Operator approach to Harmonic Oscillator and the super-symmetric quantum mechanics.

Transcription

1 Operator approach to Harmonic Oscillator and the super-symmetric quantum mechanics Apart from Schrödinger Equation approach and path-integral formalism, there is yet another powerful way of dealing with quantum mechanical problems Well, to begin with, talking of an operator in a finite dimensional Hilbert space is equivalent to talking about a matrix, so matrix quantum mechanics is not surprising at all It turns out that even in continuous space problems, eg such as particle in the harmonic oscillator potential, matrix/operator approach is also very powerful We will see that using properly designed operators we can solve the problem with much less work, as first shown by Dirac Remarkably, creation and annihilation operators appearing in this problem are used in quantum electrodynamics to describe photons, solid state physics to described phonons, spin waves, plazmons (in fact, any type of collective bosonic excitation), and form the basis of second quantization used for efficient description of many-body states of identical bosonic particles of any kind At the formal level one can claim that there is this or that harmonic oscillator behind any bosonic particle in Nature, ie the two notions are identical We begin by defining the so-called creation and annihilation operators (do not ask yet where does the name come from) â = mω ˆx + i mω ˆp, â = mω ˆx i mω ˆp () You can see that one can be obtained by Hermitian conjugation of the other ( ˆx and ˆp are Hermitian) The occupation number operator is defined by ˆN = â â It is equal to ( ) ( ) mω ˆN = ˆx i mω ˆp mω ˆx + i ˆp = p mω mω + mωx + i [x, p] = ω Ĥ, () or Ĥ = ω( ˆN + ) (3) Since ˆN and Ĥ are straightforwardly related and share the same basis states, we focus on the eigenstates of ˆN ˆN n = n n For Hermitian operator ˆN its eigenvalues, n, are real Moreover, we see that n is non-negative because n = n ˆN n = n â â n = ân ân To proceed with manipulating creation and annihilation operators we notice that their commutation relations are (they follow directly from the commutation relation for the coordinate and momentum) With this we easily get and [â, â ] = (i[p, x] i[x, p]) = [ ˆN, â] = â ââ ââ â = [â, â]â = â, [ ˆN, â ] = â ââ â â â = â [â, â ] = â This is all we need to launch an attack on the problem

2 The state â n is an eigenstate of ˆN with eigenvalue n, ie it is proportional to n Indeed ˆN â n = (â ˆN â) n = â(n ) n = (n ) â n The normalization can be established from Thus An identical consideration shows that ân ân = n â â n = n â n = n n (4) â n = n + n + (5) Problem 3 Creation operator properties Derive Eq (5) After we apply the annihilation operator k times we get â k n = n(n ) (n k + ) n k This way we can produce eigenstates of ˆN with smaller and smaller eigenvalues The only way to reconcile this possibility with the result that all eigenvalues of ˆN are non-negative is to have integer n Then for k = n we arrive at the n k = state and further application of the annihilation operator produces null result â = This terminates the downward cascade of states Thus the spectrum of the ˆN operator is given by the set of non-negative integers, and allowed energy states are E n = ω(n + /), n =,,, The eigenstate n can be created from the ground state by applying creation operator n times n = n! (â ) n Problem 3 Commutation relations a) Using commutation relations for operators â and â demonstrate that n = n! (â ) n, is normalized to unity b) From the Schrödinger Equation solutions we know that ( mω ) /4 x n = / π n n! e mωx H n ( mωx), where H n (ξ) is the Hermite polynomial Using properties of â and â prove the recursion relation d dξ ξ H n (ξ) = H n+ (ξ)

