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18 8.04 Quantum Physics Lecture X Figure IV: Infinite well. Particle in a box with infinite walls h d ψ + V (x)ψ(x) = Eψ(x) (10-) m dx implies that ψ(x) = 0 where V (otherwise E or curvature have to by infinite)= ψ(x) = 0 for x < 0 or x > a For 0 x a we have, h d ψ = Eψ(x) (10-3) m dx or me ψ = h ψ(x) (10-4) If E < 0, the solutions are of the form e ±kx with k = m h E. We require the wavefunction ψ(x) to be continuous (no steps), otherwise ψ becomes infinite. To x make ψ(0) = 0 for ψ = Ae kx + Be kx, we need A = B, but then ψ(a) 0, and ψ is not continuous. = There are no solutions for negative eigenvalues E. me For E > 0, the solutions are of the form e ±ikx with k = (or sin kx, cos kx). h The allowed values of E or k are determined by the boundary conditions. Writing ψ(x) = Ae ikx + Be ikx, we need A = B for ψ(0) = A + B = 0. Then for ψ = 0 we need Ae ika Ae ika = Ai (e ika e ikaa ) = ia sin ka = 0. This is fulfilled only if i ka = nπ or k n = n π, n = 1,, 3,... (n = 0, ψ = 0). Therefore the eigenvalues are a h k n h π E n = = n, n = 1,, 3,... energy eigenvalues m ma with corresponding eigenfunctions ( x ) ψ n (x) = C n sin k n x = C n sin nπ a where C n is some complex constant. The above is a general feature of QM. The boundary conditions on the wave function force the quantization of energy levels, i.e. allow only discrete energy values. Massachusetts Institute of Technology X-5

19 8.04 Quantum Physics Lecture X Figure V: Most de Broglie wavelengths lead to destructive interference. Figure VI: Eigenstates of box with infinite walls. Reason. Constructive interference, stationary phase after one round trip, for other energies ψ interferes away. In order to interpret ψ n (x) as a probability we need to normalize it so the probability to find the particle anywhere in space is unity dx ψ(x) = 1 Normalization of wavefunction (10-5) ( ) Here dx ψ n (x) a = C n dx sin nπ x 1 = 0 a a C n = 1 An overall global phase of the wavefunction has no physical consequences so we choose C n real. We obtain ( x ) u n = sin nπ a a normalized eigenstates of the square well n = 1,, 3,... (0 x a, u n = 0, otherwise) Note. The minimum energy of the particle in the box is not zero, but given by the h π ground-state energy E 1 = ma. This is sometimes called the zero-point energy. Massachusetts Institute of Technology X-6

20 8.04 Quantum Physics Lecture XI Since V (x) = 0 in the region where the particle can be found, the zero-point energy must be purely kinetic in this case. We could have estimated it using the Heisenberg uncertainty relation: Confining the particle to a region Δx = a introduces h a momentum uncertainty Δp = h, and a kinetic energy (Δp) h. Since Δx a m ma V (x) = 0 for x x a, and h ψ = Eψ in this region, and E = p in CM, m x m it appears that momentum is associated with the derivative of the wavefunction, x and kinetic energy with the curvature x. The eigenfunctions have the property that a ( x ) ( x ) dxu n (x)u m (x) = dx sin nπ sin mπ (11-1) 0 a a 1 a { ( x ) a ( x )} = dx cos (n m)π cos (n + m)π (11-) a 0 a a sin((n m)π) sin((n + m)π) = (11-3) (n m)π (n + m)π { 0 for n = m, = (11-4) 1 for n = m. = δ nm (11-5) { 0 for n = m, δ nm = 1 for n = m. Kronecker delta (11-6) This is a general property, not particular to this example: Eigenfunctions belonging to different eigenvalues are orthogonal, if the eigenfunctions are normalized we call them orthonormal. dxu n(x)u m (x) = δ mn orthonormality condition (11-7) Complex conjugate not necessary for box potential, where eigenfunctions are real, but necessary in general. u n u n dx = u n dx = 1 normalization (11-8) Eigenfunctions as basis Why is knowing the eigenfunctions important? Consider box potential: Fourier s theorem states that any function ψ(x) that ( satisfies ) that boundary conditions ψ(0) = 0 = ψ(a) can be written as a sum of sin nπ x. a Massachusetts Institute of Technology XI-1

21 8.04 Quantum Physics Lecture XI ( ) Note. cos nπ x terms do not appear since they do not satisfy the boundary conditions. a ( ) Since u n sin nπ x, we can also write a ψ(x) = c n u n (x) expansion of arbitrary ψ(x) into eigenfunctions (11-9) n=0 In order to calculate the expansion coefficients c n, we use the orthonormality property of the eigenfunctions. a ( ) dxu ψ(x) = dxu (x) c n u n (x) (11-10) m m 0 n=0 a = c dxu n mu n (x) = c n δ mn (11-11) n=0 0 n=0 = c m (11-1) Thus the expansion coefficient can be calculated as the integral c m = dxu m(x)ψ(x) expansion coefficients (11-13) Again we do not need the complex conjugate here, since u m is real, but we have written it in the correct general form. We call the set of functions {u n (x)} complete if an arbitrary function ψ(x) can be written as a superposition of functions of the set. A complete, orthonormal set of functions is called a basis. The above properties, derived for the particles in the box, are true in general in QM: 1. The energy eigenfunctions u n of a Hamiltonian h H + V (x) (11-14) = m x form a basis, an arbitrary wavefunction ψ(x) can be expanded as superposition of eigenstates ψ(x) = c n u n (x) (11-15) n with complex coefficients c n. If the spectrum of eigenvalues, or part of the spectrum of eigenvalues is continuous, the expansion contains an integral ψ(x) = n c n u n (x) + dec(e)u E (x). Massachusetts Institute of Technology XI-

22 8.04 Quantum Physics Lecture XI. The expansion coefficients c n are given by c n = dxu n(x)ψ(x), (11-16) i.e., they can be calculated once the eigenfunctions u n are known explicitly. Analogy to vector analysis Consider an n-dimensional vector space. A set of n mutually orthogonal unit vectors {ê i } i=1,...,n forms a basis, i.e. an arbitrary vector v can be expanded into unit vectors: n v = c i ê i (11-17) }{{}}{{} i=1 complex basis number vector with suitably chosen coefficients c i. The c i are uniquely determined, and given by c i = ê i v. (11-18) In vector analysis terms, the wavefunctions ψ(x) form a vector space, called the Figure I: projection of v onto ê i Hilbert space, the energy eigenfunctions u n (x) form a basis. The dimension of the Hilbert space is the number of independent energy eigenfunctions; if that number is infinite, the Hilbert space is infinite-dimensional. We have the following correspondences: Massachusetts Institute of Technology XI-3

