takes the form

M1 =

λ1 0 . . . 0

0... M ′

0

,

whereM ′ is an (n−1)-by-(n−1) matrix. SinceM1 is normal, one can quickly see thatM ′ isalso normal (matrices multiply in blocks!). By the induction hypothesisM ′ can be unitarily

6.6. THE SPECTRAL THEOREM 137

diagonalized so that U ′†M ′U ′ is diagonal for some (n − 1)-by-(n − 1) unitary matrix U ′.The matrix U ′ can be extended to an n-by-n unitary matrix U as follows

U =

1 0 . . . 0

0... U ′

0

. (6.6.13)

It follows by explicit calculation that U †M1U is diagonal. But then

U †M1U = U †U †1M U1U = (U1U)†M (U1U) .

Since the product of unitary matrices is unitary, we have shown that U †MU is diagonalwith U = U1U unitary. This is the desired result. �

This theorem implies that Hermitian and unitary operators are unitarily diagonalizable.In other words their eigenvectors can be chosen to form an orthonormal basis. This is truewhether or not there are degeneracies in the spectrum. The proof did not require separatediscussion of this special case. If an eigenvalue of M is degenerate and appears k times,then there are k orthonormal eigenvectors associated with the corresponding k-dimensionalM -invariant subspace of the vector space.

We conclude this section with a description of the general situation that we can en-counter when diagonalizing a normal operator T . In general, we expect degeneracies in theeigenvalues so that each eigenvalue λk is repeated dk ≥ 1 times. An eigenvalue λk is degen-erate if dk > 1. It follows that V has T -invariant subspaces of different dimensionalities.Let Uk denote the T -invariant subspace of dimension dk ≥ 1 spanned by eigenvectors witheigenvalue λk:

Uk ≡ {v ∈ V |T v = λkv} , dim Uk = dk . (6.6.14)

By the spectral theorem Uk has a basis comprised by dk orthonormal eigenvectors

(u(k)1 , . . . , u

(k)dk

) .

Note that while the addition of eigenvectors with different eigenvalues does not give eigen-vectors, in the subspace Uk all vectors are eigenvectors with the same eigenvalue, and that’swhy addition makes sense, Uk as defined is a vector space, and adding eigenvectors inUk gives eigenvectors. The full space V is decomposed as the direct sum of the invariantsubspaces of T :

V = U1 ⊕ U2 ⊕ . . . Um , dim V =

m∑

i=1

di , m ≥ 1 . (6.6.15)

All Ui subspaces are guaranteed to be orthogonal to each other. In fact the full list ofeigenvectors is a list of orthonormal vectors that form a basis for V is conveniently orderedas follows:

(u(1)1 , . . . , u

(1)d1, . . . , u

(m)1 , . . . , u

(m)dm

) . (6.6.16)


The matrix T is manifestly diagonal in this basis because each vector above is an eigenvectorof T and is orthogonal to all others. The matrix representation of T reads

T = diag(λ1, . . . , λ1︸︷︷︸

d1 times

, . . . , λm, . . . , λm︸︷︷︸

dm times

). (6.6.17)

This is is clear because the first d1 vectors in the list are in U1, the second d2 vectors are inU2, and so on and so forth until the last dm vectors are in Um.

Let us now consider how unique is the basis (6.6.16). If we had no degeneracies in thespectrum the basis (with di = 1 for all i) would be rather unique if we require the matrixrepresentation of T to be unchanged. Each vector could be multiplied by a phase. On theother hand, with degeneracies, the list can be changed considerably without changing thematrix representation of T . Let Vk be a unitary operator on Uk, namely, Vk : Uk → Ukfor each k = 1, . . . ,m. We claim that the following basis of eigenvectors leads to the samematrix T :

(V1u

(1)1 , . . . V1u

(1)d1, . . . . . . , Vmu

(m)1 , . . . , Vmu

(m)dm

). (6.6.18)

This is still a collection of orthonormal T eigenvectors because the first d1 vectors are stillorthonormal eigenvectors in U1, the second d2 vectors are still orthonormal eigenvectors inU2 and so on and so forth. More explicitly, we can calculate the matrix elements of T withinUk in the new basis:

⟨Vku

(k)i , T (Vku

(k)j )

⟩= λk 〈 Vku(k)i , Vku

(k)j 〉 = λk〈u(k)i , u

(k)j 〉 = λkδij . (6.6.19)

This shows that in the Uk subspace the matrix for T is still diagonal with all entries equalto λk.

6.7 Simultaneous Diagonalization of Hermitian Operators

We say that two operators S and T in a vector space V can be simultaneously diago-

nalized if there is some basis of V in which both the matrix representation of S and thematrix representation of T are diagonal. It then follows that each vector in this basis is aneigenvector of S and an eigenvector of T . There is a basis of shared eigenvectors.