3 c) Use the recursion relation and H (ξ) = to generate H, H, H 3 polynomials) Formally we have achieved a complete solution Remarkably, we do not even need to know the real-space or momentum-space form of the eigenstates It is perfectly OK to use the occupation number basis for all calculations By expressing coordinate and momentum in terms of creation and annihilation operators x = mω (â + â ), mω p = i (â â ), we can compute all properties directly in the n basis Say, what is the value of n x n? Answer: mω n (â + â ) n = mω n ââ + â â n = n + mω Using Hermite polynomial would be far more messy Nevertheless, let us determine the real-space form of solutions To this end act with the annihilation operator on the ground state = ψ (x) to get an equation mω xψ + dψ mω dx = dψ (x) = mωxdx = (mω/)dx, ψ with the solution ( ψ e mωx / mω ) /4 ψ = e mωx / π wavefunctions of excited states can be obtained by acting with the creation operator on the ground state: ψ n = (â ) n ψ = n/ ( mωx d ) n (mω ) /4 e mωx / = n! n! mω dx π ( mω ) /4 π n n! H n( mωx) e mωx / Quantization of electromagnetic waves Let us see how this operator approach naturally leads to the quantization of electromagnetic waves and the notion of photons We start by noting that the energy of a system of electric and magnetic fields is given by the volume integral over the energy density which contains electric and magnetic contributions (velocity of light, dielectric constant, and magnetic permittivity not mentioned for clarity; otherwise rescale fields accordingly) H = V ( E d 3 ) (r, t) r + B (r, t) Next we introduce the vector potential A (here we consider E&M waves in the absence of electric charges and thus neglect the scalar potential ϕ) and express physical fields in terms of A E = A t, B = A 3

4 In the Fourier representation A(r, t) = d 3 k (π) 3 eik r A(k, t) leading to H = d 3 r d 3 k d eik r 3 k eik r (Ȧ(k, t) (π) 3 (π) 3 Ȧ(k, t) [k A(k, t)] [k A(k, t)]) Integration over d 3 r results in the delta function (π) 3 δ(k + k ) which, in turn, removes one of the momentum integrals: d 3 k ( H = (π) 3 Ȧ(k, t) + k A(k, t)] ) [For waves in free space A(k, t) k; also, for real A(k, t) its Fourier transform is such that A( k, t) = A (k, t))] For each momentum k we introduce two polarizations e α=, (k) perpendicular to k, and observe that the total energy of the field is given in terms of a sum over independent harmonic oscillator modes, one with energy ω k = k and polarization e α=, (k) (recall, that the velocity of light is set to unity) We have two options here (i) to consider only half of the values of k as independent because A α ( k, t) = A α(k, t), and then treat real and imaginary parts of A α (k, t) as two independent fields (ii) to consider all values of k as independent, and treat A α (k, t) as one independent field Either way the final results will be identical The common choice (more convenient for algebraic manipulations) is the second one So, we look at each mode as a standard harmonic oscillator expression described by the amplitude x = A α, mass m = and frequency ω k = k Since the analogy is exact, we immediately state that the energy of each mode can only take values E k,α = ω k (n k,α + /), where n k,α is interpreted either as the label of the excited state of the harmonic oscillator, or a number of photons in the mode (k, α) Note that the energy of the mode is non-zero even when no photons occupy it This term is often called the zero-point, or vacuum energy In the ground state E G = d 3 k (π) 3 ω k = d 3 k (π) 3 ω k (6) Are there some yet to be discovered microscopic degrees of freedom which vibrate in the harmonic oscillator potential and we observe these vibrations as photons? I do not know There is no way to tell the difference unless we probe Nature at far smaller length-scales and higher energies The analogy with crystal vibrations is due here at wavelength λ a, where a is the crystal lattice constant, acoustic waves behave just like electromagnetic waves Without knowing anything about atoms one can study sound wave properties and ultimately discover that they carry quanta of energy ω k This would leads to the notion of a bosonic particle, a phonon, and one would be able to construct the full quantum theory of phonons by treating them as elementary particles In this particular case we know that they are collective harmonic excitations in the system of interacting atoms, but so what? If we do not resolve the lattice constant in the experiment we may comfortably stay solely at the level of phonon description 4

