Advanced Quantum Mechanics 2 lecture 5 …Symmetries in quantum mechanics Symmetries in quantum...

Advanced Quantum Mechanics 2lecture 5

Symmetries in Quantum Mechanics

Yazid Delenda

Departement des Sciences de la matiereFaculte des Sciences - UHLB

http://theorique05.wordpress.com/f411

Batna, 14 December 2014

(http://theorique05.wordpress.com/f411) Advanced Quantum Mechanics 2 - lecture 5 1 / 33

Symmetries in quantum mechanics

A symmetry operator is one which when acts on the Hamiltonian itrenders it unchanged:

OH = H

where O is a symmetry operator.Consider for example the Hamiltonian:

2m= − ~2

for a free particle.

H(t+ I) = H(t), so OH = H, where O here is the time translationsymmetry operator

H(x+ a) = H, so TaH = H, where Ta here is the space translationsymmetry operator

H(−x) = H(x), so P H = H, where P parity operator

OH = H

2m= − ~2

OH = H

2m= − ~2

OH = H

2m= − ~2

OH = H

2m= − ~2

OH = H

2m= − ~2

A symmetry operator commutes with the Hamiltonian:

[O, H] = 0

[O, H]ψ(x) = OHψ(x)− HOψ(x) = HOψ(x)− HOψ(x) =

where we note that OH = H implies that HO = H.

Recall that if two operators A and B are diagonal in somerepresentation:

a1 0 · · · 00 a2 · · · 0...

.... . .

...0 0 0 an

b1 0 · · · 00 b2 · · · 0...

.... . .

...0 0 0 bn

[O, H] = 0

a1 0 · · · 00 a2 · · · 0...

.... . .

...0 0 0 an

b1 0 · · · 00 b2 · · · 0...

.... . .

...0 0 0 bn

[O, H] = 0

a1 0 · · · 00 a2 · · · 0...

.... . .

...0 0 0 an

b1 0 · · · 00 b2 · · · 0...

.... . .

...0 0 0 bn

a1b1 0 · · · 00 a2b2 · · · 0...

.... . .

...0 0 0 anbn

so if two operators are diagonal in some representations then theycommute:

[A, B] = 0

If two hermitian operators commute, [A, B] = 0 with A† = A andB† = B then there exists a basis |n〉 in which:

A|n〉an|n〉B|n〉bn|n〉

which means these operators are diagonal(http://theorique05.wordpress.com/f411) Advanced Quantum Mechanics 2 - lecture 5 4 / 33

a1b1 0 · · · 00 a2b2 · · · 0...

.... . .

...0 0 0 anbn

[A, B] = 0

a1b1 0 · · · 00 a2b2 · · · 0...

.... . .

...0 0 0 anbn

[A, B] = 0

a1b1 0 · · · 00 a2b2 · · · 0...

.... . .

...0 0 0 anbn

[A, B] = 0

a1b1 0 · · · 00 a2b2 · · · 0...

.... . .

...0 0 0 anbn

[A, B] = 0

If three operators A, B and C mutually commute,[A, B] = [B, C] = [A, C] = 0, then there exists a set |n〉 in which alloperators A, B and C are diagonal.

If A, B and C commute with the Hamiltonian then there eigenvectorscan be used to classify the eigenstates of the Hamiltonian.Forexample, if the Hamiltonian of a free particle H = p2/2m commuteswith momentum, p, [H, p] then the eigenstates of the momentum |p〉can be used to characterise the Hamiltonian:

H|p〉 = const|p〉

Another example: the Hamiltonian of identical particles

H =∑j

εj a†j aj

H|p〉 = const|p〉

H =∑j

εj a†j aj

H|p〉 = const|p〉

H =∑j

εj a†j aj

H|p〉 = const|p〉

H =∑j

εj a†j aj

H|p〉 = const|p〉

H =∑j

εj a†j aj

H|p〉 = const|p〉

H =∑j

εj a†j aj

commutes with the operator of particle transposition T . The eigenstatesof the transposition operators are symmetric or anti-symmetric states witheigenvalues ±1:

T |ψ〉 = +|ψ〉, bose case

T |ψ〉 = −|ψ〉, fermi case

Thus, eigenstates of the Hamiltonian are either symmetric oranti-symmetric, corresponding to either bosons or fermions.Furthermore The Hamiltonian of the harmonic oscillator:

2mω2x2

commutes with the parity operator P since P H(x) = H(−x) = H(x).Butwe know that the eigenstates of the parity operator are either symmetric oranti-symmetric states:

P |ψ〉 = +|ψ〉P |ψ〉 = −|ψ〉

Therefore, wavefunctions of the harmonic oscillator are symmetric oranti-symmetric.

If a time-independant operator A commutes with the Hamiltonian[A, H] = 0, the mean value of A is conserved: 〈A〉 = const, since:

dt〈A〉 = i

~〈[A, H]〉 = 0⇒ 〈A〉 = const

P |ψ〉 = +|ψ〉P |ψ〉 = −|ψ〉

dt〈A〉 = i

~〈[A, H]〉 = 0⇒ 〈A〉 = const

P |ψ〉 = +|ψ〉P |ψ〉 = −|ψ〉

dt〈A〉 = i

~〈[A, H]〉 = 0⇒ 〈A〉 = const

Conclusions:

If O is a symmetry operator that commutes with the Hamiltonian[H, O] = 0 then they can be diagonalised in the same basis:

O|ψ〉 = λ|ψ〉, H|ψ〉 = E|ψ〉

The average value of the symmetry operator is conserved in time:〈ψ|O|ψ〉 = 0.

The set of operators which commute with the Hamiltonian provide“good” quantum numbers which are conserved with time.

Conclusions:

O|ψ〉 = λ|ψ〉, H|ψ〉 = E|ψ〉

Conclusions:

O|ψ〉 = λ|ψ〉, H|ψ〉 = E|ψ〉

Symmetries in quantum mechanics Translation operator

Translation operator

The translation operator is defined by:

Taψ(x) = ψ(x+ a)

in multi-dimensional space a = (a1, a2, · · · ) in general, andTaψ(x1, x2, · · · ) = ψ(x1 + a1, x2 + a2, · · · ).One may show that the translation operator may be written as:

Ta = exp

)such that:

Taψ(x) = exp

)ψ(x) =

∞∑n=0

dxnψ(x)

Taψ(x) = ψ(x+ a)

Ta = exp

)such that:

Taψ(x) = exp

)ψ(x) =

∞∑n=0

dxnψ(x)

Taψ(x) = ψ(x+ a)

Ta = exp

)such that:

Taψ(x) = exp

)ψ(x) =

∞∑n=0

dxnψ(x)

Taψ(x) = ψ(x+ a)

Ta = exp

)such that:

Taψ(x) = exp

)ψ(x) =

∞∑n=0

dxnψ(x)

Translation operatorProof: Consider the Taylor expansion of the function ψ(x) about a:

ψ(x) = ψ(a) + (x− a) dψ(x)dx

∣∣∣∣x=a

2!(x− a)2 d

2ψ(x)

∣∣∣∣x=a

+ · · ·

Let us introduce x′ = x− a, so x = x′ + a, then:

ψ(x′ + a) = ψ(a) + x′dψ(x′)dx′

∣∣∣∣x′=a

2!x′2

d2ψ(x′)dx′2

∣∣∣∣z=a

+ · · ·

where we note that

dψ(x)

∣∣∣∣x=a

=dψ(x′)dx′

∣∣∣∣x′=a

=dψ(z)

∣∣∣∣x′=a

is a fixed number independent of x and x′.By symmetry of the variables x′

and a we can consider a as the variable and x′ as a constant about whichwe expand, so we also have (removing the irrelevant prime):