23 8.04 Quantum Physics Lecture XI QM wavefunction: ψ(x) energy eigenfunction: u i (x) vector: v basis vector: ê i vector analysis # of independent u i (x) dimension of vector space dxu i (x)ψ(x) dot product: ê i v dxψ 1 (x)ψ (x) dot product: v 1 v dxu i (x)u j (x) = δ ij orthonormality: ê i ê j = δ ij ψ(x) = i c iu i (x) for any ψ(x) Completeness: v = i c iê i for any v Eigenvalue equation: Ĥψ E (x) = Eψ E (x) Set of eigenfunction of certain type of operators (Hermitian operators) forms basis. Hamiltonian is such an operator. Eigenvalue equation: Mˆ v = mv Set of eigenfunctions of certain types of matrices forms basis. (self-adjoint matrices M = M.) Physical interpretation of expansion coefficients Assume that we have prepared some arbitrary wavefunction ψ(x) (that is consistent with the coundary conditions) inside the box. Figure II: Pictorial expansion of wavefunction in a box in terms of eigenfunctions. Massachusetts Institute of Technology XI-4

24 8.04 Quantum Physics Lecture XI Expansion into eigenfunctions: ψ(x) = c i u i (x) (11-19) i=1 = c 1 u 1 (x) + c u (x) +... (11-0) phase (sign) of c i important, detemines whether more amplitude on left/right etc. The larger c i, the more the wavefunction ψ(x) is like u i ( x), (the larger the projection of ψ(x) onto u v) c i = dxu i, given by (analogy to ê i i (x)ψ(x)). Calculate i=1 c i : c i = c i c i (11-1) i=1 i=1 = c i=1 i dxu i (x)ψ(x) = dxψ(x) ci u i (x) (11-) ( ) = dxψ(x) c i u i (x) (11-4) i=1 = dxψ(x)ψ (x) (11-5) = dx ψ(x) (11-6) i=1 c Since i = 1, the quantity c i can be interpreted as the probability to find the particle in the state i, if a measurement of the particle s energy eigenstate is made, given that the particle has been initially prepared in a state characterized by an arbitrary wavefunction ψ(x). How is a measurement of the energy eigenstate of the particle made? = Measure energy of particle. h π E n = ɛn, ɛ = E 1 = ma i=1 (11-3) = 1 (11-7) Massachusetts Institute of Technology XI-5

25 8.04 Quantum Physics Lecture XI Figure III: Measurement of particle energy in state Ψ that is not an energy eigenstates can yield different values E i with probabilities c i. Massachusetts Institute of Technology XI-6

26 8.04 Quantum Physics Lecture XII After the energy measurement After the energy measurement with outcome E i, the particle will be in the energy eigenstate u i, and all subsequent energy measurements will yield the energy E i. 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: E = c n E n n=1 generally valid (1-1) Using the definition of the expansion coefficients c n, we can also write this as E = c n E n n=1 ( ) = c n dxu n(x)ψ(x) n=1 = c n dxψ (x)e n u n (x) = c n dxψ (x) Hu ˆ n (x) ( ) = dxψ (x)ĥ c n u n (x) = dxψ (x)ĥψ(x) Expectation value of energy in state ψ(x): E = dxψ (x)ĥψ(x) valid for any potential, not only box potential (1-) n E n Hamiltonian operator and energy If we postulate that a particle of momentum p is associated with a debroglie wavelength λ db = h, then it is represented by a plane wave e ikx with a wavevector p Massachusetts Institute of Technology XII-1

27 8.04 Quantum Physics Lecture XII k = λdb = = h h, or p = hk. Then, since the Fourier transform φ(k) of ψ(x) gives the probability amplitude for the plane wave with wavevector k π πp p ( ) 1 ikx ψ(x) = dkφ(k)e, (1-3) π the expectation value of momentum is given by p = dkhk φ(k) (1-4) p = dk( hk) φ(k) (1-5) Note. In PS5 you show using the properties of Fourier transforms that these expectation values can also be expressed as It follows that the expectation value of the KE is h p = dxψ (x) ψ(x) (1-6) i x ( ) h p = dxψ (x) ψ(x) (1-7) i x ( ) p h T = m = dxψ (x) m x ψ(x) (1-8) 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) = dxψ (x)v (x)ψ(x) (1-9) Since E = V + T it follows that ( ) h E = dxψ (x) + V (x) ψ(x) (1-10) m x = dxψ (x) Hψ(x) ˆ (1-11) 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 E 0, i.e. if Massachusetts Institute of Technology XII-

28 8.04 Quantum Physics Lecture XII Ĥψ E0 (x) = E 0 ψ E0 (x), then E = dxψ (x)hψ ˆ E 0 E 0 (x) = dxψe 0 (x)e 0 ψ E0 (x) = E 0 dx ψ E0 (x) = E 0, (1-1) where we have used the fact that the wavefunction is normalized. This shows that the constant E 0 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 Ô of an operator Ô acting on a wavefunction ψ(x) via the sandwich form Ô = dxψ (x) Oψ(x) ˆ (1-13) Then we have E = Ĥ = dxψ (x) Hψ(x) ˆ (1-14) The mean energy of the state described by the wave-function ψ(x) is the expectation value of the Hamiltonian operator Ĥ. We say that the Hamiltonian Ĥ is the operator associated with the measurable quantity energy. The operator Tˆ associated with the kinetic energy is h Tˆ = (1-15) m x with T = Tˆ = dxψ (x) T ˆ ψ(x), while the operator V ˆ for the potential energy is simply a multiplicative factor Vˆ = V (x) (1-16) with V = V ˆ = dxψ (x) V ˆ ψ(x) Why is potential energy associated with a simple multiplicative 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

29 8.04 Quantum Physics Lecture XII Another 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 to weigh p for each p with the probability to find the particle momentum between m p and p + dp: p = m = dp p m φ(p) (1-17) dpφ (p) p φ(p) (1-18) m We see that in momentum space the KE operator is simply a multiplicative factor pˆ ˆ p T = = in momentum space (1-19) m m 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) (1-0) p ( ) n n x = dpφ (p) ih φ(p) (1-1) p Consequently, for any potential function V (x) = a n x n (1-) n=0 we can calculate the expectation value V as ( ) V = dpφ (p)v ih φ(p) p = dpφ (p) V ˆ φ(p) (1-3) Consequently, the representation of the operator for the PE in momentum space is ( ) Vˆ = V ih, (1-4) p Massachusetts Institute of Technology XII-4

30 8.04 Quantum Physics Lecture XII where a function V of an operator is defined in terms of its Taylor expansion, Eq. (1-). It follows that the Hamiltonian is Ĥ = Tˆ + Vˆ (1-5) 1 ( h ) = + V (x) in position space (1-6) m i x ( = p m + V i h ) in momentum space (1-7) p The SE equation is always the same: ih Ψ(x, t) = ĤΨ(x, t) time-dependent SE (1-8) t ih Φ(p, t) = ĤΦ(p, t) t time-dependent SE (1-9) Ĥψ(x) = Eψ(x) time-independent SE (1-30) Ĥφ(p) = Eφ(p) time-independent SE (1-31) Example. For the harmonic oscillator, the SE (in appropriately chosen units) looks the same in position and momentum space: 1. linear potential V (x) = Ax h ψ (x) + Axψ(x) = Eψ(x) m p φ(p) + i m haφ (p) = Eφ(p) in position space (1-3) simpler equation in momentum space (1-33). harmonic oscillator: V (x) = 1 mω x h 1 ψ (x) + mω x ψ(x) = Eψ(x) (1-34) m p φ(p) 1 h mω φ (p) = Eφ(p) (1-35) m 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

