One of the most important problems in quantum mechanics is the simple harmonic oscillator, in part because its properties are directly applicable to field theory. The treatment in Dirac notation is particularly satisfying. 1 Hamiltonian 1 2 q k Writing the potential 2 kx in terms of the classical frequency, ! = m , puts the classical Hamiltonian in the form p2 m!2x2 H = + 2m 2 Since there are no products of non-commuting operators, there is no ambiguity in the resulting Hamiltonian operator, 1 m!2 H^ = P^2 + X^ 2 2m 2 We make no choice of basis. 2 Raising and lowering operators Notice that ip ip p2 x + x − = x2 + m! m! m2!2 2 1 p2 = m!2x2 + m!2 2 2m so that we may write the classical Hamiltonian as m!2 ip ip H = x + x − 2 m! m! We can write the quantum Hamiltonian in a similar way. Choosing our normalization with a bit of foresight, we define two conjugate operators, rm! i a^ = X^ + P^ 2~ m! rm! i a^y = X^ − P^ 2~ m! The operator a^y is called the raising operator and a^ is called the lowering operator. Notice that they are not Hermitian. In taking the product of these, we must be careful with ordering since X^ and P^ ! ! m! iP^ iP^ a^ya^ = X^ − X^ + 2~ m! m! ! m! i i P^2 = X^ 2 + X^P^ − P^X^ + 2~ m! m! m2!2 ! m! i h i P^2 = X^ 2 + X;^ P^ + 2~ m! m2!2 1 h i Using the commutator, X;^ P^ = i~1^, this becomes ! 1 1 P^2 a^ya^ = m!2 X^ 2 − ~ + ~! 2 m! m2!2 ^2 ! 1 1 2 2 1 P = m! X^ − ~! + ~! 2 2 2m 1 1 = H^ − ~! ~! 2 and therefore, 1 H^ = ! a^ya^ + ~ 2 3 The number operator This turns out to be a very convenient form for the Hamiltonian because a^ and ay have very simple properties. First, their commutator is simply " ! !# m! iP^ iP^ a;^ a^y = X^ + ; X^ − 2~ m! m! m! i i = X;^ − P^ + P;^ X^ 2~ m! m! 2i m! h i = − X;^ P^ m! 2~ i = − i~ ~ = 1 Consider one further set of commutation relations. Defining N^ ≡ a^ya^ = N^ y, called the number operator, we have h i N;^ a^ = a^ya;^ a^ =a ^y [^a; a^] + a^y; a^ a^ = −a^ and h i N;^ a^y = a^ya;^ a^y =a ^ a^y; a^y +a ^y a;^ a^y =a ^y ^ ^ ^ 1 Notice that N is Hermitian, hence observable, and that H = ~! N + 2 . 4 Energy eigenstates 4.1 Positivity of the energy Prove that expectation values of Hamiltonian, hence all energies, are positive. 2 Q1: Consider an arbitrary expectation value of the Hamiltonian, 1 h j H^ j i = h j ! a^ya^ + j i ~ 2 1 = ! h j a^ya^ j i + h j i ~ 2 Use the fact that the norm of any state is positive, h j i > 0 to show that the energy of any SHO state is positive: h j H^ j i = E > 0 Hint: It helps to define jβi ≡ a^ j i. 4.2 The lowest energy state Suppose that jEi is any energy eigenket with H^ jEi = E jEi Consider the new ket formed by acting on jEi with the lowering operator a^. Q2: Show that a^ jEi is also an energy eigenket with energy E − }!, H^ (^a jEi) = (E − }!) (^a jEi) Since a^ jEi is an energy eigenket, we may repeat this procedure to show that a^2 jEi is an energy eigenket with energy E − 2~!. Continuing in this way, we find that a^k jEi will have energy E − k~!. This process cannot continue indefinitely, because the energy must remain positive. Let k be the largest integer for which E − k~! is positive, k k H^ a^ jEi = (E − k~!)a ^ jEi with corresponding state a^k jEi. Then applying the lowering operator one more time cannot give a new k state. The only other possibility is zero. Rename the lowest energy state j0i = A0a^ jEi, where we choose A0 so that j0i is normalized. We then must have a^ j0i = 0 This is the lowest energy state of the oscillator, called the ground state. 