A necessary condition for simultaneous diagonalization is that the operators S and Tcommute. Indeed, if they can be simultaneously diagonalized there is a basis where bothare diagonal and they manifestly commute. If the operators don’t commute, this is a basis-independent statement and therefore a simultaneous diagonal presentation cannot exist.Since arbitrary linear operators S and T on a complex vector space cannot be diagonalized,the vanishing of [S, T ] does not guarantee simultaneous diagonalization. But if the operatorsare Hermitian it does, as we show now.

Theorem. If S and T are commuting Hermitian operators they can be simultaneouslydiagonalized.

6.7. SIMULTANEOUS DIAGONALIZATION OF HERMITIAN OPERATORS 139

Proof. The main complication is that degeneracies in the spectrum require some discussion.Either both operators have degeneracies or at least one has no degeneracies. Without lossof generality we can assume that there are two cases to consider

1. There is no degeneracy in the spectrum of T or,

2. Both T and S have degeneracies in their spectrum.

First case. Since T is non-degenerate there is a basis (u1, . . . un) of eigenvectors of T withdifferent eigenvalues

Tui = λiui , i not summed , λi 6= λj for i 6= j . (6.7.1)

We now want to understand what kind of vector is Sui. For this we act with T on it

T (Sui) = S(Tui) = S(λiui) = λi(S ui) , (6.7.2)

It follows that Sui is also an eigenvector of T with eigenvalue λi, thus it must equal ui, upto scale,

Sui = ωiui , (6.7.3)

showing that ui is also an eigenvector of S, this time with eigenvalue ωi. Thus any eigen-vector of T is also an eigenvector of S, showing that these operators are simultaneouslydiagonalizable.

Second case. Since T has degeneracies, as explained in the previous section, we have adecomposition of V in T -invariant subspaces Uk spanned by eigenvectors:

Uk ≡ {u |Tu = λku} , dim Uk = dk , V = U1 ⊕ . . . Um ,

orthonormal basis for V : (u(1)1 , . . . , u

(1)d1, . . . , u

(m)1 , . . . , u

(m)dm

) .

T = diag(λ1, . . . , λ1︸︷︷︸

d1 times

, . . . , λm, . . . , λm︸︷︷︸

dm times

)in this basis.

(6.7.4)

We also explained that the alternative orthonormal basis of V

(V1u

(1)1 , . . . V1u

(1)d1, . . . , Vmu

(m)1 , . . . , Vmu

(m)dm

). (6.7.5)

leads to the same matrix for T when each Vk is a unitary operator on Uk.We now claim that the Uk are also S-invariant subspaces! To show this let uk ∈ Uk and

examine the vector Suk. We have

T (Suk) = S(Tuk) = λkSuk → Suk ∈ Uk . (6.7.6)

It follows that in the basis (6.7.4) the matrix for S takes block-diagonal form, with blockson each of the Uk subspaces. We cannot guarantee, however, that S is diagonal within each

square block; we only know that Su(k)i ∈ Uk.


Since S restricted to each S-invariant subspace Uk is hermitian we can find an orthonor-mal basis of Uk in which the matrix S is diagonal. This new basis is unitarily related to the

original basis (u(k)1 , . . . , u

(k)dk

) and thus takes the form (Vku(k)1 , . . . , Vku

(k)dk

) with Vk a unitaryoperator in Uk. Note that the eigenvalues of S in this block need not be degenerate. Doingthis for each block, we find a basis of the form (6.7.5) in which S is diagonal. But T is stilldiagonal in this new basis, so both S and T have been simultaneously diagonalized. �

Remarks:

1. The above proof gives an algorithmic way to produce the common list of eigenvectors.One diagonalizes one of the matrices and constructs the second matrix in the basisof eigenvectors of the first. These second matrix is block diagonal, where the blocksare organized by the degeneracies in the spectrum of the first matrix. One must thendiagonalize within the blocks of the second matrix and is guaranteed that the newbasis that works for the second matrix also works for the first.

2. If we had to simultaneously diagonalize three different commuting Hermitian operatorsS1, S2 and S3, all of which have degenerate spectra, we would proceed as follows. Wediagonalize S1 and fix a basis in which S1 is diagonal. In this basis we must find thatS2 and S3 have exactly the same block structure. The corresponding block matricesare simply the matrix representations of S2 and S3 in each of the invariant spaces Ukappearing in the diagonalization of S1. Since S2 and S3 commute, their restrictionsto Uk commute. These restrictions can be diagonalized simultaneously, as guaranteedby our theorem which works for two matrices. The new basis in Uk that makes therestriction of S2 and S3 diagonal, will not disturb the diagonal form of S1 in thisblock. This is repeated for each block, until we get a common basis of eigenvectors.

