� � � �� � � � � � � � � � � 8.04 Quantum Physics Lecture XIIAfter the energy measurement After the energy measurement with outcome Ei, the particle will be in the energy eigenstate ui, and all subsequent energy measurements will yield the energy Ei. What is the energy before measurement is made? If ψ(x) is not an eigenstate, the energy is “uncertain”. A measurement can yield different energy values, only probabilities can be predicted. However, an average value of the energy can be calculated: ∞→ generally valid (12-1) cn�E� = ||2En n=1 Using the definition of the expansion coefficients cn, we can also write this as ∞cn�E� = ||2En n=1 ∞∗ = cn dxu∗ n(x)ψ(x) En n=1 = cn dxψ∗(x)Enun(x) = cn Hundxψ∗(x)ˆ(x) = dxψ∗(x)Hˆcnun(x) � n = dxψ∗(x)ˆHψ(x) Expectation value of energy in state ψ(x): �E� = dxψ∗(x)ˆHψ(x) valid for any potential, not only box potential → (12-2) Hamiltonian operator and energy If we postulate that a particle of momentum p is associated with a deBroglie wavelenngt λdB = h , then it is represented by a plane wave eikx with a wavevector p Massachusetts Institute of Technology XII-1 � � � � � � � � � � � � � � 8.04 Quantum Physics Lecture XII2π 2πp pk = λdB = h = h¯, or p = ¯Then, since the Fourier transform φ(k) of ψ(x)hk. gives the probability amplitude for the plane wave with wavevector k → 1 ikx ψ(x)= dkφ(k)e, (12-3) √2π the expectation value of momentum is given by �p� = dkhk¯|φ(k)|2 (12-4) p 2 = dk(¯hk)2|φ(k)|2 (12-5) Note. In PS5 you show using the properties of Fourier transforms that these expectattio values can also be expressed as h¯∂ �p� = dxψ∗(x) i ∂x ψ(x) (12-6) � ��2�� h¯∂ p 2 = dxψ∗(x) ψ(x) (12-7) i ∂x It follows that the expectation value of the KE is ��� �� p2 h¯2 ∂2 �T � =2m = dxψ∗(x) −2m ∂x2 ψ(x) (12-8) How large is the expectation value of the potential energy? Potential V (x) should be weighted by probability to find particle between x and x + dx, hence �V � = dxV (x)|ψ(x)|2 = dxψ∗(x)V (x)ψ(x) (12-9) Since �E� = �V � + �T � it follows that h2 ∂2¯�E� = dxψ∗(x) −2m ∂x2 + V (x) ψ(x) (12-10) = dxψ∗(x) ˆ(12-11) Hψ(x) This is the so-called “sandwich form” for calculating the mean value (expectation value) of the energy. If ψ(x) is an energy eigenfunction with eigenvalue E0, i.e. if Massachusetts Institute of Technology XII-2 � � � � � � � � � � � � 8.04 Quantum Physics Lecture XIIˆHψE0 (x)= E0ψE0 (x), then �E� = dxψ∗ (x)HψˆE0 (x)E0 = dxψ∗ (x)E0ψE0 (x)E0 = E0 dx|ψE0 (x)|2 = E0, (12-12) where we have used the fact that the wavefunction is normalized. This shows that the constant E0 appearing when we make the product ansatz Ψ(x,t)= T (t)ψ(x�) to � solve the SE is really the energy of the system. We define the expectation value Oˆof an operator Oˆacting on a wavefunction ψ(x) via the sandwich form Oˆ= Oψ(x)dxψ∗(x)ˆ(12-13) Then we have �� � �E� = Hˆ= Hψ(x)dxψ∗(x) ˆ(12-14) The mean energy of the state described by the wave-function ψ(x) is the expectation value of the Hamiltonian operator Hˆ. We say that the Hamiltonian Hˆis the operator associated with the measurable quantity energy. The operator Tˆassociated with the kinetic energy is h2 ∂2¯Tˆ(12-15) = −2m ∂x2 with �T � = Tˆ= dxψ∗(x)ˆV for the potential energy is Tψ(x), while the operator ˆsimply a multiplicative factor Vˆ= V (x) (12-16) with �V � =ˆ= Vψ(x)V dxψ∗(x)ˆWhy is potential energy associated with a simple multiplicatiiv factor while kinetic energy is associated with a second derivative? Because we are working with wavefunctions in real space