ψ(x+ a) = ψ(x) + adψ(a)

∣∣∣∣a=x

d2ψ(a)

∣∣∣∣a=x

+ · · ·

ψ(x) = ψ(a) + (x− a) dψ(x)dx

∣∣∣∣x=a

2!(x− a)2 d

2ψ(x)

∣∣∣∣x=a

+ · · ·

ψ(x′ + a) = ψ(a) + x′dψ(x′)dx′

∣∣∣∣x′=a

2!x′2

d2ψ(x′)dx′2

∣∣∣∣z=a

+ · · ·

where we note that

dψ(x)

∣∣∣∣x=a

=dψ(x′)dx′

∣∣∣∣x′=a

=dψ(z)

∣∣∣∣x′=a

ψ(x+ a) = ψ(x) + adψ(a)

∣∣∣∣a=x

d2ψ(a)

∣∣∣∣a=x

+ · · ·

ψ(x) = ψ(a) + (x− a) dψ(x)dx

∣∣∣∣x=a

2!(x− a)2 d

2ψ(x)

∣∣∣∣x=a

+ · · ·

ψ(x′ + a) = ψ(a) + x′dψ(x′)dx′

∣∣∣∣x′=a

2!x′2

d2ψ(x′)dx′2

∣∣∣∣z=a

+ · · ·

where we note that

dψ(x)

∣∣∣∣x=a

=dψ(x′)dx′

∣∣∣∣x′=a

=dψ(z)

∣∣∣∣x′=a

ψ(x+ a) = ψ(x) + adψ(a)

∣∣∣∣a=x

d2ψ(a)

∣∣∣∣a=x

+ · · ·

ψ(x) = ψ(a) + (x− a) dψ(x)dx

∣∣∣∣x=a

2!(x− a)2 d

2ψ(x)

∣∣∣∣x=a

+ · · ·

ψ(x′ + a) = ψ(a) + x′dψ(x′)dx′

∣∣∣∣x′=a

2!x′2

d2ψ(x′)dx′2

∣∣∣∣z=a

+ · · ·

where we note that

dψ(x)

∣∣∣∣x=a

=dψ(x′)dx′

∣∣∣∣x′=a

=dψ(z)

∣∣∣∣x′=a

ψ(x+ a) = ψ(x) + adψ(a)

∣∣∣∣a=x

d2ψ(a)

∣∣∣∣a=x

+ · · ·

ψ(x) = ψ(a) + (x− a) dψ(x)dx

∣∣∣∣x=a

2!(x− a)2 d

2ψ(x)

∣∣∣∣x=a

+ · · ·

ψ(x′ + a) = ψ(a) + x′dψ(x′)dx′

∣∣∣∣x′=a

2!x′2

d2ψ(x′)dx′2

∣∣∣∣z=a

+ · · ·

where we note that

dψ(x)

∣∣∣∣x=a

=dψ(x′)dx′

∣∣∣∣x′=a

=dψ(z)

∣∣∣∣x′=a

ψ(x+ a) = ψ(x) + adψ(a)

∣∣∣∣a=x

d2ψ(a)

∣∣∣∣a=x

+ · · ·

but since

dψ(a)

∣∣∣∣a=x

=dψ(x′)dx′

∣∣∣∣x′=x

=dψ(x)

∣∣∣∣x=x

=dψ(x)

we find:

ψ(x+ a) =ψ(x) + adψ(x)

2!a2d2ψ(x)

dx2+ · · · =

∞∑n=0

dxnψ(x)

( ∞∑n=0

)ψ(x) = exp

)ψ(x)

Taψ(x) = ψ(x+ a) = exp

)ψ(x)⇒ Ta = exp

)(http://theorique05.wordpress.com/f411) Advanced Quantum Mechanics 2 - lecture 5 11 / 33

but since

dψ(a)

∣∣∣∣a=x

=dψ(x′)dx′

∣∣∣∣x′=x

=dψ(x)