3. An inductive algorithm is clear. If we know how to simultaneously diagonalize n com-muting Hermitian operators we can diagonalize n+ 1 of them, call them S1, . . . Sn+1,as follows. We diagonalize S1 and then consider the remaining n operators in thebasis that makes S1 diagonal. We are guaranteed a common block structure for the noperators. The problem becomes one of simultaneous diagonalization of n commutingHermitian block matrices, which is assumed known by the induction argument.

Corollary. If {S1, . . . , Sn} is a set of mutually commuting Hermitian operators they canall be simultaneously diagonalized.

6.8 Complete Set of Commuting Observables

We have discussed the problem of finding eigenvectors and eigenvalues of a Hermitian op-erator S. This hermitian operator is thought as a quantum mechanical observable. Theeigenvectors of S are physical states of the system in which the observable S can be mea-sured without uncertainty. The result of the measurement is the eigenvalue associated withthe eigenvector.

6.8. COMPLETE SET OF COMMUTING OBSERVABLES 141

If the Hermitian operator S has a non-degenerate spectrum, all eigenvalues are differentand we have a rather nice situation in which each eigenvector can be uniquely labeled withthe corresponding eigenvalue of S. The physical quantity associated with the observable isbeing used to distinguish the various eigenstates. Moreover, these eigenstates provide anorthonormal basis for the full vector space. In this case the operator S provides a “completeset of commuting observables” or a CSCO, in short. The set here has just one observable,the operator S.

The situation is more nontrivial if the Hermitian operator S exhibits degeneracies in itsspectrum. This means that V has an S-invariant subspace of dimension d > 1, spannedby orthonormal eigenvectors (u1, . . . , ud) all of which have S eigenvalue λ. This time, theeigenvalue of S does not allow us to label uniquely the basis eigenstates of the invariantsubspace. Physically this is an uncomfortable situation, as we have different basis states– the various ui’s – that we can’t tell apart by the measurement of S alone. This timeS does not provide a CSCO. Labeling eigenstates by the S eigenvalue does not suffice todistinguish them.

We are thus physically motivated to find another Hermitian operator T that is com-patible with S. Two Hermitian operators are said to be compatible observables if theycommute, since then we can find a basis of V comprised by simultaneous eigenvectors of theoperators. These states can be labeled by two observables, namely, the two eigenvalues. Ifwe are lucky, the basis eigenstates in each of the S-invariant subspaces of dimension higherthan one can be organized into T eigenstates of different eigenvalues. In this case T breaksthe spectral degeneracy of S and using T eigenvalues as well as S eigenvalues we can labeluniquely a basis of orthonormal states of V . In this case we say that S and T form a CSCO.

We have now given enough motivation for a definition of a complete set of commutingobservables. Consider a set of commuting observables, namely, a set {S1, . . . , Sk} of Hermi-tian operators acting on a complex vector space V that represents the physical state-spaceof some quantum system. By the theorem in the previous section, we can find an orthonor-mal basis of vectors in V such that each vector is an eigenstate of every operator in the set.Assume that each eigenstate in the basis is labeled by the eigenvalues of the Si operators.The set {S1, . . . , Sk} is said to be a complete set of commuting observables if no twostates have the same labels.

It is a physically motivated assumption that for any physical quantum system there isa complete set of commuting observables, for otherwise there is no physical way to distin-guish the various states that span the vector space. It would mean having states that aremanifestly different but have no single observable property in which they differ! In anyphysical problem we are urged to find such complete set of commuting observables, and wemust include operators in such set until all degeneracies are broken. A CSCO need not beunique. Once we have a CSCO, adding another observable causes no harm, although it isnot necessary. Also, if the pair (S1, S2) form a CSCO, so will the pair (S1 + S2, S1 − S2).Ideally, useful CSCO’s have the smallest possible set of operators.

The first operator that is usually included in a CSCO is the Hamiltonian H. For bound


state problems in one dimension, energy eigenstates are non-degenerate and thus the energycan be used to label uniquely the H-eigenstates. A simple example is the infinite squarewell. Another example is the one-dimensional harmonic oscillator. In such cases H formsthe CSCO. If we have, however, a two-dimensional isotropic harmonic oscillator in the (x, y)plane, the Hamiltonian has degeneracies. At the first excited level we can have the firstexcited state of the x harmonic oscillator or, at the same energy, the first excited state ofthe y harmonic oscillator. We thus need another observable that can be used to distinguishthese states. There are several options, as you will discuss in the homework.

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Uncertainty Principle and Compatible Operators