31 8.04 Quantum Physics Lecture XII Time 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) = c 1 u 1 (x) + c u (x) + = n=1 c n u n (x). Since each eigenstate u n (x, t) evolves at a rate given by its eigenenergy E n, u n (x, t) = u n (x, t = 0)e ient/ h (1-36) = u n (x)e ient/ h (1-37) the wavefunction Ψ(x, t) at later time t is simply given by the linear superposition Ψ(x, t) = c n u n (x)e ient/ h (1-38) n=1 where the expansion coefficients c n are calculated at t = 0: c n = dxu n(x)ψ(x, t = 0) (1-39) Hence the importance of energy eigenstates and eigenvalues: The eigenvalues represent 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) = 1 (u 1(x) + u (x)). Particle in equal superposition of ground and first excited state. 1 [ ] Ψ(x, t) = u 1 (x)e ie 1t/ h + u (x)e ie t/h 1 [ ] = e ie 1t/ h u 1 (x) + u (x)e i(e E 1 )t/h (1-40) (1-41) 1 Ψ(x, t) = u 1 (x) + u (x)e i(e E 1 )t/ h (1-4) At any fixed position, interference term between u 1 and u oscillates from constructive to destructive interference with angular frequency E E 1 ω 1 = (1-43) h The energy difference determines the oscillation of the particle between the halves of the box. Massachusetts Institute of Technology XII-6

32 8.04 Quantum Physics Lecture XII Figure 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) = u n (x), then Ψ(x, t) = Ψ(x, 0), 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 eigenstates. An electron in a Bohr atom that emits a Lyman α photon is in a superposition of the ground (E 1 ) and the first excited state (E ). It oscillates in space at the fre E E quency 1, exactly the frequency of the emitted Lyman α photon. h 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 evolution will be (there will be interference terms oscillating at (E E 1 )/h, (E 3 E 1 )/h, (E 3 E )/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 function, x is coordinate, not particle position in the SE. dt Massachusetts Institute of Technology XII-7

33 8.04 Quantum Physics Lecture XII dx d m = m (1-44) dt dt x d = m dxψ (x, t)xψ(x, t) (1-45) dt { } Ψ Ψ = m dx xψ + Ψ x (1-46) t t [ { } { }] 1 h Ψ 1 h Ψ = m dx + V (x)ψ xψ + Ψ x ih + V (x)ψ m x ih m x (1-47) [ ] h Ψ Ψ 1 = dx i x xψ Ψ x dx [Ψ V (x)xψ Ψ xv (x)ψ] x ih The second term is zero, the first term can be integrated by parts: (1-48) ( ) Ψ Ψ Ψ x xψ = x x xψ x x (xψ) ( ) Ψ Ψ Ψ Ψ = xψ Ψ x (1-49) x x x x x Similarly, ( ) Ψ Ψ Ψ x = xψ (1-50) x x [ ] Ψ Ψ Ψ Ψ A = Ψ Ψ x Ψ + Ψ x x x x x [ ] ψ Ψ Ψ = xψ Ψ x (Ψ Ψ) + Ψ x x x x x (1-51) (1-5) ( ) Ψ h ih = + V (x) Ψ SE (1-53) t m x For the wavefunction to be normalizable, it has to vanish at ± faster than 1 x. Consequently, the integral over the first two terms in A yields zero and we are left Massachusetts Institute of Technology XII-8

34 8.04 Quantum Physics Lecture XII with d h Ψ m dxψ (1-54) dt x = i x ( ) h = dxψ (x, t) Ψ(x, t) (1-55) i x = dxψ (x, t)ˆpψ(x, t) = p (1-56) 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 x = p (1-57) dt This equation follows from the SE in combination with the position representation of the momentum operator ˆp = h 1. Does the appearance of mean that momentum is i x i 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

35 8.04 Quantum Physics Lecture XIII p = pˆ (13-1) ( ( ) ) h = dxψ Ψ (13-) i x ( ) h = dxψ Ψ (13-3) i x [ ] h = dx (ΨΨ ) Ψ Ψ (13-4) i x x ( ) h = dxψ Ψ (13-5) i x = p, (13-6) where again we have used integration by parts and the fact that Ψ vanishes at ±. Consequently, p = p, i.e. all expectation values of ˆp are real. Since an eigenvalue is the expectation value for the corresponding stationary state, all eigenvalues of the momentum operator must be real. An operator whose eigenvaluse are real (or equivalently, whose expectation value for all admissible wavefunctions is real) is called a Hermitian operator. Physically measurable quantities are represented by Hermitian operators. Similarly, one can show that E = E for any state, so all energy eigenvalue are real: The Hamiltonian operator Ĥ is a Hermitian operator. Can we derive Newton s F = ma from the SE? dv CM: F = = ma = dp Let us calculate the expectation value of dp dx dt dt : dp d = p (13-7) dt dt ( ) d h = dxψ (x, t) Ψ(x, t) (13-8) dt i x ( ) h Ψ Ψ = dx + Ψ (13-9) i t x x t [ ( )] h Ψ ψ Ψ h Ψ = dx m x x + V Ψ x Ψ x + V ψ (13-10) m x }{{} A Massachusetts Institute of Technology XIII-1

36 8.04 Quantum Physics Lecture XIII The integrand is ( ) h Ψ Ψ + Ψ h Ψ A = V Ψ + Ψ m x x x m x x ( ) h Ψ Ψ V Ψ Ψ (13-11) m x x Ψ Ψ V x x [ ] ( ) h Ψ Ψ Ψ V = Ψ m x x Ψ Ψ (13-1) x x x Again the integral over the first term vanishes since Ψ 0 for x ±, and we are left with ( ) dp V dv = dxψ (x, t) (x) Ψ(x, t) = (13-13) dt x dx or d dv dt p = dx (13-14) It follows from the SE that the expectation values obey the classical equations of motion. d m x = p dt d dv dt p = dx (13-15) (13-16) Average momentum changes due to average force ( ) dv V = dxψ Ψ = dxf (x) Ψ(x, t), (13-17) dx x i.e. position-dependent force F (x) is weighted by probability density Ψ(x, t) for dv d finding the particle at position x at time t. Note, however, that = V ( x ) dx d x Example 1. Double-peaked distribution. The probability to find the particle at the average position x is small, so the force there cannot be of much consequence for the particle s motion. Example. Force varying quickly on wavepacket scale. Classical calculation d V ( x ) d x would predict very large (and quickly varying force as x changes), actual QM force Massachusetts Institute of Technology XIII-

