� � � �8.04 Quantum Physics Lecture XVIII12 2multiplied by u0(y)= e− yat infinity. Consequently, we need the series to terminate, which requires �m =2m + 1 for some m. Thus, ¯1hω En = �n =¯hω n + → HO energy levels (18-1) 22 Quantized energy levels of a harmonic oscillator. The ground state (zeropoiint energy is E0 = 12 hω, the energy levels are equidistant. ¯Note. This feature allows us to identify the HO not only with a particle in potential V (x)= 12 mω2x2, 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, �x2��= 0 corresponds to �E� = 0, �E2= 0 etc. For given � =2n + 1, the recursion relation (∗)(m + 1)(m + 2)cm+2 = (2m − �n + 1)cm = (2m − 2n)cm (18-2) yields 2n • c2 = −12 c0 ·4−2n (4−2n)(0−2n)• c4 = 34 c2 = 1234 c0 ····and in general c2k =(−2)k n(n − 2) ··· (n − 2k + 4)(n − 2k + 2) c0,(2k)! 0 ≤ 2k ≤ n, n even (18-3) for the even coefficients. For the odd coefficients we have c3 = (2−2n) c1 =(−2)n−1 c1• 23 23··• c5 = 6−2n c3 =(−2)n−3 c3 =(−2)2(n−3)(n−1) c145 45 12345······and in general c2k+1 =(−2)k (n − 1)(n − 3) · (n − 2k + 3)(n − 2k + 1) c1,(2k + 1)! 0 ≤ 2k +1 ≤ n, n odd. (18-4) Massachusetts Institute of Technology XVIII-1 � � � � � � � � � 8.04 Quantum Physics Lecture XVIIIThe eigenfunction un(x) for energy level n with energy En =¯hω n + 12 is given by n/212un(y)= e− y c2ky 2k for even n, (18-5) n=0 (n�−1)/2 un(y)= e− 1 y c2k+1y 2k+1 for odd n, (18-6) 2 n=0 The coefficient c0 or c1 has to be chosen such that the wavefunction is normalized, and y is related to the position coordinate x via y = mω x. The quantity ¯h ¯mωh has units of length and defines the natural quantum length scale for the harmonic oscillator. Apart from the normalization, the polynomials n/2hn(y)= c2ky 2k (18-7) k=0 (n�+1)/2 hn(y)= c2k+1y 2k+1 (18-8) k=0 are the Hermite polynomials Hn(y). The Hermite polynomials obey the following relations: Hn��(y) − 2yHn�(y)+2nHn(y)=0 → (defining equation) (18-9) Hn+1(y) − 2yHn(y)+2nHn−1(y) = 0 (18-10) Hn+1(y) − Hn�(y)+2yHn(y) = 0 (18-11) � n∞z2 Hn(y)= e 2zy−z(18-12) n! n=0 ��2 �� d alternative definition Hn(y)=(−1)n ey2 dy e−y2 → of Hermite polynomials (18-13) Since the wavefunction belonging to level n is un(y)= Ce− 212Hn(y), in order to normalize it, we need to calculate ∞∞ 2 dy|un(y)|2 = dy|C|2 e−yHn 2(y) (18-14) −∞ −∞ The Hermite polynomials are real. One can show that ∞ dye−y2 Hn 2(y)=2n n!