SUPERPOSITION, MEASUREMENT, NORMALIZATION, EXPECTATION VALUES Readig: QM course packet Ch 5 up to 5. 1
ϕ (x) = E = π m( a) =1,,3,4,5 for x<-a ad x>a (x) = πx si L L * = πx L si L.5 ϕ' -.5 z 1 (x) = L si πx L * = L si πx L.5 ϕ' -.5 z 1
ϕ (x) = E = π m( a) =1,,3,4,5 for x<-a ad x>a ϕ 4 (x) = L si 4πx L ϕ 4 *ϕ 4 = L si 4πx L.5 ϕ' -.5 z 1 (x) = L si 3πx L * = L si 3πx L.5 ϕ' -.5 z 1 3
Draw a measured eergy from the quatum state preseted to you. Put the paper back ad pass it o. The ext perso does the same. Oe by oe, we call out our results ad record them. Do we all measure the same eergy? What iformatio do we have about the quatum state? How ca we ifer what the quatum state is, ad coversely, give a particular quatum state, ca we make predictios about what we might measure i a give measuremet? 4
The quatum state is a superpositio of (eergy) eigestates. The result of a (eergy) measuremet may be ONLY a eigevalue of the (eergy) operator. The coefficiets of the eigestates allow us to ifer the probability that a particular eigevalue will be measured. Do we all measure the same eergy? What iformatio do we have about the quatum state? How ca we ifer what the quatum state is, ad coversely, give a particular quatum state, ca we make predictios about what we might measure i a give measuremet? 5
The quatum state is a superpositio of eigestates. Φ = c 1 + c + c 3 +... = ϕ geeral state coefficiet eigestate Φ(x) = c 1 (x) + c (x) + c 3 (x) +... = ϕ (x) 1 1 1.5.5.5 1 -.5 -.5 -.5-1 -1 -.1 -.5.5.1-1 -.1 -.5.5.1-1 -.1 -.5.5.1 - -.1 -.5.5.1
1 Iitial state.5 -.5-1 -.1 -.5.5.1 Result of sigle eergy measuremet: E State after sigle eergy measuremet 1.5 -.5-1 -.1 -.5.5.1 Φ(x) = 1 5 (x) + 3 5 (x) + 1 5 (x) Probability that the result of sigle eergy measuremet is E is c * c = c = 3 5 7
The act of measuremet has somethig to do with projectio. Projectio of a geeral state oto a eigestate of the quatity beig measured yields the AMPLITUDE coefficiet. Φ = c 1 + c + c 3 = ϕ = c This is because eigestates of the Hamiltoia (eergy) operator are orthogoal (also true for ay other Hermitia operator): ϕ i ϕ j = δ ij = 1 i = j i j The PROBABILITY that the result of a particular measuremet yields E is give by the modulus SQUARED of the amplitude coefficiet or projectio: = Φ Class to show that: m = ϕ m Φ = c m * c m 8
The probability that a measuremet o a state will yield SOME EIGENSTATE must be uity, so What if = 1 1? Φ(x) = 1 5 (x) + 3 5 (x) + 1 5 (x) Φ(x) = (x) + 3 (x) + (x) Same RELATIVE probabilities. Always possible to scale all coefficiet values to make sum of squares =1 ( ormalizig the coeffs). Discuss: Φ(x) = i (x) + (x) (x) 9
1.5 -.5-1 1.5 -.5-1 1.5 -.5-1 (x) -.1 -.5.5.1 (x) -.1 -.5.5.1 -.1 -.5.5.1 ϕ * (x) (x)dx = δ,3 = Orthogoality ad projectios usig fuctios β α = β 1 * α 1 ( * * β β ) 3 α α 3 = δ,3 = β α = * β m α m m β * (x)α(x)dx 1