∣∣∣∣x=x

=dψ(x)

we find:

2!a2d2ψ(x)

dx2+ · · · =

∞∑n=0

dxnψ(x)

( ∞∑n=0

)ψ(x) = exp

)ψ(x)

)ψ(x)⇒ Ta = exp

but since

dψ(a)

∣∣∣∣a=x

=dψ(x′)dx′

∣∣∣∣x′=x

=dψ(x)

∣∣∣∣x=x

=dψ(x)

we find:

2!a2d2ψ(x)

dx2+ · · · =

∞∑n=0

dxnψ(x)

( ∞∑n=0

)ψ(x) = exp

)ψ(x)

)ψ(x)⇒ Ta = exp

In position representation we can write the translation operator as:

Ta =∑xx′

(Ta)xx′ |x〉〈x′|

(Ta)xx′ = 〈x|Ta|x′〉

To find the matrix elements we consider:

Taψ(x) = ψ(x+ a)⇒ 〈x|Ta|ψ〉 = 〈x+ a|ψ〉⇒ 〈x|Ta = 〈x+ a| ⇒ 〈x|Ta|x′〉 = 〈x+ a|x′〉 = δx′,x+a

⇒ 〈x|Ta|x′〉 =(Ta

)xx′

= δx+a,x′

where in the last relation we multiplied by the ket |x′〉 both sides of theequation. Substituting back we find:

Ta =∑xx′

)xx′|x〉〈x′| =

∑xx′

δx+a,x′ |x〉〈x′| =∑x

|x〉〈x+ a|

Ta =∑xx′

⇒ 〈x|Ta|x′〉 =(Ta

)xx′

= δx+a,x′

Ta =∑xx′

)xx′|x〉〈x′| =

∑xx′

δx+a,x′ |x〉〈x′| =∑x

|x〉〈x+ a|

Ta =∑xx′

⇒ 〈x|Ta|x′〉 =(Ta

)xx′

= δx+a,x′

Ta =∑xx′

)xx′|x〉〈x′| =

∑xx′

δx+a,x′ |x〉〈x′| =∑x

|x〉〈x+ a|

Ta =∑xx′

⇒ 〈x|Ta|x′〉 =(Ta

)xx′

= δx+a,x′

Ta =∑xx′

)xx′|x〉〈x′| =

∑xx′

δx+a,x′ |x〉〈x′| =∑x

|x〉〈x+ a|

and indeed:

Taψ(x) = 〈x|Ta|ψ〉 =∑x′

〈x|x′〉〈x′+ a|ψ〉 =∑x′

δx,x′ψ(x′+ a) = ψ(x+ a)

We can also write:Ta =

a†x+aax

which is equivalent to saying ax destroys a particle at position x and a†x+acreates it at position x+ a.Note that the operator Ta is not hermitian, but is unitary. We have:

T †a =∑x

|x+ a〉〈x| =∑x

|x〉〈x− a| = T−a

where we made a change of variable x→ x− a.So the Ta 6= T †a , butT †a = T−a.Besides it is straightforward to show that:

T †a Ta = 1(http://theorique05.wordpress.com/f411) Advanced Quantum Mechanics 2 - lecture 5 13 / 33

and indeed:

〈x|x′〉〈x′+ a|ψ〉 =∑x′

δx,x′ψ(x′+ a) = ψ(x+ a)

a†x+aax

T †a =∑x

|x+ a〉〈x| =∑x

|x〉〈x− a| = T−a

and indeed:

〈x|x′〉〈x′+ a|ψ〉 =∑x′

δx,x′ψ(x′+ a) = ψ(x+ a)

a†x+aax

T †a =∑x

|x+ a〉〈x| =∑x

|x〉〈x− a| = T−a

and indeed:

〈x|x′〉〈x′+ a|ψ〉 =∑x′

δx,x′ψ(x′+ a) = ψ(x+ a)

a†x+aax

T †a =∑x

|x+ a〉〈x| =∑x

|x〉〈x− a| = T−a

and indeed:

〈x|x′〉〈x′+ a|ψ〉 =∑x′

δx,x′ψ(x′+ a) = ψ(x+ a)

a†x+aax

T †a =∑x

|x+ a〉〈x| =∑x

|x〉〈x− a| = T−a

and indeed:

〈x|x′〉〈x′+ a|ψ〉 =∑x′

δx,x′ψ(x′+ a) = ψ(x+ a)

a†x+aax

T †a =∑x

|x+ a〉〈x| =∑x

|x〉〈x− a| = T−a

and indeed:

〈x|x′〉〈x′+ a|ψ〉 =∑x′

δx,x′ψ(x′+ a) = ψ(x+ a)

a†x+aax

T †a =∑x

|x+ a〉〈x| =∑x

|x〉〈x− a| = T−a

by just considering the action on the wave functions:

T †a Taψ(x) = T−aTaψ(x) = T−aψ(x+ a) = ψ(x+ a− x) = ψ(x)

This can also be easily proven by considering the explicit positionrepresentation of the operator Ta:

T †a Ta∑xx′

|x+ a〉〈x|x′〉〈x′ + a| =∑xx′

δxx′ |x+ a〉〈x+ a| =∑x′′

|x′′〉〈x′′| = 1

where x′′ ≡ x+ a. This means that the translation operator conserves thenorm.Also note that:

Ta|x〉 =(〈x|T †a

)†=(〈x|T(−a)

)†= (〈x− a|)† = |x− a〉

T †a Ta∑xx′

|x+ a〉〈x|x′〉〈x′ + a| =∑xx′

δxx′ |x+ a〉〈x+ a| =∑x′′

|x′′〉〈x′′| = 1

)†=(〈x|T(−a)

)†= (〈x− a|)† = |x− a〉

T †a Ta∑xx′

|x+ a〉〈x|x′〉〈x′ + a| =∑xx′

δxx′ |x+ a〉〈x+ a| =∑x′′

|x′′〉〈x′′| = 1

)†=(〈x|T(−a)

)†= (〈x− a|)† = |x− a〉

T †a Ta∑xx′

|x+ a〉〈x|x′〉〈x′ + a| =∑xx′

δxx′ |x+ a〉〈x+ a| =∑x′′

|x′′〉〈x′′| = 1

)†=(〈x|T(−a)

)†= (〈x− a|)† = |x− a〉

Now if the Hamiltonian commutes with the translation operator[H, Ta] = 0 (for example for a free particle) then the eigenstates of theHamiltonian are also eigenstates of the translation operator:

H|ψ〉 = E|ψ〉, Ta|ψ〉 = λ|ψ〉

and since the translation operator is time-independent then〈ψ|Ta|ψ〉 = const. This leads to a conservation law (conservation ofmomentum).Eigenvalues of the translation operator: Consider writing thetranslation operator in momentum representation:

Ta =∑x

|x〉〈x+ a| =∑x

1|x〉〈x+ a|1

=∑x,p,p′

|p〉〈p|x〉〈x+ a|p′〉〈p′|

H|ψ〉 = E|ψ〉, Ta|ψ〉 = λ|ψ〉

Ta =∑x

|x〉〈x+ a| =∑x

1|x〉〈x+ a|1

=∑x,p,p′

|p〉〈p|x〉〈x+ a|p′〉〈p′|

H|ψ〉 = E|ψ〉, Ta|ψ〉 = λ|ψ〉

Ta =∑x

|x〉〈x+ a| =∑x

1|x〉〈x+ a|1

=∑x,p,p′

|p〉〈p|x〉〈x+ a|p′〉〈p′|

H|ψ〉 = E|ψ〉, Ta|ψ〉 = λ|ψ〉

Ta =∑x

|x〉〈x+ a| =∑x

1|x〉〈x+ a|1

=∑x,p,p′

|p〉〈p|x〉〈x+ a|p′〉〈p′|

H|ψ〉 = E|ψ〉, Ta|ψ〉 = λ|ψ〉

Ta =∑x

|x〉〈x+ a| =∑x

1|x〉〈x+ a|1

=∑x,p,p′

|p〉〈p|x〉〈x+ a|p′〉〈p′|

where 〈x+ a|p′〉 = N exp(i(x+ a)p′/~) = N exp(ixp′/~) exp(iap′/~) =〈x|p′〉 exp(iap′/~). Hence