37 8.04 Quantum Physics Lecture XIII Figure I: Double-peaked particle distribution with vanishing probability to find particle at average position. Figure II: Particle wavepacket large compared to spatial variation of the force. The time evolution of the wavepacket will depend on the value of the force averaged over the wavepacket, not just on the force at the average particle location. dv experienced by particle is much smaller. However, if the force varies slowly dx compared to the size of a single wavepacket, then dv d = F (x) F ( x ) = V ( x ) (13-18) dx d x This is the reason why we can treat particles in macroscopic potential usually as classical particles. d dv always true (13-19) dt p = dx Massachusetts Institute of Technology XIII-3

38 8.04 Quantum Physics Lecture XIII d d dt p d x V ( x ) Ehrenfest s theorem for slowly varying potentials (13-0) Eigenfunctions of the momentum operator What are the eigenfunctions u p of the momentum operator, i.e. the eigenfunctions satisfying pu ˆ p = pu p, (13-1) where p is some (fixed) particular eigenvalue of ˆp. We know that the operator ˆp is Hermitian, so all eigenvalues p are real. In position space, we have pˆ = h and i x h u Ae ipx/ h p (x) = pu p (x), u p (x) =. The momentum eigenfunctions are (of course) i x just the plane waves. Let us check the orthonormality condition for eigenstates: dxu p (x)u p (x) = A p A p dxe ip x/h e ipx/h (13-) = A p A p dxe i(p p )x/ h (13-3) = ha p A p dye i(p p )y (13-4) dx u p (x) = A P = ha p A p δ(p p )π (13-5) = hπa p A p δ(p p ) (13-6) The momentum eigenfunction are orthogonal for p = p, but we have a normalization problem for p = p : The Dirac delta function diverfes, or equivalently, the integral dx e ipx/ h (13-7) }{{ } =1 diverges. Before looking at possibilities to deal with normalization problem, let us calculate the expansion coefficients c(p) c(p) = dxa p e ipx/ h ψ(x) = A p dxψ(x)e ipx h (13-8) }{{} π hφ(p) Massachusetts Institute of Technology XIII-4

39 8.04 Quantum Physics Lecture XIII We see that if we make the normalization choice A p = 1, then the expansion π h coefficients c(p) into momentum eigenstates are just given by the Fourier transform 1 u p (x) = e ipx/ h normalized momentum eigenstates (13-9) πh ψ(x) = dpφ(p)u p (x) (13-30) } {{} expansion into momentum eigenstates 1 = dpφ(p)e ipx/ h (13-31) } πh {{} Fourier transformation The expansion into momentum eigenstates and Fourier tranformation are one and the same. Since ψ(x) and φ(p) contain the same information about the particle, we can use either one to characterize the position and motion of the particle. A more fundamental motion is the state of the particle (a state is a vector in Hilbert space), the state can be expressed (written down) in various representations (like position representation ψ(x), momentum representation φ(p), energy representation c E ) associated with Hermitian operators (position ˆx, momentum ˆp, energy Ĥ). We call φ(p) the momentum representation of a particular state, and interpret it as the wavefunction in momentum space. The SE governs the time evolution of the wavefunction, or equivalently, the time evolution of the state of the particle in Hilbert space. For one particle in one (three) dimensions, the Hilbert space is one- (three-) dimensional, but for N particles in three dimensions the Hilbert space is 3N-dimensional. In general, it cannot be factored into a tensor product of N three-dimensional vector space V system = V 1 V V N, or equivalently, the wavefunction for N particles does not factor into a product of wavefunctions for each particle, Ψ(r 1, r,..., r 1, t) = Ψ 1 (r 1 )Ψ (r )... Ψ N (r N ) (13-3) In this case, when the wavefunction for an N-particle system cannot be written as a product of wavefunctions for the individual particles, i.e. when the particles do not evolve independently, we speak of an entangled state. Because of this possibility a quantum system of N particles is vastly (exponentially in N) richer than an classical system of N particles. However, in most cases we lose track of the particle-particle correlations associated with entanglement, and the system behaves quasi-classically. A quantum system that could preserve the correlations, and that could be manipulated externally, would constitute a quantum computer. A quantum computer could solve certain computation problems (only a handful have been discovered so far) exponentially faster than a classical computer. Because of the enormous size of the Massachusetts Institute of Technology XIII-5

40 8.04 Quantum Physics Lecture XIII Hilbert space, certain quantum mechanical problems involving many-particle correlations (e.g. high temperature superconductivity that involves correlated motion of many electrons) are very difficult to solve or simulate on a classical computer. Now back to a single particle in one dimension... Massachusetts Institute of Technology XIII-6

41 8.04 Quantum Physics Lecture XIV Normalization of wavefunctions in free space The momentum eigenstates in the position representation, u p (x) defined by and given by h pu ˆ p (x) = u p (x) = pu p (x), (14-1) i x 1 ipx/ h u p (x) = e (14-) πh cannot be normalized in free space to be interpreted as a probability density since u p (x) 1 =, and π dx u h p (x) diverges. However, they do satisfy the continuum orthonormality condition dxu p (x)u p (x) = δ(p p ). (14-3) This normalization corresponds to a uniform particle density (particle per meter) given by u p (x) 1 =. Let us calculate the probability current (particles moving π h past a point x per second), defined by [ ( ) ] h ψ ψ j(x) = ψ (x) (x) (x) ψ(x) see PS (14-4) im x x For ψ(x) = u p (x) we find [ ( )] h 1 ip ip j(x) = im πh h h (14-5) 1 p =, πh m (14-6) which is exactly what we expect for a uniform particle density u p (x) = π 1 h moving p at velocity v =. m In general, choosing a wavefunction ψ(x) = Ce ipx corresponds to particles moving p at velocity m, a particle density psi(x) = C, and a particle current j(x) = p C. Alternatives to deal with the normalization problem (wavefunction not squarem integrable) for momentum states are: 1. Wavepackets A superposition of a finite number of momentum eigenstates is not normalizable, but a wavepacket consisting of an infinite number of momentum eigenstates (Fourier components) is. Massachusetts Institute of Technology XIV-1

42 8.04 Quantum Physics Lecture XIV Figure I: A wavepacket ψ(x) in position space or φ(k) in momentum space whose wavefunction for large x (or k) falls off faster than x 1/ (k 1/ ) can be directly normalized.. Periodic boundary conditions Assume box of finite length L, require periodic boundary conditions ψ(0) = ψ(l) (14-7) For plane waves e ipx/h this implies that e ipl/h = 1 or pl = kl = nπ, n integer, h π i.e. momentum is quantized, p n = n hk o with k 0 =. The corresponding L momentum states are normalizable in the interval [0, L], L dx Ce ipx/ h = L C = 1 (14-8) 0 1 u pn (x) = L e ipnx/ h (14-9) normalized momentum eigenstates in box of size L with p n = n hk o L dxu pn (x)u pm (x) = δ nm orthonormality condition in box (14-10) 0 We perform all calculations for fixed size box, then take the limit L (i.e. 0, momentum spectrum becomes continuous). All physically sensible k 0 Massachusetts Institute of Technology XIV-