Abstract bra-ket & positio state (wavefuctio) represetatios State superpositio Eigestates & eigevalues Eigestate orthogoality Normalizatio Expectatio value Φ = ϕ ϕ m ϕ = δ,m = 1 Φ( x) = ϕ ( x) ˆQ ϕ = q ϕ ˆQϕ ( x) = q ϕ ( x) Φ Φ = 1 Q = Φ ˆQ Φ Q = q ϕ * (x)ϕ m (x)dx Φ * (x)φ(x) dx Probability desity Probability Q = Φ * x = δ,m = 1 ( ) ˆQΦ x ( )dx 11
Measuremet of a quatity Q (maybe eergy, maybe mometum.), represeted by a operator ˆQ, may yield values q 1, q etc. (suppose we kow them). The AVERAGE value measured is called the expectatio value ad is deoted by Q = Φ * x ( ) ˆQΦ x ( )dx = Φ ˆQ Φ Q Discuss : Φ(x) = i (x) + (x) (x) Class to show that i geeral Q = c 1 q 1 + c q +... = q 1
Example where Q ˆQ is the Hamiltoia or eergy operator, ad we expad quatum state i basis of eigestates of H. Bra-ket otatio: H = E = Φ Ĥ Φ Φ = i + 1 Ĥ Φ = ie 1 + E E 3 Is Φ a eigestate of the Hamiltoia or eergy operator? NO!! 13
Example where Q ˆQ is the Hamiltoia or eergy operator, ad we expad quatum state i basis of eigestates of H. Bra-ket otatio H = E = Φ Ĥ Φ Φ = i + 1 = E 1 ie + ie 3 + ie 1 + 4 E E 3 ie 1 E + E 3 E = 1 E 1 + 4 E + 1 E 3... = E 14
Example where Q is the Hamiltoia or eergy operator, ad we expad quatum state i basis of eigestates of H. Wave fuctio represetatio Q = Φ * x ( ) ˆQΦ x ( )dx Φ(x) = i (x) + (x) 1 (x) ĤΦ(x) = i Ĥ (x) + Ĥ (x) 1 Ĥ (x) = i E 1 (x) + E (x) 1 E 3 (x) Is Φ(x) a eigestate of the Hamiltoia or eergy operator? NO!! 15
E = Φ * ( x )ĤΦ( x)dx E = i ϕ * 1 + ϕ * 1 ϕ * 3 = E 1 + ie 1 ie 1 = Φ Ĥ Φ Φ(x) = i ϕ (x) + ϕ (x) 1 ϕ (x) 1 3 Φ * (x) = i ϕ * 1 (x) + ϕ * (x) 1 ϕ * 3(x) i E 1 + E 1 E 3 dx ϕ * 1 dx ie ϕ * 1 dx + ie 3 ϕ * 1 dx ϕ * dx + 4 E ϕ * dx E 3 ϕ * dx ϕ * 3 dx E ϕ * 3 dx + E 3 ϕ * 3 dx 1
E = Φ * ( x )ĤΦ( x)dx 1 = E 1 * dx + ie 1 ie 1 ie * dx ϕ * dx + 4 E ϕ * dx 1 = Φ Ĥ Φ + ie 3 * dx E 3 * dx ϕ * 3 dx E ϕ * 3 dx + E 3 ϕ * 3 dx 1 E = 1 E 1 + 4 E + 1 E 3... = E 17
Example where Q is NOT the Hamiltoia or eergy operator, ad we expad quatum state i basis of eigestates of H (ad they are NOT eigestates of the Q operator). Wave fuctio represetatio: Q = Φ * x ( ) ˆQΦ x ( )dx Well, ow there are fewer shortcuts. You have to put i the specific wave fuctios, put i the specific operator, ad evaluate a itegral. Sometimes, there are symmetry argumets that help you evaluate the itegral. 18
If geeral state is ormalized: Φ Φ = 1 ϕ *(x)ϕ(x)dx =1 So Φ Φ = 1 = * c m ϕ m ϕ = c * m ϕ m ϕ m=1 m=1 =1 =1 = c m * δ m, =1 = m=1 =1 19
Expectatio value of eergy: E = ϕ ˆ H ϕ ϕ Ĥ ϕ = c * ϕ m m Ĥ ϕ m=1 =1 m=1 = c m * m=1 =1 ϕ m =1 = c m * E δ m, =1 = E E ϕ Coefficiets: the modulus squared (probability of measuremet) is a weightig factor to determie average.
Projectios: = ϕ Φ ϕ Φ = ϕ = c m δ,m m=1 c m ϕ m = c m ϕ ϕ m m=1 = =1 Coefficiets: They are the projectio of the geeral state oto the eigestate. Do specific example o board Remember Fourier coefficiets? 1