Ta =∑x,p,p′

|p〉〈p|x〉〈x|p′〉 exp(iap′/~)〈p′|

=∑p,p′

|p〉〈p|p′〉 exp(iap′/~)〈p′|

=∑p,p′

|p〉δpp′ exp(iap′/~)〈p′|

exp(iap/~)|p〉〈p|

Thus we clearly see that the operator Ta is diagonal in momentumrepresentation,which means automatically that Ta and p commute andthat the eigenstates of the momentum operator |p〉 are also eignstates ofthe translation operator. We have:

Ta =∑x,p,p′

=∑p,p′

Ta =∑x,p,p′

=∑p,p′

Ta =∑x,p,p′

=∑p,p′

Ta =∑x,p,p′

=∑p,p′

Ta =∑x,p,p′

=∑p,p′

Ta =∑x,p,p′

=∑p,p′

Ta|p〉 =∑p′

exp(iap′/~)|p′〉〈p′|p〉 =∑p′

exp(iap′/~)δpp′ |p′〉 = exp(iap/~)|p〉

So the eigenstates of the translation operator are |p〉 with eigenvalueseiap/~.In fact this can be seen by returning to the exponential form of theoperator Ta:

Ta =exp

)= exp

)Ta|p〉 =exp

)|p〉 = exp

)|p〉

Ta =exp(ia

~p)⇒ T †a = exp

(−ia

Ta|p〉 =∑p′

exp(iap′/~)|p′〉〈p′|p〉 =∑p′

Ta =exp

)= exp

)Ta|p〉 =exp

)|p〉 = exp

)|p〉

Ta =exp(ia

~p)⇒ T †a = exp

(−ia

Ta|p〉 =∑p′

exp(iap′/~)|p′〉〈p′|p〉 =∑p′

Ta =exp

)= exp

)Ta|p〉 =exp

)|p〉 = exp

)|p〉

Ta =exp(ia

~p)⇒ T †a = exp

(−ia

Ta|p〉 =∑p′

exp(iap′/~)|p′〉〈p′|p〉 =∑p′

Ta =exp

)= exp

)Ta|p〉 =exp

)|p〉 = exp

)|p〉

Ta =exp(ia

~p)⇒ T †a = exp

(−ia

Ta|p〉 =∑p′

exp(iap′/~)|p′〉〈p′|p〉 =∑p′

Ta =exp

)= exp

)Ta|p〉 =exp

)|p〉 = exp

)|p〉

Ta =exp(ia

~p)⇒ T †a = exp

(−ia

Generators of the translation operator: To find the generators of asymmetry operator we just consider transformations of infinitesimalparameter.In the case of the translation operator let us consider atranslation with a small parameter δa→ 0, so the translation operator willbe very close to the unit operator:

Tδa = exp

)≈ 1 + i

We say that p is the generator of the translation group:

p = −i~ ∂

∂aTa

∣∣∣∣a=0

Now if the Hamiltonian has a translational invariance than〈T 〉 = const,and this then automatically means 〈p〉 = const since thetranslation operator is expanded in terms of the momentum operator.