43 8.04 Quantum Physics Lecture XIV Figure II: Wavefunction in box of length L with periodic boundary conditions. results will be independent of the initially chosen box size L as long as L is large compared to distances of interest. Time evolution of free-particle wavepackets In free space we often work with normalized Gaussian wavepackets 1 x 4w Ψ(x, t = 0) = 0 (π) 1/4 w 1/ e (14-11) 0 Written in this form we have x Ψ(x, 0) 1 = (π) 1 / w 0 e w 0 dx Ψ(x, 0) = 1 x = 0 x = w 0 (δx) = x x = w 0 Δx = w 0 is the uncertainty or rms width (root-mean-square width) of the wavepacket. Why do we prefer this Gaussian form of wavepacket? 1. Particularly simple and symmetric, the Fourier transform is also a Gaussian wavepacket: 1 4k φ(k) = k 0 1/ e (14-1) (π) 1/4 k 0 with k 0 = w 1 0. (Δk) = k k = k 0. This is a wavepacket with the minimum uncertainty ΔxΔk = 1 (ΔxΔp = h ) allowed by QM 3. Physical system after give rise to Gaussian broadening in momentum or position, e.g., thermal distribution of atomic momenta in a gas is a Gaussian distribution. Massachusetts Institute of Technology XIV-3

44 8.04 Quantum Physics Lecture XIV How do we make a wavepacket move at velocity v 1? We displace the distribution in momentum space from p = hk = 0 to p = hk = hk 1 = mv 1 (see Fig. III). φ(k) = 1 (π) 1/4 k 0 (k k 1 ) 4k 0 1/ e. (14-13) The inverse Fourier transform, i.e., the spatial wavefunction 1 Ψ(x, t = 0) = (π) 1/ x 4wo 4 w 1/ e e ik 1x 0 (14-14) is still a Gaussian, but now with a phase variation e ik 1x, rather than a constant phase over the wavepacket (compare Eq. (14-11). This phase variation e ik 1x in position Figure III: Moving Gaussian wavepacket with average velocity v 1 = hk 1 /m and spatial 1 wavefunction ψ 1 (x) = (π) 1/ e x 4w eik 1 x 0. 4 w 1/ 0 hk 1 space encodes the motion of the wavepacket at velocity v 1 = m : The dominant de Broglie wavelength in the wavepacket corresponds to a wavevector k 1, or a momentum hk 1. Massachusetts Institute of Technology XIV-4

45 8.04 Quantum Physics Lecture XIV How does a free-space Gaussian wave packet evolve in time? In general, we expand a wavefunction Ψ(x, 0) into energy eigenfunctions u E (x), and then evolve the energy eigenfunctions as e iet/ h. In free space, there is only KE. Then the momentum eigenstates u p (x) are simultaneous eigenstates of energy: or ˆ pˆ Hu p (x) = u p (x) (14-15) m ( ) 1 1 = hi e ipx/ h (14-16) m x πh p = u p (x) (14-17) m Hu ˆ (x) = p p u p (x) m (14-18) = E p u p (x) (14-19) in free space. The energy eigenstates are said to be doubly degenerate: For each eigenvalue of energy E > 0 there are two different momentum states (namely u ±p (x) with p = mh ) that have the same energy. It follows that a momentum eigenstate with eigenvalue p evolves in time as e iept/ h, so that the wavefunction in momentum space evolves in time as Φ(p, t) = Φ(p, 0)e i p t/ h m (14-0) time evolution of momentum eigenfunctions in free space. The wavefunction in real space is given by the inverse Fourier transform Ψ(x, t), or equivalently, as the superposition of energy eigenfunctions with their corresponding phase evolution factors e iept/ h : 1 ipx/h Ψ(x, t) = dpφ(p, t)e (14-1) πh ( = dpφ(p, t)u p (x)) (14-) 1 p = dpφ(p, 0)e ipx/ h e m t/h (14-3) πh = dpφ(p, 0)U p (x, t) (14-4) = dpφ(p, t)u p (x) (14-5) Massachusetts Institute of Technology XIV-5

46 8.04 Quantum Physics Lecture XIV eipx/ h e i p π h where U p (x, t) = u p (x)e p t/ h m = 1 t/ h m are the time-dependent momentum eigenfunctions in free space. The above equation shows that the phases of different Fourier components u p (x) = 1 eipx/ h evolve in time at different speeds, the π h running out of phase of different Fourier components leads to a spreading of the wavepacket in position space. In the problem sets you will show that the rms width Δx(t) = w(t) of the wavepacket grows in time as h k w(t) = w (14-6) m w 0 Since a wavepacket contains different momentum components, it changes in time in free space even though there are non external forces acting. For long times t t 0 = mwo h h the wavepacket spreads as w(t) t, i.e. at a speed v 0 = that h mw 0 mw 0 is inversely proportional to its initial size. That speed is negligible for macroscopic wavepacket size, but can be appreciable for initially well-localized microscopic objects. The spreading of a wavepacket in free space was early evidence that the wavepacket size cannot be identified with the particle size. The spreading is due to the quadratic (i.e. not linear) dependence of the energy, and hence the phase evolution rate, on momentum. Note that the wavepacket of a massless particle, e.g. a photon, with E = pc would not spread. (The SE is non-relativistic and does not apply to photons.) Motion of wave packets, group velocity, and stationary phase Why is it that a wavefunction 1 Ψ(x, 0) = (π) 1/ x 4 w 1/ e 4w0 e ik 1x 0 (14-7) hk 1 represents a particle moving at velocity v 1 = m? Since a crest of a single momentum 1 hk π m component u k1 (x, t) = π e k1x e i t moves forward a distance λ = k1 in a time hk T = π (remember that ω 1 ω1 1 = and e ie 1t/ h = e iω1t ), the velocity of the crest is m λ π ω ω 1 hk v 1 ph = T = = k1 π k 1 = m ω 1 hk 1 p 1 v ph = = = (14-8) k 1 m m This is the phase velocity of a momentum component. The particle does not move at the phase velocity v ph = ω 1 k 1 at which the plane wave associated with a single momentum moves forward. At what velocity then? Massachusetts Institute of Technology XIV-6

47 8.04 Quantum Physics Lecture XIV Look at exponent and write: hk E 1 1 = E 1 (k 1 ) = hω 1 = hω(k 1 ) = m e ( ) x 4w0 +ik 1x iω(k 1 )t Remember Fermat s principle of stationary phase: path is defined by region of space x where phasors point mostly in one direction, i.e. where the phase φ(k) = 4wo + ikx + iω(k 1 )t does not vary between different momentum components k to lowest order ( ) ( ( )) φ ω ω 0 = = ix i t = i x t, (14-9) k k x or ( ) Fermat s principle leads us to the concept of group velocity ω x(t) = t. (14-30) k ω hk 1 p 1 v gr = (k 1 ) = = (14-31) k m m Group velocity of the wavepacket at which the wavepacket, i.e. the region of constructive interference, propagates. The difference between group and phase velocity is due to the fact the ω = ω, or E = ( hω) = E, i.e. the quadratic dependence of KE k k p ( hk) p on momentum in free space. This is in contrast to photons with a linear dispersion relation ω = ω = c in vacuum, where group and phase velocity are the same. k k Massachusetts Institute of Technology XIV-7

