THEOREMS OF QUANTUM MECHANICS
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1 THEOREMS OF QUANTUM MECHANICS In order to develop methods to treat many-electron systems (atoms & molecules), many of the theorems of quantum mechancs are useful. Useful Notaton The matrx element A mn s defned by A mn = φ m A φ n dτ = <φ m A φ n > bracket notaton = (φ m A φ n ) = <m A n> The overlap ntegral between two functons s φ m φ n dτ = <φ m φ n > = <m n>. The complex conjugate of the overlap ntegral s [ φ m φ n dτ] = φ n φ m dτ = <n m> = [<m n>] Also <m m> = <m m>. Theorems The average value of a physcal quantty must be a real number.
2 Let A be the lnear operator for the property A. So f A s real, then <A> = <A> and A s sad to be a Hermtan Operator. For a Hermtan Operator: <A> = ψ Aψ dτ = <A> = ( ψ Aψ dτ) = ψ (Aψ) dτ Usng the above relaton, prove f Ag dτ = g (Af) dτ. If ψ = f + cg & A s a Hermtan operator, then (f + cg) A(f + cg) dτ = (f + cg)[ A(f + cg)] dτ. Left-hand sde = (f + cg) A(f + cg) dτ = (f + c g ) A(f + cg) dτ = (f + c g ) Af dτ + (f + c g ) Acg dτ = f Afdτ + c g Afdτ + c f Ag dτ + cc g Ag dτ By symmetry, the rght-hand sde = f(af) dτ + c g(af) dτ + c f ( Ag) dτ + cc g(ag) dτ Snce A s Hermtan & f Afdτ = f(af) dτ & g Ag dτ = g(ag) dτ, then, from the left- & rght-hand sdes:
3 c g Afdτ + c f Ag dτ = c g(af) dτ + c f ( Ag) dτ. Snce ths must hold for all c s, t must hold for c= or 1. Set c=1 to gve one eqn., set c= to gve a second eqn. Then add. c=1: g Afdτ + f Ag dτ = g(af) dτ + f ( Ag) dτ c=: - g Afdτ + f Ag dτ = g(af) dτ - f ( Ag) dτ; = - Add eqns. f Ag dτ = <φ A φ j > = <φ j A φ > = g(af) dτ. So for a Hermtan operator, <φ A φ j > = <φ j A φ > or < A j> = <j A > or A j = A j What Operators Are Hermtan? Is V(x), the potental energy operator, Hermtan? <φ j V(x) φ > = - φ j (Vφ ) dx V = V & V s just a multplcatve operator (no square roots, etc). So - φ j V φ dx = - φ j Vφ dx = - φ Vφ j dx <φ V(x) φ j > Hermtan Is p x = -h / x Hermtan? (p x = h / x) <φ p x φ j > = - φ (-h / x)φ j dx = -h - φ φ j dx
4 Integraton by parts: uv = uv - vu So -h - φ φ j dx = -h[φ φ j φ j (φ ) dx] Assume φ & φ j are well-behaved (.e. =0) at +, so that the frst term on the rght equals 0. Then <φ p x φ j > = h - φ j (φ ) dx = - φ j ( h / x)φ dx = - φ j (p x ) φ dx = <φ j p x φ > Hermtan (Prove: T, the knetc energy operator, s Hermtan). Then H = T + V s Hermtan. PROVE: The egenvalues of a Hermtan operator are real. (Ths means they represent a physcal quantty.) For A φ = b φ, show that b = b (b s real). If A s Hermtan, then φ Aφ dτ = φ (Aφ ) dτ. Or, Then φ bφ dτ = φ (bφ ) dτ = φ b φ dτ b φ φ dτ = b φ φ dτ = b φ φ dτ So b = b PROVE: The egenfunctons of a Hermtan operator can be chosen to be orthogonal. Show that, f B F = s F & B G = t G & t s not equal to s, then <F G> = 0. Snce B s Hermtan
5 <F B G> = <G B F> Or <F t G> = <G s F> So t <F G> = s <G F> = s <F G> = s<f G> (t-s) <F G> = 0 t not equal to s <F G> = 0 The requrement that t s not equal to s means that F & G are ndependent egenfunctons that have dfferent egenvalues (.e. they are non-degenerate) PROVE: That n the case of degenerate egenfunctons, we can construct from these egenfunctons a new egenfuncton that wll be orthogonal. Remember: We have shown that any lnear combnaton of degenerate egenfunctons correspondng to the same egenvalue s also an egenfuncton wth the same egenvalue. Let B F = s F & B G = s G Let φ 1 & φ 2 be the new egenfunctons that wll be orthogonal. Set φ 1 = F, φ 2 = G + c F. Fnd c such that <φ 1 φ 2 > = 0. Procedure: Schmdt Orthogonalzaton <φ 1 φ 2 > = <F G + cf> = <F G > + c <F F > If <φ 1 φ 2 > = 0, then c = - <F G >/<F F >
6 (Unless otherwse noted, assume all egenfunctons are orthogonal & normalzed: <φ φ j > = δ j = 0 unless = j; δ = 1) Expanson n terms of egenfunctons: We can use the egenfunctons of a Hermtan operator to descrbe an arbtrary well-behaved functon. We can expand the arbtrary functon n terms of all (or a complete set) of egenfunctons of the operator. Let f be an arbtrary well-behaved functon that obeys the same boundary condtons as the complete set φ f = Σ a φ a s an expanson coeffcent To fnd a (formal soluton): <φ j f> = <φ j Σ a φ > = Σ a <φ j φ > = Σ a δ j = a j Or f = Σ <φ f> φ So, the egenfunctons of a Hermtan operator form a complete orthonormal set wth real egenvalues Egenfunctons of Commutng Operators: In Chapter 5 we stated that a wavefuncton can be smultaneously an egenfuncton of two dfferent operators f those operators commute. Or, more exactly, a necessary condton for the exstence of a complete set of smultaneous egenfunctons of two operators s that the operators commute wth each other. Ths means that the physcal propertes assocated wth the operators can be measured smultaneously.