Tδa = exp

)≈ 1 + i

p = −i~ ∂

∂aTa

∣∣∣∣a=0

Tδa = exp

)≈ 1 + i

p = −i~ ∂

∂aTa

∣∣∣∣a=0

Tδa = exp

)≈ 1 + i

p = −i~ ∂

∂aTa

∣∣∣∣a=0

Tδa = exp

)≈ 1 + i

p = −i~ ∂

∂aTa

∣∣∣∣a=0

Tδa = exp

)≈ 1 + i

p = −i~ ∂

∂aTa

∣∣∣∣a=0

Tδa = exp

)≈ 1 + i

p = −i~ ∂

∂aTa

∣∣∣∣a=0

Therefore the translation symmetry implies conservation ofmomentum.This is Noether theorem: for every symmetry of theHamiltonian there is a conserved quantity, which is just the generator ofthe symmetry group.We note that the operator p = −i~ ∂

∂x is hermitian,p† = p,which means

the operator ∂∂x is not hermitian, but instead anti-hermitian:(

)†= − ∂

Passive and active transformations:An active transformation is one which keeps the origin of the frame andchanges points of measurement,while a passive transformation is onewhich changes the origin of the frame of reference and keeps the samepoint of measurement.

Active Translation

Passive Translation

O O O′

For an active translation we have:

Taψ(x) = ψ(x+ a)

and for a passive translation we have

Taψ(x) = ψ(x− a)

We can generalise the above discussion to general symmetries of theHamiltonian which are not hermitian but unitary.Consider a generaltransformation symmetry U(α) which is a symmetry of the Hamiltonian:

U(α)H = H

which means that [U , H] = 0.Now if the symmetry operator is unitary butnon-hermitian (and thus conserves the norm of the wavefunction):

U †U = 1

Taψ(x) = ψ(x+ a)

U(α)H = H

U †U = 1

Taψ(x) = ψ(x+ a)

U(α)H = H

U †U = 1

Taψ(x) = ψ(x+ a)

U(α)H = H

U †U = 1

Taψ(x) = ψ(x+ a)

U(α)H = H

U †U = 1

Taψ(x) = ψ(x+ a)

U(α)H = H

U †U = 1

Taψ(x) = ψ(x+ a)

U(α)H = H

U †U = 1

then we can construct “good” quantum numbers by constructinghermitian operators from this non-hermitian operator.We do this byexpanding the operator U for infinitesimal transformation parameter α:

U(α) = 1+ αdU(α)

∣∣∣∣∣α=0

Now we define the operator:

Q =dU(α)

∣∣∣∣∣α=0

such that U(α) = 1+ αQ+O(α2).Then we have from unitarity of theoperator U

U †U = 1 =(1+ αQ†

)(1+ αQ

)= 1+ α(Q† + Q) +O(α2)

U(α) = 1+ αdU(α)

∣∣∣∣∣α=0

Q =dU(α)

∣∣∣∣∣α=0

U †U = 1 =(1+ αQ†

)(1+ αQ

)= 1+ α(Q† + Q) +O(α2)

U(α) = 1+ αdU(α)

∣∣∣∣∣α=0

Q =dU(α)

∣∣∣∣∣α=0

U †U = 1 =(1+ αQ†

)(1+ αQ

)= 1+ α(Q† + Q) +O(α2)

U(α) = 1+ αdU(α)

∣∣∣∣∣α=0

Q =dU(α)

∣∣∣∣∣α=0

U †U = 1 =(1+ αQ†

)(1+ αQ

)= 1+ α(Q† + Q) +O(α2)

U(α) = 1+ αdU(α)

∣∣∣∣∣α=0

Q =dU(α)

∣∣∣∣∣α=0

U †U = 1 =(1+ αQ†

)(1+ αQ

)= 1+ α(Q† + Q) +O(α2)

U(α) = 1+ αdU(α)

∣∣∣∣∣α=0

Q =dU(α)

∣∣∣∣∣α=0

U †U = 1 =(1+ αQ†

)(1+ αQ

)= 1+ α(Q† + Q) +O(α2)

It then follows that:Q = −Q†

i.e. the operator Q is anti-hermitian.That is if U is unitary than dU(α)dα

∣∣∣α=0

is anti-hermitian.Thus we can construct a hermitian operator from thisanti-hermitian operator by defining:

R = −i~ dU(α)

∣∣∣∣∣α=0

and the operator R is hermitian and corresponds to a conserved measuredphysical quantity (by Noether theorem).