√π (18-15) −∞ Massachusetts Institute of Technology XVIII-2 � � 8.04 Quantum Physics Lecture XVIIINormalization actually requires ∞ dx(x)2 = 1, but since x and y = mω x are � −∞ |un|�� ¯h 2¯h 2related by a constant factor, ∞ dxun(x)= mω ∞ dyun(y). −∞ ||−∞ ||H0(y) = 1 (18-16) H1(y)=2y (18-17) H2(y)=4y 2 − 2 (18-18) H3(y)=8y 3 − 12y (18-19) H4(y) = 16y 4 − 48y 2 + 12 (18-20) H5(y) = 32y 5 − 160y 3 + 120y (18-21) 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 h¯∂ xˆ= x, pˆ= in position space ψ(x) (18-22) i ∂x ∂ xˆ= ih¯,pˆ= p in momentum space φ(p) (18-23) ∂p Massachusetts Institute of Technology XVIII-3 � � 8.04 Quantum Physics Lecture XVIIIThe harmonic oscillator ground state, being a Gaussian function with no spatial dependdenc of the complex phase, has minimum uncertainty allowed by the Heisenberg relation: h¯ΔxΔp = for the ground state (18-24) 2h¯ΔxΔp> for any excited state (18-25) 2 • Show Bose-Einstein condensate expansion Thermal cloud • • isotropic expansion for anistropic trap 2 (2pm = 21 kT ) condensate: 2 • p1 mω anisotropic expansion 22m ∝ x0 ∝ ¯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 p2 mω 2H =¯hω + x (18-26) 2mhω¯2¯h ⎡⎤ ��2 ��2 px =¯hω ⎣ √2mhω¯+ � 2¯h/mω ⎦ (18-27) �� �2 ��2 � px =¯hω + (18-28) p0 x0 with p02 =2m¯2 = 2¯h . Classically we can write, hω, x0 mω � �� � xpxpHcl =¯hω − i + i, (18-29) x0 p0 x0 p0 however, since in QM pˆand ˆx do not commute, we have � �� ���2 ��2 xˆpˆxˆpˆxˆpˆ− i + i = + (18-30) x0 p0 x0 p0 x0 p0 ��2 ��2 xˆpˆi = + +[ˆx,pˆ] . (18-31) x0 p0 x0p0 Massachusetts Institute of Technology XVIII-4 � � � � � � � � � � � 8.04 Quantum Physics Lecture XVIIIUsing the commutator [ˆx,pˆ] = ih¯∂ p − pih¯∂ = ih¯and 1 = mω 1 = 1 we ∂p ∂p 2¯√2m¯2¯λ0p0 hhω h have �� �2 ��2 � ˆxpˆH =¯hω + (18-32) x0 p0 ���� � � xˆpˆxˆpˆ1 =¯hω − i + i − i [ˆx,pˆ] (18-33) x0 p0 x0 p0 x0p0�� �� �� xˆpˆxˆpˆ1 =¯hω x0 − ip0 x0 + ip0 + 2 (18-34) We can define a new, non-Hermitian operator by xˆpˆaˆ:= + i (18-35) x0 p0 Consequently, the Hermitian conjugate operator is aˆ† = xˆ+ ipˆ† = xˆ− ipˆ(18-36) x0 p0 x0 p0 since ˆp† = pˆ, ˆx† = xˆNote. The Hermitian conjugate operator O† of any operator is defined by the relation ∞ ∞ dxψ2 ∗(x)O†ψ1(x)= dx (Oψ2(x))∗ ψ1(x) (18-37) −∞ −∞ for any well-behaved wavefunctions ψ1(x), ψ2(x). Consequently, for any operator Oˆ= c1Oˆ1 + c2Oˆ2, where c1 and c2 are complex numbers, Oˆ† = c1Oˆ1 + c2Oˆ2 † = c1 ∗Oˆ1†+ c2 ∗Oˆ2†(18-38) and for any operator Oˆ= Oˆ1Oˆ2†we have Oˆ† = Oˆ1Oˆ2 † = Oˆ2†Oˆ1†Proof. See problem set. Using the operators ˆa,ˆa†, we can write the Hamiltonian for the HO in the particulaarl simple form 1 Hˆ=¯hω aˆ†aˆ+ (18-39) 2 Massachusetts Institute of Technology XVIII-5 � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 8.04 Quantum Physics Lecture XVIIIRather 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 ˆa,ˆa†: �� xˆpˆxˆpˆˆa† + i, − i (18-40) a, ˆ= x0 p0 x0 p0 pˆxˆxˆpˆ= i, − i, (18-41) p0 x0 x0 p0 i = ([ˆp,xˆ] − [ˆx,pˆ]) (18-42) p0x0 i = 2[ˆp,xˆ] (18-43) 2¯hih¯= (18-44) h¯i =1, (18-45) where we have