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86 8.04 Quantum Physics Lecture XVIII 1 multiplied by u 0 (y) = e y at infinity. Consequently, we need the series to terminate, which requires ɛ m = m + 1 for some m. Thus, ( ) hω 1 E n = ɛ n = hω n + HO energy levels (18-1) Quantized energy levels of a harmonic oscillator. The ground state (zeropoint) energy is E 0 = 1 hω, the energy levels are equidistant. Note. This feature allows us to identify the HO not only with a particle in potential V (x) = 1 mω x, but also with a system of noninteracting (bosonic) particles. Therefore, a mode of an electromagnetic field of frequency ω can be viewed as a HO with frequency ω; n photons in that mode correspond to the n-th occupied state of the HO. The uncertainty in x and p of the HO ground state corresponds to the vacuum fluctuations of the electromagnetic field x = 0, x = 0 corresponds to E = 0, E = 0 etc. yields For given ɛ = n + 1, the recursion relation ( ) n c = 1 c 0 (m + 1)(m + )c m+ = (m ɛ n + 1)c m = (m n)c m (18-) 4 n (4 n)(0 n) c 4 = c 3 4 = c and in general c k = ( ) k n(n ) (n k + 4)(n k + ) c 0, (k)! 0 k n, n even (18-3) for the even coefficients. For the odd coefficients we have c 3 = ( n) c 1 = ( ) n 1 c c 5 = 6 n c 3 = ( ) n 3 c 3 = ( ) (n 3)(n 1) c and in general c k+1 = ( ) k (n 1)(n 3) (n k + 3)(n k + 1) c 1, (k + 1)! 0 k + 1 n, n odd. (18-4) Massachusetts Institute of Technology XVIII-1

87 8.04 Quantum Physics Lecture XVIII ( ) The eigenfunction u n (x) for energy level n with energy E n = hω n + 1 is given by n/ 1 u n (y) = e y c k y k for even n, (18-5) u n (y) = e 1 y n=0 (n 1)/ n=0 c k+1 y k+1 for odd n, (18-6) The coefficient c 0 or c 1 has to be chosen such that the wavefunction is normalized, mω h and y is related to the position coordinate x via y = x. The quantity h mω has units of length and defines the natural quantum length scale for the harmonic oscillator. Apart from the normalization, the polynomials n/ h n (y) = c k y k (18-7) k=0 (n +1)/ h n (y) = c k+1 y k+1 (18-8) k=0 are the Hermite polynomials H n (y). The Hermite polynomials obey the following relations: H n (y) yh n (y) + nh n (y) = 0 (defining equation) (18-9) H n+1 (y) yh n (y) + nh n 1 (y) = 0 (18-10) H n+1 (y) H n (y) + yh n (y) = 0 (18-11) n z H n (y) = e zy z (18-1) n! n=0 ( ) ( ) d alternative definition H n (y) = ( 1) n e y e y (18-13) dy of Hermite polynomials Since the wavefunction belonging to level n is u n (y) = Ce 1 H n (y), in order to normalize it, we need to calculate dy u n (y) = dy C e y H n (y) (18-14) The Hermite polynomials are real. One can show that dye y H n (y) = n n! π (18-15) Massachusetts Institute of Technology XVIII-

88 8.04 Quantum Physics Lecture XVIII Normalization actually requires dx (x) mω u = 1, but since x and y = x are n h h related by a constant factor, dx u n (x) = dy u mω n (y). H 0 (y) = 1 (18-16) H 1 (y) = y (18-17) H (y) = 4y (18-18) H 3 (y) = 8y 3 1y (18-19) H 4 (y) = 16y 4 48y + 1 (18-0) H 5 (y) = 3y 5 160y y (18-1) Consequently, the lowest eigenfunctions look like (Fig. I). The eigenfunctions of the Figure I: HO eigenfunctions. HO look the same in momentum space, since the Hamiltonian is symmetric in x and p, and xˆ = x, h pˆ = i x in position space ψ(x) (18-) xˆ = ih, p pˆ = p in momentum space φ(p) (18-3) Massachusetts Institute of Technology XVIII-3

89 8.04 Quantum Physics Lecture XVIII The harmonic oscillator ground state, being a Gaussian function with no spatial dependence of the complex phase, has minimum uncertainty allowed by the Heisenberg relation: h ΔxΔp = for the ground state (18-4) h ΔxΔp > for any excited state (18-5) Show Bose-Einstein condensate expansion Thermal cloud isotropic expansion for anistropic trap p ( = 1 kt ) m condensate: p 1 mω anisotropic expansion m x 0 h HO: operator method There is an elegant and instructive way to derive the HO eigenstates without directly solving the SE. Instead, we use commutation relation between operators. We start by writing the Hamiltonian in dimensionless form [ ] p mω H = hω + x (18-6) mhω h ( ) ( ) p x = hω + (18-7) mhω h/mω [ ( ) ( ) ] p x = hω + (18-8) p 0 x 0 with p 0 = m hω, x = h 0. Classically we can write, mω ( ) ( ) x p x p H cl = hω i + i, (18-9) x 0 p 0 x 0 p 0 however, since in QM pˆ and ˆx do not commute, we have ( ) ( ) ( ) ( ) xˆ pˆ xˆ pˆ xˆ pˆ i + i = + (18-30) x 0 p 0 x 0 p 0 x 0 p 0 ( ) ( ) xˆ pˆ i = + + [ˆx, pˆ]. (18-31) x 0 p 0 x 0 p 0 Massachusetts Institute of Technology XVIII-4

90 8.04 Quantum Physics Lecture XVIII Using the commutator [ˆx, pˆ] = ih p pih 1 mω 1 = ih and = = 1 we p p λ m 0 p 0 h hω h have ( ( ) ( ) ) ˆ x pˆ H = hω + (18-3) x 0 p 0 (( ) ( ) ) xˆ pˆ xˆ pˆ 1 = hω i + i i [ˆx, pˆ] (18-33) x 0 p 0 x 0 p 0 x 0 p (( ) ( ) ) 0 xˆ pˆ xˆ pˆ 1 = hω i + i + (18-34) x 0 p0 x 0 p0 We can define a new, non-hermitian operator by Consequently, the Hermitian conjugate operator is since ˆp = pˆ, ˆx = xˆ xˆ pˆ â := + i (18-35) x 0 p 0 ( ) â xˆ pˆ = + i = xˆ i pˆ (18-36) x 0 p 0 x 0 p 0 Note. The Hermitian conjugate operator O of any operator is defined by the relation dxψ (x)o ψ 1 (x) = dx (Oψ (x)) ψ 1 (x) (18-37) for any well-behaved wavefunctions ψ 1 (x), ψ (x). Consequently, for any operator Ô = c 1 Ô 1 + c Ô, where c 1 and c are complex numbers, ( ) Ô = c 1 Ô 1 + c Ô = c1 Ô 1 + c Ô (18-38) ( ) and for any operator Ô = Ô 1 Ô we have Ô = Ô 1 Ô = Ô Ô 1 Proof. See problem set. Using the operators â,â, we can write the Hamiltonian for the HO in the particularly simple form ( ) 1 Ĥ = hω â â + (18-39) Massachusetts Institute of Technology XVIII-5