ψ(x). We say that we are working with wavefunctions in position space or in the position representation. Massachusetts Institute of Technology XII-3 � � � � � � � � � � 8.04 Quantum Physics Lecture XIIAnother possibility is to work in momentum space (the momentum representation). Then the wavefunction should be the probability amplitude in momentum space, which is just the Fourier transform φ(p) of ψ(x). Then to calculate the KE we have 2 to weigh pfor each p with the probability to find the particle momentum between 2m p and p + dp: � p2 2m � = � dp p2 2m|φ(p)|2 (12-17) = � dpφ∗(p) p2 φ(p) (12-18) 2m We see that in momentum space the KE operator is simply a multiplicative factor ˆpˆ2 p2 T == → in momentum space (12-19) 2m 2m How to calculate the potential energy V (x) in terms of the wavefunctions in momentum space φ(p)? Note. In PS5, you have shown that ∂ �x� = dpφ∗(p) ih¯φ(p) (12-20) ∂p � ��n∂ n�x � = dpφ∗(p) ih¯∂p φ(p) (12-21) Consequently, for any potential function ∞V (x)= anx n (12-22) n=0 we can calculate the expectation value �V � as ∂ �V � = dpφ∗(p)V ih¯φ(p)∂p = dpφ∗(p) ˆ(12-23) Vφ(p) Consequently, the representation of the operator for the PE in momentum space is Vˆ= V ih¯∂, (12-24) ∂p Massachusetts Institute of Technology XII-4 8.04 Quantum Physics Lecture XIIwhere a function V of an operator is defined in terms of its Taylor expansion, Eq. (12-22). It follows that the Hamiltonian is Hˆ= Tˆ+ Vˆ (12-25) 1 � ¯h ∂ �2 = 2m i ∂x + V (x) in position space (12-26) = p2 2m + V � i¯h ∂ ∂p � in momentum space (12-27) The SE equation is always the same: ih¯∂ Ψ(x,t)= HˆΨ(x,t) time-dependent SE (12-28) ∂tih¯∂ Φ(p,t)= HˆΦ(p,t) time-dependent SE (12-29) ∂tˆHψ(x)= Eψ(x) time-independent SE (12-30) ˆHφ(p)= Eφ(p) time-independent SE (12-31) Example. For the harmonic oscillator, the SE (in appropriately chosen units) looks the same in position and momentum space: 1. linear potential V (x)= Axh2¯−2mψ��(x)+ Axψ(x)= Eψ(x) in position space (12-32) 2pφ(p)+ i¯ simpler equation in momentum space hAφ�(p)= Eφ(p)2m (12-33) 2. harmonic oscillator: V (x)= 12 mω2x2h¯2 1−2mψ��(x)+ 2 mω2 x 2ψ(x)= Eψ(x) (12-34) p2 φ(p) − 1 h¯2mω2φ��(p)= Eφ(p) (12-35) 2m 2 If we know the solutions in one space, we know the solutions in the other. The HO is symmetric in position and momentum. Massachusetts Institute of Technology XII-5 � � � 8.04 Quantum Physics Lecture XIITime evolution of the wave function Consider a particle in the infinite box with a wavefunction at t = 0, Ψ(x,t = 0). Expansion into eigenstates Ψ(x,t = 0) = c1u1(x)+ c2u2(x)+ = ∞ cnun(x).··· n=1 Since each eigenstate un(x,t) evolves at a rate given by its eigenenergy En, un(x,t)= un(x,t = 0)e−iEnt/¯h (12-36) = un(x)e−iEnt/¯h (12-37) the wavefunction Ψ(x,t) at later time t is simply given by the linear superposition ∞Ψ(x,t)= cnun(x)e−iEnt/¯h (12-38) n=1 where the expansion coefficients cn are calculated at t = 0: ∞ cn = dxu∗ n(x)Ψ(x,t = 0) (12-39) −∞ Hence the importance of energy eigenstates and eigenvalues: The eigenvalues repressen not only the only possible outcomes of individual energy measurements, but the combination of eigenstates and eigenvalues allows one to write down the time evolution of an arbitrary initial state. How does a particle move? Example. Ψ(x,t = 0) = √12 (u1(x)+ u2(x)). Particle in equal superposition of ground and first excited state. 