QM Review and Spin-12Matthew ZhangModern PhysicsSeptember 4, 2008Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 1 / 17State function (vector) of quantum mechanicsStates of a quantum particle is described by a complex-valued:Ifunctionψ2,1,−1(~r) =1√24a−3/2rae−r/2a38π1/2sin θe−iφIn-tuplet (vector)φspin=1 + i1 − iIall of the above:Ψ = ψspace(~r) φspinQuantum mechanics “lives” in a complex linear vector space calledHilbert space.Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 2 / 17Physical observablesPhysical observable are identified with a differential operator ormatrix, e.g.ˆx = x , ˆpx= −i~∂∂x,ˆH = −~22m∇2+ U(ˆx)States with deterministic value for certain observable are eigenstatesof its corresponding operator, e.g.State with x-momentum qx:ψqx(x) = e+iqxx~: ˆpxψqx(x) = ˆpxe+iqxx~= qxe+iqxx~Stationary states are states that solve the time-independentschrödinger’s equation:ˆHφ = EφX Eigenstates of HamiltonianX deterministic energyX no time dependence in probability distributionMatthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 3 / 17Example: Orbital angular momentumOperatorsˆL2= −~2"∂2∂θ2+cos θsin θ∂∂θ+1sin2θ∂2∂φ2#,ˆLz= −i~∂∂φEigenstates, spherical harmonics: Yml(θ, φ)e.g. Y−11=12q32π· sin θ ·e−iφˆLzΘ(θ)e−iφ= −1 ~ Θ(θ)e−iφˆL2# · sin θ ·e−iφ= −~2−sin θ +cos θsin θcos θ −sin θsin2θ# · e−iφ= 2~2· # · sin θ ·e−iφ2 = l(l + 1), confirms l = 1.Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 4 / 17Physical intepretationSuppose the system is in some state ψ1The probability distribution of this state: ψ∗1ψ1.e.g.ψ1= Ce−x2/a2ψ∗1ψ1= C2e−2x2/a2ZC2e−2x2/a2dx = 1The expected value of a physical observable:Rψ∗1ˆOψ1dx.e.g.ψ1= Ce−x2/a2hxi = C2Zxe−2x2/a2dxhpi =Z−i~C2e−x2/a2ddxe−x2/a2dx = i2~a2C2Zxe−2x2/a2dxThe probability of the system to be in another state ψ2: |Rψ∗2ψ1dx|2Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 5 / 17Example: average positionThe n = 2, l = 0 state has (normalized) radial wavefunction:R20(r) =12a03/22 −ra0e−r/2a0The average distance to nucleus:hri = C2Z∞0r2 −ra02e−r/a0r2dr= C2Z∞0 4r3−4r4a0+r5a0!e−r/a0dr= C2a40(4 × 3! − 4 × 4! + 5!)=18a30· a40· 48 = 6a0Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 6 / 17Example: average potential energyAny function of position can be similarly averaged:hUi = C2Z∞0 −k e2r!2 −ra02e−r/a0r2dr= −k e2C2Z∞0 4r −4r2a0+r3a0!e−r/a0dr= −k e2C2a20(4 × 1! − 4 × 2! + 3!)= −k e28a30· a20· 2 = −k e24a0Note that hU (r)i 6= U (hri).hKi + hUi = E. Now you can also find average kinetic energy.Computing it directly from operator is hard!Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 7 / 17General stateGeneral quantum mechanical states ψ (stationary or not) can be buildout of orthongonal and normalized set φjlike this:ψ = c1φ1+ c2φ2+ ··· + cnφncj’s are in general complex number. The modulus square |cj|2givesthe probability that the state ψ is measured as state φjNormalization requires: |c1|2+ |c2|2+ ···|cn|2= 1To find cj, do the inner product: cj= ψ · φ.Iinner product for n-tuple is just good old vector dot product.IFor function, generalize to integral with the appropriate limit:cj=Zψ∗φjdxMatthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 8 / 17Example: particle-in-boxOne-dimensional particle in a box. Suppose the particle is in the n = 1state: φ1(x) =q2LsinπxL.What is the probability that it is in the n = 2 state:φ2(x) =q2Lsin2πxL?2LZL0sin2πxLsinπxLdx =2πZπ0sin (2u) sin (u) du = 0For particle-in-box, different n states are Orthogonal: If you are in oneof the state, the probability to be found in a different n state is zero.What is the probability that it has momentum px= +π~L:ψpx=1√LeiπxL?Zψ∗pxφ1(x) dx =√2LZL0e−iπxLsinπxLdx = −i√2→ ans:12Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 9 / 17ExcerciseShow that the probabilities that an electron in hydrogen atom state(n, l, m) = (2, 0, 0) to be found in states (n, l, m) = (1, 0, 0) or(n, l, m) = (2, 1, 0) are both 0. Hydrogen stationary states withdifferent quantum numbers are orthogonal to each other.Again for particle in box of size L, How is the result from an integralin last slide is equivalent to:r2LsinπxL=12i1√LeπxL−12i1√Le−πxL?Suppose the a PiB state has wavefunctionψ = C8 sin3πxL− 6 sinπxLICheck that this is not a stationary state.IIt could be either in state n = 3 or n = 1 with probability 0.64 and0.36, respectively. VERIFY THIS.IFind C.Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 10 / 17Angular momentumSomething is fundamentally weird about angular momentum becauseit is product of two incompatible quantities:~L = ~r × ~pWe can therefore never measure the angular momentum vector withcertainty because of space-momentum uncertainty.However, its magnitude and one of its components are fair game:Lz= xpy− ypx.Turns out, for algebrae reason, (generalized) angular momentum isquantized:I|~L|2can only take on values l(l + 1)~ with the contain that 2l is aninteger.IIf system is in a state of a particular l then there are (2l + 1) possiblevalues for one component (e.g. Jz) goes from −l, −l + 1, . . . , l − 1, l.Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 11 / 17Spin-1/2The simplest quantum non-trivial angular momentum system is thespin-1/2 system: l =12For spin-1/2, all angular momentum operators are 2×2 matrices:Lx=~2 0 11 0!, Ly=~2 0 −ii 0!, Lz=~2 1 00 −1!The two states with deterministic anuglar momentum Lzare: 10!with Lz=~2:~2 1 00 −1! 10!=~2 10!, 01!with Lz= −~2:~2 1 00 −1! 01!=~2 0−1!They don’t have deterministic angular momentum in neither the x ory direction.Matthew Zhang (Physics 2D) QM Review and Spin-12September 4, 2008 12 / 17What about x direction?What about states with deterministic angular momentum in the xdirection (Lx)?~2 0 11 0! 11!=~2 11!~2 0 11 0! 1−1!=~2 −11!Here’s a trivial question: If a particle is in spin state (1, 1)Tand hasSx= ~/2 what is the probability to find it in the state Sx= ~/2?ILet’s check:1 111= 2?!Forgot to normalize! Actual states are:|Lx=