91 8.04 Quantum Physics Lecture XVIII Rather than being explicitly defined in terms of ˆx, ˆp,..., and operator can be defined through its commutation relations with other operators. Let us look at â, â : [ ] [ ] xˆ pˆ xˆ pˆ a, ˆ a ˆ = + i, i (18-40) x [ 0 p ] 0 x 0 p [ 0 ] pˆ xˆ xˆ pˆ = i, i, (18-41) p 0 x 0 x 0 p 0 i = ([ˆp, xˆ] [ˆx, pˆ]) (18-4) p 0 x 0 i = [ˆp, xˆ] (18-43) h i h = h i (18-44) = 1, (18-45) where we have used [ˆx, xˆ] = 0 = [ˆp, pˆ]. So we have [ ] a, ˆ a ˆ = 1 (18-46) [ ] [â, â] = â, a = 0 (18-47) As will be elaborated on in 8.06, this defines a commutation relation for bosonic (quasi)-particles, i.e. particles whose wavefunction is symmetric under the exchange of two particles. For the commutators with the Hamiltonian, we have [ ] [ ] H, ˆ â = hω a ˆ ˆ a, a ˆ (18-48) ( ) = hω â ââ ââ â (18-49) ( ( ) ) = hω â ââ 1 + â â â (18-50) and = hωâ (18-51) [ ] [ ] H, ˆ â = hω a ˆ ˆ a, a ˆ (18-5) ( ) = hω â ââ â â â (18-53) ( ( )) = hω â ââ â ââ 1 (18-54) = hωâ (18-55) We are now in the situation to calculate the spectrum of eigenenergies of the HO simply using those commutation relations. Let us first note that since Ĥ is quadratic in x and p, all eigenvalues must be positive: E = Ĥ = T + V (18-56) 1 1 = dpφ (p)p φ(p) + mω dxψ (x)x ψ(x) > 0 (18-57) m Massachusetts Institute of Technology XVIII-6

92 8.04 Quantum Physics Lecture XVIII for any wavefunction ψ(x) and its Fourier transform φ(p). Before determining the eigenspectrum, let us define a convenient notation. State vector notation (Dirac notation) We have already argued that a physical state, (i.e., a physical system whose initial conditions have been prepared to the maximum extent allowed by QM), is described by a vector in an abstract vector space (Hilbert space), and that a wavefunction in position space is only one possible representation of the state. Alternatively, the state can be described in the momentum representation (wavefunction in momentum space), or by specifying the expansion coefficients when expanding the basis of energy eigenstates. Using a notation introduced by Paul Dirac, one of the creators of QM, we write the state as ψ (18-58) and define φ ψ := dxφ (x)ψ(x) (18-59) for any two states ψ, φ whose wavefunctions are by ψ(x), φ(x). Dirac introduced φ ψ (18-60) }{{}}{{}}{{} bra -c- ket So ψ is called a ket, and φ a bra. You can think of the bra as the transpose of the ket vector ( ) w1 v 1 v w = c-number (complex number), (18-61). but possibly for infinite-dimensional vectors. In this sandwich or Dirac notation, the expectation value of any operator  is given by  = ψ A ψ ˆ = dxψ (x) Aψ(x). ˆ (18-6) In Dirac notation, ( ) ψ φ = dxψ (x)φ(x) = dxφ (x)ψ(x) = φ ψ (18-63) An operator  acting on a state produces another state, symbolically A ˆ ψ = Aψ ˆ (18-64) Massachusetts Institute of Technology XVIII-7

93 8.04 Quantum Physics Lecture XVIII Consequently, ( ) φ A ψ ˆ = φ Aψ ˆ = dxφ (x) Aψ(x) ˆ = dxφ (x) Aψ(x) ˆ (18-65) The Hermitian conjugate operator  is defined by ( ) φ  ψ = dx (Aφ(x)) ψ(x) = dxψ (x) (Aφ(x)) (18-66) = ψ Aφ (18-67) = Aφ ψ (18-68) In Dirac notation, the orthonormality condition for eigenstates n, m reads n m = dxu n (x)u m (x) = δ nm, (18-69) where the expansion coefficients are c n = dxu n ψ(x) = n ψ. (18-70) A bracket like a b is a complex number, but a ketbra like b a is an operator since acting on a state it produces another state b }{{} a ψ. (18-71) }{{} state c-number One can show that the sum over all eigenstates n n n of a Hermitian operator is the unity operator n n = ˆ1, (18-7) and n ( ) expansion into ψ = ˆ1 ψ = n n ψ = c n n eigenstates n (18-73) Back to the operator treatment of the HO: Let us assume that we have found an energy eigenstate with eigenenergy E and let us denote that state by E. Let us define a new state ψ by having the operator â act on E, ψ := â E. What happens if we act with the Hamiltonian ψ? Ĥ ψ = Ĥâ ([ E (18-74) ] ) = H, ˆ â + âh ˆ E (18-75) = ( hω â + âe) E (18-76) = (E hω ) â E (18-77) = (E hω) ψ (18-78) Massachusetts Institute of Technology XVIII-8

94 8.04 Quantum Physics Lecture XVIII [ ] Here we have used the previously calculated result H, ˆ â = hωâ for the commu tator, and the fact that complex (here real) numbers commute with everything. The above formula signifies that ψ is also an energy eigenstate, but with lower energy E = hω. Since starting from any eigenstate E we can repeat the procedure any number of times, â n E = E n hω, (18-79) and the eigenenergy has to remain positive, (we have shown ψ Ĥ ψ > 0 for any state), there must exist a state 0 such that i.e., a state whose energy cannot be lowered further. â 0 = 0 (18-80) Note. It is important to distinguish between 0 (lowest energy eigenstate, vector in Hilbert space) and 0 (zero of the Hilbert space, vector of zero length). Note. Nothing implies that the state 0 has zero energy. in fact, ( ) 1 1 Ĥ 0 = hω â â + 0 hω 0, (18-81) = â 0 =0 1 so the ground state has eigenenergy E 0 = hω, this is the zero-point energy. In the context of identifying a HO at frequency ω with an electromagnetic mode at frequency ω, the ground state 0 is also called the vacuum (ground state has no excitations, 1 photon number is zero): the vacuum has finite vacuum energy E 0 = hω 0. What happens if â acts on ground state? Let us define 1 := â 0. (18-8) The tilde is there to remind us that this state is not necessarily normalized, even if 0 is chosen to be normalized. Massachusetts Institute of Technology XVIII-9