7 PROVE: If there exsts a common complete set of egenfunctons for two lnear operators, then the operators commute. Let φ be the complete set of egenfunctons of the operators A & B. A φ = s φ & B φ = t φ Show that [A,B] = 0 or (AB - BA)f = 0 where f s an arbtrary functon. We can expand f n terms of the complete set of egenfunctons of A & B: f = Σ c φ So (AB - BA) f = (AB -BA) Σ c φ = Σ c (AB -BA) φ = Σ c (ABφ -BAφ ) = Σ c (A t φ -B s φ ) I = Σ c (t A φ - s B φ ) = Σ c (t s φ - s t φ ) = Σ c t s (φ - φ ) = 0 The mportant pont here s that both operators must have a common complete set of egenfunctons. The exstence of just one egenfuncton n common s not enough to guarantee that [A,B] = 0.
8 Look over the proofs for: If A & B commute, we can select a common complete set of egenfunctons for them. If A s a Hermtan operator wth egenfuncton φ such that Aφ = s φ & [A,B] = 0, then B j = <φ B φ j > = 0 (s not = s j ). PARITY OPERATOR - a quantum mechancal operator that has no classcal mechancal equvalent Π f(x,y,z) = f(-x,-y,-z) The party operator, Π, replaces the Cartesan coordnates wth ther negatve values. Example: Π (x 2 - z e Ay ) = (x 2 + z e -Ay ) [In Cartesan coordnates, Π (x,y,z) = (-x,-y,-z). What about sphercal polar coordnates? The allowed ranges for the varables are: 0 < r <, 0 < θ < π, 0 < φ < 2π To move nto the quadrant of (-x,-y,-z), r r, φ π + φ, θ π - θ] Fnd the egenvalues of the party operator: Π g = c g Frst, fnd Π 2 :
9 Π 2 f(x,y,z) = Π [Π f(x,y,z)] = Π f(-x,-y,-z) = f(x,y,z) So Π 2 = 1 (unt operator) Then Π 2 g = Π [Πg ] = Π c g = c Π g = c c g = c 2 g So c 2 = 1 & c = +1 Fnd the egenfunctons of the party operator: Π g = c g Or Π g (x,y,z) = + g (x,y,z) And Π g (x,y,z) = g (-x,-y,-z) If c = +1, g (x,y,z) = g (-x,-y,-z) & g s an even functon If c = -1, g (x,y,z) = -g (-x,-y,-z) & g s an odd functon So the egenfunctons of Π are all the possble well-behaved even & odd functons. The party relatonshps are useful n constructng varatonal wavefunctons & molecular wavefunctons (later chapters). If Π & H commute, we can select a common set of egenfunctons. H = -h 2 /(2m) ( 2 / x / y / y 2 ) + V [H, Π] = -h 2 /(2m)[ 2 / x 2, Π] - h 2 /(2m)[ 2 / y 2, Π] - h 2 /(2m)[ 2 / z 2, Π] + [V, Π] Consder [ 2 / x 2, Π] = 2 / x 2 Π - Π 2 / x 2 :
10 Π 2 / x 2 φ(x,y,z) = / (-x) / (-x) φ(-x,-y,-z) = [- / x][- / x] φ(-x,-y,-z) = 2 / x 2 φ(-x,-y,-z) = 2 / x 2 Π φ(x,y,z) So [ 2 / x 2, Π] = 0 = [ 2 / y 2, Π] = [ 2 / z 2, Π] Consder [V, Π] = VΠ - ΠV ΠV(x,y,z) φ(x,y,z) = V(-x,-y,-z) φ(-x,-y,-z) If V s an even functon, V(x,y,z) = V(-x,-y,-z) ΠV(x,y,z) φ(x,y,z) = V(x,y,z) φ(-x,-y,-z) = V(x,y,z) Πφ(x,y,z); V & Π commute. Otherwse they don t commute. So [H, Π] = 0 f V s an even functon. When V s even, we can choose the ψ so that they are even or odd,.e. have defnte party. Ths s used n the Varaton Method to construct the approprate wavefuncton. MEASUREMENT & SUPERPOSITION OF STATES: The basc method s a scheme for calculatng the probabltes of varous possble outcomes of a measurement. Example: If the state functon, Ψ(x,t) s known, the probablty of fndng the partcle between x & x+dx s Ψ(x,t) 2 dx.