1 �� Ψ(x,t)= √2 u1(x)e−iE1t/¯h + u2(x)e−iE2t/h¯(12-40) 1 �� = √2 e−iE1t/¯h u1(x)+ u2(x)e i(E2−E1)t/h¯(12-41) 1 |Ψ(x,t)|2 =2|u1(x)+ u2(x)e−i(E2−E1)t/¯h|2 (12-42) At any fixed position, interference term between u1 and u2 oscillates from constructive to destructive interference with angular frequency E2 − E1ω21 = (12-43) h¯The energy difference determines the oscillation of the particle between the halves of the box. Massachusetts Institute of Technology XII-6 � � 8.04 Quantum Physics Lecture XIIFigure I: A particle in a superposition of the ground state an the first excited state oscillates from left to right at the frequency corresponding to the energy difference between the two states. Note. If Ψ(x,t = 0) is an eigenstate, Ψ(x,t = 0) = un(x), then |Ψ(x,t)|2 = |Ψ(x, 0)|2 , i.e. the probability density does not change in time: Bohr’s stationary states are energy eigenstates. An oscillating electron (particle) is in a superposition of at least two energy eigenstaates An electron in a Bohr atom that emits a Lyman α photon is in a superposition of the ground (E1) and the first excited state (E2). It oscillates in space at the freEE2E1quency h , exactly the frequency of the emitted Lyman α photon. ¯Our box example also shows: The more localized the initial spatial distribution Ψ(x, 0), the more eigenstates are involved, and the more complicated the time evolutiio will be (there will be interference terms oscillating at (E2 − E1)/h¯, (E3 − E1)/h¯, (E3 − E2)/h¯, . . . ) All motion of particles involves oscillating interference. What is the relation between the SE and CM QM should reproduce CM as limiting case dx • CM p = mv = m dt • We expect this (and other) classical equation(s) to hold for the QM expectation values (average position, momentum), at least in some limiting case. Calculate m dx : the only time variation arises from the time variation of wave • dt function, x is coordinate, not particle position in the SE. Massachusetts Institute of Technology XII-7 � � � � � � � � � � � � � � � � � � � � � � 8.04 Quantum Physics Lecture XIIdx d m = m (12-44) dt dt �x� d = m dxΨ∗(x,t)xΨ(x,t) (12-45) dt ∂Ψ∗ ∂Ψ = mdx xΨ+Ψ∗x (12-46) ∂t ∂t ��� �� �� 1 h¯2 ∂2Ψ∗ 1 h¯2 ∂2Ψ = m dx −ih¯−2m ∂x2 + V (x)Ψ∗ xΨ+Ψ∗xih¯−2m ∂x2 + V (x)Ψ (12-47) h¯∞ ∂2Ψ ∂2Ψ1 =2i dx ∂x2 xΨ − Ψ∗x ∂x2 − ih dx [Ψ∗V (x)xΨ − Ψ∗xV (x)Ψ] −∞ (12-48) The second term is zero, the first term can be integrated by parts: ∂2Ψ∗ ∂∂Ψ∗ ∂Ψ∗ ∂ ∂x2 xΨ= ∂x ∂x xΨ − ∂x ∂x (xΨ) ∂∂Ψ∗ ∂Ψ∗ ∂Ψ∗ ∂Ψ = ∂x ∂x xΨ − ∂x Ψ − ∂x x∂x (12-49) Similarly, ∂2Ψ ∂2Ψ∗∗ Ψ∗x = xΨ (12-50) ∂x2 ∂x2 ∂∂Ψ∗ ∂Ψ ∂Ψ∗ ∂Ψ A = ∂x ∂x Ψ − Ψ∗x ∂x − ∂x Ψ+Ψ∗ ∂x (12-51) ∂ ∂ψ∗ ∂Ψ ∂∂Ψ = ∂x ∂x xψ − Ψ∗x ∂x − ∂x (Ψ∗Ψ) + 2Ψ∗ ∂x (12-52) ∂Ψ h¯2 ∂2 ih¯= + V (x)Ψ → SE (12-53) ∂t −2m ∂x2 For the wavefunction to be normalizable, it has to vanish at ±∞ faster than √1x . Consequently, the integral over the first two terms in A yields zero and we are left Massachusetts Institute of Technology XII-8 � 8.04 Quantum Physics Lecture XIIwith dh¯∞ ∂Ψ m dx2Ψ∗ (12-54) dt �x� =2i ∂x � −∞ �� ∞ h¯∂ = dxΨ∗(x,t) Ψ(x,t) (12-55) i ∂x �−∞∞ = dxΨ∗(x,t)ˆpΨ(x,t)= �p� (12-56) −∞ So it follows from the SE that the expectation value of momentum is equal to the particle mass times the rate of change of the expectation value of particle position: d m (12-57) dt �x� = �p� This equation follows from the SE in combination with the position representation of ¯∂the momentum operator ˆp = hi ∂x . Does the appearance of 1 i mean that momentum is complex (imaginary)? Let us calculate the complex conjugate �p�∗ of the expectation values of �p� in some arbitrary state Ψ(x,t) ... Massachusetts Institute of Technology XII-9