95 8.04 Quantum Physics Lecture XIX We then have Ĥ 1 = Ĥâ 0 ( [ = ] ) Ĥ, â + â Ĥ 0 ( = hωâ + â 1 ) hω 0 0 = 3 hωâ 0 = 3 hω 1, = a ˆ 0 is also an energy eigenstate, but with eigenenergy (19-1) (19-) (19-3) (19-4) (19-5) i.e., hω instead of hω for 0. Similarly, we can show that = aˆ 1 is also an energy eigenstate, but with 5 energy hω etc. Consequently, we can construct a ladder of (yet to be normalized) energy eigenstates ñ by ( ) ñ = â n 0 (19-6) with ( ) 1 E n = n + hω. (19-7) â (â ) is called the lowering (raising) operator, it lowers (raises) the energy by hω. Figure I: â, â are sometimes called ladder operators since they take us up and down the ladder of energy eigenstates. When describing a monochromatic electromagnetic field quantum mechanically, we can associate the frequency ω with a harmonic oscillator of that frequency. For noninteracting particles (such as photons) a state with n photons can be associated with the n-th eigenstate of the HO with n. The ground state then corresponds to an Massachusetts Institute of Technology XIX-1

96 8.04 Quantum Physics Lecture XIX empty mode (no photons, n = 0), however there is still a finite energy 1 hω that we associate with vacuum fluctuations of the electromagnetic field. In this context, â and â are called creation and annihilation operators, respectively, since they create and annihilate photons, or more generally, arbitrary non-interacting bosonic particles. Normalization of HO energy eigenstates Let us assume that the ground state 0 is already chosen to be properly normalized: 0 0 = 1. Note. Remember that 0 0 denotes 0 0 = dxu 0 (x)u 0 (x). How long is the state 1 = â 0? 1 1 = â 0 â 0 (19-8) The state 1 is already normalized, so we can write: = 0 â â 0 (19-9) = 0 a ˆâ 0 (19-10) [ ] = 0 a, ˆ a ˆ + a ˆ ˆ 0 a (19-11) ( ) = â â 0 â 0 = 0 (19-1) = 1 (19-13) 1 = â 0 normalized eigenstate (19-14) What about = â 1 = â 1? = â 1 â 1 (19-15) = 1 ââ 1 (19-16) ( ) = 1 â â (19-17) ( ) = 1 â â 1 = 0 (19-18) = (19-19) = (19-0) Massachusetts Institute of Technology XIX-

97 8.04 Quantum Physics Lecture XIX Then the properly normalized second excited state is 1 1 ( ) = = â 0. (19-1) ( ) We can show, in general, (see PS) that the length squared of the state ñ = â n 0 is ñ ñ = n!. Consequently, the n-th normalized eigenstate is 1 ( ) n n := n! â 0. (19-) We can also show (see PS) that â n = n n 1, (19-3) â n = n + 1 n + 1. (19-4) From operators back to spatial wavefunctions The condition on the ground state 0, â 0 = 0, reads in position space using our definition of the annihilation operator, xˆ pˆ mω i â = + i = xˆ + p, ˆ (19-5) x 0 p 0 h hmω ( ) mω i h au ˆ h x + 0 (x) = u 0 (x) = 0 (19-6) hmω i x ( ) mωx + h u 0 (x) = 0. (19-7) x The simple DE has the solution u 0 (x) = ce x with normalization 1 = c π h. mω Consequently, the normalized ground-state wavefunction is mω h The normalized n-th eigenstate can be obtained from ( ) 1 mω 4 mω u 0 (x) = e h x. (19-8) πh 1 ( ) n = n! â n 0 (19-9) or ( ) n 1 mω i h u n (x) = n! h x u 0 (x). (19-30) hmω i x Massachusetts Institute of Technology XIX-3

98 8.04 Quantum Physics Lecture XIX Commutators, Heisenberg uncertainty, and simultaneous eigenfunctions The fact that ˆp = h in the position representation (or ˆx = ih in the momentum i x p representation) implies ( ) ( ) pˆˆ xψ(x) = p ˆ xψ(x) = xˆ ˆpψ(x) = x pψ(x) ˆ, (19-31) i.e., ˆx and ˆp do not commute. Define the difference between ˆpxˆ and ˆxpˆ as the commutator [ ] p, ˆ xˆ = pˆxˆ xˆp. ˆ (19-3) Here: [ ] h p, ˆ xˆ = (c-number) (19-33) i [ ] In general, A, ˆ B ˆ = A ˆB ˆ BˆÂ is an operator. The commutator is linear. Other useful relations [ ] [ ] [ ] c 1  1 + c Â, Bˆ = c 1  1, Bˆ + c Â, Bˆ (19-34) [ ] [ ] B, ˆ  = A, ˆ Bˆ (19-35) [ ] [ ] [ ] ÂB, ˆ C ˆ = A ˆ B, ˆ C ˆ + A, ˆ C ˆ B ˆ (19-36) Simultaneous eigenfunctions Consider a free particle. The plane waves ψ(x) = e ±ikx are simultaneous eigenfunc tions of energy with eigenvalue h k, m h h k He ˆ ±ikx = m x e±ikx = e ±ikx, (19-37) m and of momentum with eigenvalue ± hk, pe ±ikx h e ±ikx ˆ = = ± hke ±ikx. (19-38) i x Note. If we had chosen cos(kx), sin(kx), these would have also been energy eigenh k functions with eigenvalue m, but not momentum eigenfunctions. Massachusetts Institute of Technology XIX-4

99 8.04 Quantum Physics Lecture XIX However, since cos(kx) and sin(kx) are degenerate (i.e., have the same energy eigenvalue), it is possible to choose linear combinations of degenerate eigenstates e ±ikx = cos(kx) ± i sin(kx) that are simultaneous eigenstates of momentum. In the potential well, on the other hand, the energy eigenstates were not simultaneous eigenstates of momentum. In general, we have: Theorem Two Hermitian operators Â, Bˆ have a set of simultaneous eigenfunctions if and only if they commute. Proof. Assume a complete set {u ab } of simultaneous eigenfunctions is found, i.e., Âu ab = au ab (19-39) Bu ˆ ab = au ab (19-40) a, b, eigenvalues. Then [ A, ˆ Bˆ]uab = (ab ba)u ab = 0 for all eigenfunctions [ A, ˆ B] ˆ = 0. See Gasiorowicz, 5-4. Since only an eigenstate of  will have a definite outcome when a measurement of  is made, this means that ΔA and ΔB can always be simultaneously made zero only when  and Bˆ commute. Theorem 19.. One can prove that in any chosen state ψ, (ΔA) ψ(δb) ψ i[ A, ˆ Bˆ] ψ (19-41) for any two Hermitian operators Â, Bˆ. Proof. see Gasiorowicz, online supplement SA. For ˆx, ˆp, we have (Δx) ψ(δp) 1 ψ iih 4 ψ = h 4, (19-4) where the RHS does not depend on the state ψ. This is another derivation of the Heisenberg uncertainty relation ΔxΔp h. The Schrödinger equation in three dimensions with Ĥψ(r) = Eψ(r) SE in 3D (19-43) pˆ = pˆ + ˆp + ˆp (19-44) x y z Massachusetts Institute of Technology XIX-5