MODULE 5MODULE 5HERMITICITY, ORTHOGONALITY, AND THE SPECIFICATION

OF STATES

we have stated that we need Hermitian operators because their eigenvalues are real

This is so they can be related to experimentally determined observables (always real)

A definition of Hermiticity (see Barrante, chap 10) is

To prove the real property consider the eigenvalue equation with the eigenket normalized

*ˆ ˆm n n m

ˆ

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Now form the complex conjugate of both sides

From the Hermiticity condition the two LHS are equal and therefore the two RHS are equal

i.e.

which is only possible if is real

* *ˆ

ˆ

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More about OrthogonalityWe have stated, and used a symmetry argument to show, that

“eigenfunctions corresponding to different eigenvalues of the same operator are orthogonal.“

Now we can be a little more rigorous and prove the condition

According to the orthogonality statement, if we have two eigenkets According to the orthogonality statement, if we have two eigenkets of the Hermitian operator of the Hermitian operator ^ having eigenvalues ^ having eigenvalues and and

where where =/= then then

1 2 0

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Now form the complex conjugate of the RH equation and subtract it from the LH one

1 2; Suppose we have two eigenstates

That satisfy the two EV equations

1 2 21 21ˆ ; ˆ

2 1 1 2 1ˆ 1 2 2 1 2

ˆ

2 1

2 1 1 2 1 2 1 2 1 2ˆ ˆ * *

LH is zero (Hermitian condition)

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Our initial condition was that the two eigenvalues are different. The only way to satisfy the last equation is for

1 2 1 2 1 2 * 0

1 2 2 1*

1 2 1 2 2 1 0

And since then

1 2 2 1( ) 0

1 2 0 Thus different eigenfunctions of a Hermitian operator having

different eigenvalues are orthogonal.

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The Specification of StatesThree questions:

Can a state be simultaneously an eigenstate of all possible observables, A, B, C, … ?

Are there restrictions in the number or type of variables that can be simultaneously specified?

And if so, how can we identify such things?

If If simultaneitysimultaneity is possible then if we measure the observable is possible then if we measure the observable represented by the operator represented by the operator A^ we shall get exactly we shall get exactly a as the as the

outcome (P 4) and likewise for the other observables. outcome (P 4) and likewise for the other observables.

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We first find the conditions under which two observables may be specified simultaneously with arbitrary precision.

we need to establish the conditions whereby a given ket can be simultaneously an eigenket of two Hermitian operators

We assume that the property is true and find conditions that allow it to be so.

Thus we assume that ket Iq> is an eigenket of A^ and B^

ˆ

ˆ

B

A q a q

q b q

MODULE 5MODULE 5Write the following chain:

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆAB q Ab q bA q ba q ab q aB q Ba q BA q

ˆ ˆˆ ˆAB q BA q

ˆ ˆˆ ˆ 0AB BA Thus the two operators commute (Barrante, chap 10)

This is the condition that is necessary for Iq> to be a simultaneous eigenstate of the two operators.

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However, we need to find out whether commutation of the operators is a sufficient condition for simultaneity.

Or, if

is it certain that the ket is also an eigenstate of B^?

ˆ ˆ[ˆ , ] 0A q a q and A B

ˆˆ ˆ ˆBA q Ba q aB q A q a q

ˆ ˆˆ ˆBA q AB qˆ ˆ[ , ] 0A B

ˆ ˆ ˆ( ) ( )A B q a B q

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This is an eigenvalue equation (with eigenvalue a) and by comparing it with with

where b is a constant of proportionality.

Thus, is an eigenstate of B^, as we set out to prove.

A q a q

B q q

B q b q

ˆ ˆ ˆ( ) ( )A B q a B q

q

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Thus if we wish to know whether a pair of observables can be specified simultaneously we need to inspect whether the

corresponding operators commute.

The converse is also true, i.e., if a pair of operators does not commute, then their corresponding observables will not have

simultaneously precisely defined values.

We can extend this to more than two observables by taking them successively in pairs in which one of the pair is common to all

pairs.

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For example Is the simultaneous specification of the position and linear

momentum of a particle allowed?

The three operators for the x, y, and z coordinates commute with each other (these operators tell us to multiply by the variable

and multiplication is always commutative).

Also the three operators for the components of momentum

commute with each other.

ˆ ˆ ˆ, ,x y zp p p

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There are no constraints on the complete specification of position or of momentum as individual observables.

What about the position and momentum pairs?

To proceed we need to find the commutator of

ˆ ˆ, xx p

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SimilarlyAnd we see that momentum and position cannot be simultaneously

specified with arbitrary precision

ˆ ˆ ˆˆ ˆ ˆ ˆˆ ˆ ˆ, ( )

( )

{ ( )

( )

x x x x x x x x x

x x

x x x

x x

x p f xp p x f xp f p xf

d dx f xfi dx i dxd dx f f x f

i dx dx

f i fi

ˆ ˆ, 0xx p i

ˆ ˆ[ , ] 0yy p

ˆ ˆ[ , ] 0yy p ˆˆ[ , ] 0zz p

MODULE 5MODULE 5In general we can see that the operator for any component of momentum does not commute

with its own position operator,

z

pypz

x

y

px

pz py

z

But it does with the other position operators.

Lines link those observables that can be specified

simultaneously; those that cannot be so specified are not

linked.

MODULE 5MODULE 5Observables that cannot be simultaneously specified are said to

be complementary.

This is completely against the tenets of classical physics which presumed that no restrictions existed on the simultaneous

determination of observables (no complementarity).

Quantum mechanics tells us there can be restrictions on the extent that we can specify a state.

Refer to the final segment of Module 5 about Complementarity and Uncertainty