5 On immediately notices that the integral in Eq (6) is infinite which is definitely troubling There are two (related) ways of thinking about it: (i) ignore (!) Say that the counting zero for energy is an abstract notion and one is free to choose any Simply count all energies starting from the ground state (ii) Assume that there is yet to be discovered physics at short wavelength (high momentum) which makes this integral finite So, introduce some formal high energy cut-off in momentum space and proceed with calculating all properties with this cut-off You d better do not see this cut-off appearing explicitly in any of the final answers Suppose we choose the second option Is there any physical effect attached to the fact that the vacuum energy in the absence of any photons is non-zero? The answer is Yes with reservations Zero-point fluctuations of the electromagnetic field lead to the famous Lamb shift of atomic levels The other prominent effect is the Casimir force Consider an electromagnetic field between two conducting plates Due to complete reflection of low-energy waves from the metal the allowed wavenumbers in this configuration in the direction perpendicular to the metal surface, let it be ˆx-direction, are quantized and the minimal one is set by L This replaces dk x /π with the sum n x= for low-energy photons and introduces dependence Metal L Metal of E G on the distance between the plates Since the number of allowed modes increases with L we have de/dl > and an attractive Casimir force F = de/dl < between the two plates One may object that this is not really a vacuum energy effect because one needs material plates to be present to see it Moreover, it is possible to reformulate the calculation in such a way that reflection from the boundaries is treated rather as resulting from electromagnetic waves interacting with matter and mediating coupling between the walls A valid philosophical objection would be who is going to see any effect in true vacuum when there is nothing around? An observer and tools used to observe, by definition, violate the notion true of vacuum To complete the discussion, let us go one step further and complete the formulation of quantum electrodynamics It takes no time with identification A α x, mass m and frequency ω k k, just plug in the expression for x in terms of creation and annihilation operators: A α (k) = e ) α(k) (â k,α + â k,α ωk We conclude with the general second-quantized form of the electromagnetic field (the plane-wave factors are different for creation and annihilation operators to ensure that A(r) is Hermitian) dk e α (k) A(r) = (π) 3 (â k,α e ik r + â e ik r) k,α ωk α 5

6 It can be used right away to study scattering of light, radiative transitions, renormalization of the particle charge and mass, etc = QED problems Real vs momentum space for Harmonic oscillators It is common to discuss harmonic oscillators in real space But by simply looking at the dimensionless Hamiltonian written in terms of dimensionless momentum ˆ p = ˆp/ mω and coordinate ˆ x = ˆx mω which obey the same commutation relation [ˆ x, ˆ p] = i : Ĥ/ω = (ˆ p + ˆ x ), we see a complete symmetry between the real and momentum spaces It means that we do not have to work hard in performing Fourier transforms of the harmonic oscillator states derived previously since the result will be identical to replacing dimensionless combinations x mω with dimensionless combinations p/ mω The only care should be taken for normalization factors since the probability density ψ n (x) has a dimension of [x ] while the dimension of ψ n (p) is [p ] This brings about a factor of / mω in front of the properly normalized wave function in addition to the π factor required to compensate /π in momentum integrals Thus ψ n (p) = π mω ( mω ) /4 /mω π n n! e p H n (p/ 4π /4 mω) = /mω mω n n! e p H n (p/ mω) (in doing Fourier transforms you may find additional factors of ( i) n attached to the momentum states, but they can be dropped because any basis state is defined only up to a phase factor) You decide which representation you like more Supersymmetric Quantum Mechanics Dirac s handling of the harmonic oscillator is very impressive in terms of simplicity, elegance, and efficiency of calculating physical properties It is interesting to ask if other problems may be addressed in a similar fashion The answer is yes, though some of the simplicity is gone Suppose we have solved for the ground state wavefunction of the Schrödinger Equation in some potential V (x) ( this is a little unpleasant start, but let s continue) Now define ie ( m d ) dx + V (x) ψ (x) = E ψ (x) V (x) = V (x) E, so that d ψ (x) m dx + V (x)ψ (x) = From this we see that V (x) can be expressed as V (x) = m ψ (x) ψ (x), 6

7 and the Hamiltonian Ĥ = p /m + V (x) as Ĥ = m ( d dx + ψ (x) ) ψ (x) It has all its eigenvalues being non-negative because its ground state energy is zero by construction (recall non-negativity of eigenvalues of ˆN for harmonic oscillator) Now define operators  = ( d m dx ψ (x) ) ψ (x),  = ( d m dx ψ (x) ) ψ (x) which form a Hermitian conjugated pair (if p = i(d/dx) is Hermitian, then (d/dx) is anti- Hermitian = changes sign under hermitian conjugation) Then (   = d m dx + d [ ] [ ] ) ψ (x) ψ + (x) = ( d dx ψ (x) ψ (x) m dx + ψ (x) ) = ψ (x) Ĥ You may notice a close connection between non-negativity of eigenvalues and representation Ĥ =   which was already employed before: Ĥ =  ψ  ψ At his point it is convenient to introduce W (x) = ψ (x) [ ln m ψ (x) = ψ (x) ] m to write A-operators in a form reminiscent of the creation and annihilation operators for harmonic oscillator  = d m dx + W (x),  = d m dx + W (x) The W (x)-function is known under the name of superpotential It is easy to check that A-operators satisfy the following commutation relation [Â,  ] = dw (x) m dx We use this property to define yet another Hamiltonian where = m Ĥ + =   = Ĥ + [Â,  ] = Ĥ + dw (x) = m dx ( d dx + ψ (x) ψ (x) d dx [ ]) ψ (x) = ψ (x) m V + (x) = V (x) + W (x), d dx + V +(x), which corresponds to some other potential energy (derived from properties of the same ground state), 7