∣∣∣α=0

R = −i~ dU(α)

∣∣∣∣∣α=0

∣∣∣α=0

R = −i~ dU(α)

∣∣∣∣∣α=0

∣∣∣α=0

R = −i~ dU(α)

∣∣∣∣∣α=0

∣∣∣α=0

R = −i~ dU(α)

∣∣∣∣∣α=0

For example cf. U(α) = eiα,which corresponds to a (global) phase shift –or global gauge transformation.Then clearly U †U = 1 (i.e. unitaryoperator) but U † 6= U (i.e. not hermitian).Then the associated conservedquantity is:

R = −i~ ∂eiα

∣∣∣∣α=0

The corresponding conserved quantity is actually the norm of thewavefunction since:

〈ψ|R|ψ〉 = ~〈ψ|ψ〉

In the case of the translation operator Q = ∂∂x and R = p.

R = −i~ ∂eiα

∣∣∣∣α=0

〈ψ|R|ψ〉 = ~〈ψ|ψ〉

R = −i~ ∂eiα

∣∣∣∣α=0

〈ψ|R|ψ〉 = ~〈ψ|ψ〉

R = −i~ ∂eiα

∣∣∣∣α=0

〈ψ|R|ψ〉 = ~〈ψ|ψ〉

R = −i~ ∂eiα

∣∣∣∣α=0

〈ψ|R|ψ〉 = ~〈ψ|ψ〉

R = −i~ ∂eiα

∣∣∣∣α=0

〈ψ|R|ψ〉 = ~〈ψ|ψ〉

The general rule is that if we have continuous symmetry which is notHermitian (i.e., does not correspond to the measured quantity), but isunitary (does conserve the system state norm), like the operator Ta, thenwe can construct a Hermitian operator by evaluating the symmetryoperator for small values of the transformation.The new operator will beconserved during system evolution and gives us quantum numbers tocharacterise the energy states of the system.The famous symmetries thatlead to conservation laws are:

translational invariance leads to momentum conservation

rotational invariance leads to angular momentum conservation

transposition invariance leads to conservation of number of particles.

Symmetries in quantum mechanics Rotation symmetry and angular momentum

Rotation symmetry and angular momentum

Let us consider a rotation in 3D, about the z axis through an angle θ:

z′ =z

x′ =x cos θ − y sin θy′ =x sin θ + y cos θ

which we can express as: x′

cos θ − sin θ 0sin θ cos θ 00 0 1

Then a general rotation operator U(θ) may actually be expressed by itsaction on the wave-function as follows (active transformation):

U(θ)ψ(x, y, z) = ψ(x′, y′, z′)

z′ =z

U(θ)ψ(x, y, z) = ψ(x′, y′, z′)

z′ =z

U(θ)ψ(x, y, z) = ψ(x′, y′, z′)

where x′, y′ and z′ are defined above.For rotations about the z axis

U(θ)ψ(x, y, z) = ψ(x cos θ − y sin θ, x sin θ + y cos θ, z).

It is clear that the operator U is actually unitary but not hermitian,sinceU † 6= U , and U †U = 1.This is true for arbitrary rotation through any angleabout any axis.In fact it may be easy to show that (U(θ))† = U(−θ).For infinitesimal rotations of angles δθ � 1 we have: x′

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

So we clearly see that the matrix:

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

1 −δθ 0δθ 1 00 0 1

= 1+ δθ

0 −1 01 0 00 0 0

is actually the generator for this rotation about the z axis.We can alsoconsider rotations about the other two axis x and y and find that thegenerators are:

0 0 00 0 −10 1 0

, Xy =

0 0 −10 0 01 0 0

which satisfy the commutation relations