11 In general, consder the property, G: How can we calculate the probablty for each possble result of the measurement of G? (Assume there are N partcles & three coordnates; Let q represent the poston coordnates.) G φ (q) = g φ (q) The egenfunctons of any Hermtan operator form a complete set (.e. they are all the lnearly ndependent egenfunctons). The φ form a complete set, so we can expand any arbtrary wavefuncton n terms of them: Requre Ψ(q,t) = Σ c (t)φ (q) Ψ Ψ dτ = 1, where dτ s the volume element for the spatal coordnates (not tme). Then 1 = Σc (t)φ (q) Σ c j(t)φ j (q) dτ j = Σc Σ c j φ (q)φ j (q) dτ = Σc Σ c jδ j j = Σ c 2 Choose the c s so that 1 = Σ c 2
12 Snce Ψ(q,t) s a normalzed state functon, we can wrte the average value of G as <G> = Ψ(q,t) G Ψ(q,t) dτ = Σc (t)φ (q) G Σ c j(t)φ j (q) dτ j = Σc Σ c j φ (q) G φ j (q) dτ j = Σc Σ c j φ (q) g j φ j (q) dτ j
13 = Σc Σ c j g j φ (q) φ j (q) dτ j = Σc Σ c j g j δ j j = Σ c 2 g We have prevously defned the average value n terms of the probablty of gettng one of the egenvalues, g, when G s measured (.e. When a property s measured, we can only get one of the egenvalues as a result. No other numbers are possble.) Then <G> = Σ Pg g, where Pg s the probablty of fndng the egenvalue g. So Pg = c 2 We can predct the result of the measurement of G wth certanty only f all the c s except one are 0: c = 0, not equal to k; c k not equal to zero. Then c k 2 = 1 & the result of the measurement wll be g k & ψ k = φ k snce all the other c s are 0. So Ψ(q,t) = Σ c (t)φ (q) gves the state functon as a superposton of egenstates, φ, of G. The coeffcent, c, of φ n the expanson s related to the probablty of fndng the egenvalue g when G s measured (.e.
14 The larger the contrbuton of φ n the expanson - ndcated by the magntude of c - the larger the probablty of measurng that egenvalue). Calculate c : Ψ(q,t) = Σ c (t)φ (q) φ j (q) Ψ dτ = φ j (q) [Σ c (t)φ (q)] dτ = Σ c (t) φ j (q) φ (q) dτ = Σ c (t) δ j = c j = <φ j Ψ> = probablty ampltude So the probablty of measurng g j as a value of G s: c j 2 = <φ j Ψ> 2 = φ j (q) Ψ (q,t) dτ 2 So, f we know Ψ, we can predct the outcome of the measurement of G. Example: G = p x lnear momentum φ = e k x/h ; g = k We prevously found the general form for the wavefuncton for a free partcle n one dmenson: Ψ = a 1 e -Et/h e (2mE) x/ h + a 2 e -Et/h - (2mE) x/ h e = c 1 e k 1 x/ h + c 2 e k 2 x/ h, where c 1 = a 1 e -Et/h, c 2 = a 2 e -Et/h, k 1 = (2mE), k 2 = - (2mE).
15 So c 1 2 = a 1 e -Et/h 2 = a 1 2 e Et/h e -Et/h = a 1 2 = the probablty of gettng k 1 when measurng G. And c 2 2 = a 2 2 = the probablty of gettng k 2 when measurng G. The probablty of gettng any other number when measurng G s zero. So Ψ n (q,t) = e -E n t/h ψ n(q) s a statonary state & Hψ n(q) = E n ψ n(q) For a statonary state, the probablty densty does not depend on tme: Ψ n (q,t) 2 = ψ n (q) 2 s ndependent of t. If we take a superposton of statonary states, Ψ n, Ψ = Σ c n Ψ n = Σ c n e -E t/h n ψ n n n s not an egenfuncton of H, but s an egenfuncton of the tmedependent Schrödnger Eq. [-h / t + H] Ψ = 0 Ψ doesn t have a defnte energy because t s a combnaton of the ψ n, each wth energy E n. The probablty of gettng E n when the energy s measured s
16 c n e -E n t/h 2 = c n 2. Ψ s a nonstatonary state because the probablty densty depends on t due to cross terms of the form e -E n t/h e E n t/h Physcal example: If a system that s n a statonary state s exposed to radaton (.e. a laser feld whose potental vares wth tme) the state changes to a nonstatonary state.
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