8 We can now show that the energy levels of Ĥ + and Ĥ are identical (!), except that the ground state of Ĥ is unique, ie E (+) n = E ( ) n+ for all n =,,, n n A A Moreover, the eigenfunctions of states with the same energy are also related via ψ (+) n = E ( ) n+ Âψ ( ) n+, (7) E ψ ( ) n+ =  ψ (+) E n (+) n (8) The proof is along the same lines as for the harmonic oscillator Consider which confirms Eq (7) Similarly Ĥ + Âψ ( ) n+ =    ψ ( ) ( ) n+ =  Ĥ ψ n+ = E( ) n+â ψ( ) n+, Ĥ  ψ (+) n =    ψ (+) n =  Ĥ + ψ (+) n = E n (+)  ψ (+) n, confirms Eq (8) The normalization factors in (7) and (8) are readily obtained from and E (+) n = ψ (+) n Ĥ+ ψ (+) n = ψ (+) n   ψ (+) n =  ψ (+) n  ψ (+) n, E ( ) n+ = ψ( ) ( ) n+ Ĥ ψ n+ = ψ( ) n+   ψ ( ) n+ =  ψ( ) n+  ψ( ) n+ This pairwise arrangement is nice, but where does it lead us? To begin with we have to know the ground state After that we find that for any potential there is a supersymmetric partner potential with identical spectrum and closely related wavefunctions, but nothing tells us what they are in terms of the found ground state Thus we still have to solve one of the two problems by other means There are, however, special cases, which are exceptional What if the potential is such that V + = V + const, ie they are identical up to a constant term? First, the constant itself must be E (+) = E ( ) E ( ) = E ( ) = C We also realize that in this case the ground state of Ĥ+ is the same as of Ĥ which allows us to get the first excited state of Ĥ using Â, which is the same as the first excited state of Ĥ+, etc In other words, knowing the ground state we start climbing the ladder and solve the entire problem This is the harmonic oscillator case, because the spectrum is equidistant Indeed, for V (x) = (mω x ω)/ and ψ ( ) = N e mωx / we get W (x) = mω ( mωx) = x, m 8

9 and Also V + = mω x ω + W (x) = mω x ω + mω x = mω x + ω  = d m dx + mωx = ω â, = V (x) + ω reproducing (up to a factor) Dirac s operators Of course, any exactly solvable problem can be used to generate another exactly solvable problem which is it s supersymmetric partner If we take an infinite square well V (x) = for x < a and V (x) = otherwise, then the ground state energy is E = π /8ma, and the ground state wavefunction is ψ = a / cos(πx/a) This leads to V = E for x < a, and the partner potential W (x) = m π a tan(πx/a) = E tan(πx/a), V + = E ( + tan (πx/a) ), for x < a The two potentials, as well as their eigenfunctions, are quite different from each other but they do share the same spectrum E E 8ma a a ( ) E a a Why not using known properties of Ĥ + to generate yet another partner Hamiltonian constructed from the ground state of Ĥ +? The idea is absolutely valid, and this way one obtains a whole hierarchy of related Hamiltonians The harmonic oscillator example is not unique in the following sense: there are other so-called shape invariant potentials which admit a complete solution using  and  algebra The table of known shape invariant potentials is attached Remarkably, the Coulomb potential is one of them The general condition for the potential to be shape invariant is V + (x; a + ) = V (x, a ) + R(a ), 9

10 where a is the set of Hamiltonian parameters, such as mass, potential strength, etc, and a + is some function of a In essence, the two potentials represent the same Hamiltonian but for a different set of parameters Further analysis of this topic as well as connections to SUSY goes beyond our course (if interested, look at arxiv:hep-th/9459 or PhysRept 5 (995): 67385)

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