used [ˆx,xˆ] = 0 = [ˆp,pˆ]. So we have ˆa† (18-46) a, ˆ=1 [ˆa,aˆ] = aˆ†,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 haveˆ=¯ˆa, ˆ(18-48) H, aˆhω a†ˆa =¯hω aˆ†aˆaˆ− aˆaˆ†aˆ(18-49) =¯hω aˆ†aˆaˆ− 1+ˆa†aˆaˆ(18-50) hωaˆ(18-51) = −¯and ˆ=¯ˆa, ˆ(18-52) H, aˆ† hω a†ˆa† =¯hω aˆ†aˆaˆ† − aˆ†aˆ†aˆ(18-53) =¯hω aˆ†aˆaˆ† − aˆ† aˆaˆ† − 1 (18-54) =¯hωaˆ† (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 Hˆis quadratic in x and p, all eigenvalues must be positive: �E� = Hˆ= �T � + �V � (18-56) 11 = dpφ∗(p)p 2φ(p)+ mω2 dxψ∗(x)x 2ψ(x) > 0 (18-57) 2m 2 Massachusetts Institute of Technology XVIII-6 � �� � � �8.04 Quantum Physics Lecture XVIIIfor 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⎟ ⎜⎟ v1 v2 ⎜w2⎟ = c-number (complex number), (18-61) ··· · ⎝⎠ . . . but possibly for infinite-dimensional vectors. In this sandwich or Dirac notation, the expectation value of any operator Aˆis given by ��� �� Aˆ= ψ|A|ψ = Aψ(x). (18-62) ˆdxψ∗(x)ˆIn Dirac notation, ∗ �ψ|φ� = dxψ∗(x)φ(x)= dxφ∗(x)ψ(x)= �φ|ψ�∗ (18-63) An operator Aˆacting on a state produces another state, symbolicallyˆ�� ˆA |ψ� = Aψ (18-64) Massachusetts Institute of Technology XVIII-7 � | | � �� � � � | �� | � � 8.04 Quantum Physics Lecture XVIIIConsequently, � ���� � �� φ|A|ψ = φ|Aψ = dxφ∗(x) Aψ(x)= Aψ(x)ˆˆˆdxφ∗(x) ˆ(18-65) The Hermitian conjugate operator Aˆ† is defined by ��� ���∗ φ|Aˆ†|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� = dxun ∗(x)um(x)= δnm, (18-69) where the expansion coefficients are cn = dxun ∗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, (18-72) n and �� expansion into |ψ� = ˆ1 |ψ� = |n��n|ψ� = cn |n�→ eigenstates (18-73) n 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 aˆact on |E�, |ψ� := ˆa |E�. What happens if we act with the Hamiltonian |ψ�? Hˆψ� = HˆaˆE� (18-74) ˆˆ= H, aˆ+ˆaH |E� (18-75) =(−hω¯aˆ+ˆaE) |E� (18-76) =(E − hω¯)ˆa |E� (18-77) =(E − ¯|ψ� (18-78)hω) Massachusetts Institute of Technology XVIII-8 � � � � � � 8.04 Quantum Physics Lecture XVIIIˆHere we have used the previously calculated result H, aˆ= hωaˆfor 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, aˆn E − n¯(18-79) |E� = |hω� , �� and the eigenenergy has to remain positive, (we have shown ψ|Hˆ|ψ> 0 for any state), there must exist a state |0� such that aˆ|0� = 0 (18-80) i.e., a state whose energy cannot be lowered further. 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, 11 Hˆ|0� =¯hω aˆ†aˆ+ 2 |0� aˆ|0= �=0 2¯|0� , (18-81) hω so the ground state has eigenenergy E0 = 21 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 E0 = 2 hω0. What happens if ˆa† acts on ground state? Let us define �˜1 :=ˆa† |0� . (18-82) 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