4.3 The complete spectrum Q3: Now that we have the ground state, we reverse the process, acting instead with the raising operator. Show that, acting any energy eigenket, jEi, the raising operator gives another energy eigenket, a^y jEi, with energy E + ~!, y y H^ a^ jEi = (E + ~!) a^ jEi 3 Q4: Find the energy of the state a^y j0i. y Q5: Define the normalized state to be j1i ≡ A1a^ j0i. Find the normalization constant, A1. There is nothing to prevent us continuing this procedure indefintely. Continuing n times, we have states yn jni = An a^ j0i satisfying 1 H^ jni = n + ! jni 2 ~ This gives the complete set of energy eigenkets. 4.4 Normalization th yn Now, consider the expectation of N^ in the n state, jni = An a^ j0i. Q6: Develop a recursion relation for the normalization constants, An, starting from 1 = hn j ni. 2 2 Specifically, show that jAn−1j = n jAnj . Iterating the recursion relation gives 1 jA j2 = jA j2 n n n−1 1 = jA j2 n (n − 1) n−2 . 1 = jA j2 n! 1 so that, choosing all of the coefficients real, we have the complete set of normalized states 1 n jni = p a^y j0i n! p Q7: Show that a^y jni = n + 1 jn + 1i and find the corresponding relation for a^ jni. 5 Wave function Choosing a coordinate basis to express the condition a^ j0i = 0 gives 0 = hxj a^ j0i rm! i = hxj X^ + P^ j0i 2~ m! rm! i = hxj X^ j0i + P^ j0i 2~ m! rm! i = x hx j 0i + hxj P^ j0i 2~ m! 4 ^ d and since we have shown that hxj P j0i = −i~ dx hx j 0i this becomes a differential equation, d m!x hx j 0i + hx j 0i = 0 dx ~ Setting 0 (x) = hx j 0i and integrating, d m!x 0 = − dx 0 ~ (x) m!x2 ln 0 = − 0 (0) 2~ 2 − m!x 0 (x) = 0 (0) e 2~ we find that the wave function of the ground state is a Gaussian, with the magnitude of the remaining coefficient 0 (0) determined by normalizing the Gaussian, 1=4 2 m! − m!x 0 (x) = e 2~ π~ Now consider the wave function, n (x), for the eigenstates. We have already found that the ground state is given by a normalized Gaussian,To find the wave functions of the higher energy states, consider n (x) = hx j ni 1 n = hxj p a^y j0i n! 1 1 n−1 = p hxj a^y a^y j0i n p(n − 1)! 1 rm! i = p hxj X^ − P^ jn − 1i n 2~ m! 1 rm! i d = p x hx j n − 1i − −i~ hx j n − 1i n 2~ m! dx r m! ~ d = x − n−1 (x) 2n~ m! dx Therefore, we can find all states by iterating this operator, n n 0 m! 2 ~ @ n (x ) = x − 0 (x) 2n~ m! @x The result is a series of polynomials, the Hermite polynomials, times the Gaussian factor. Q8: Find 1 (x) and 2 (x). 6 Time evolution of a mixed state of the oscillator Consider the time evolution of the most general superposition of the lowest two eigenstates j i = cos θ j0i + ei' sin θ j1i 5 Applying the time translation operator with t0 = 0, j ; ti = U (t) j i − i Ht^ = e ~ j i − i Ht^ i' − i Ht^ = cos θe ~ j0i + e sin θe ~ j1i − i !t i' − 3 i!t = cos θe 2 j0i + e sin θe 2 j1i − i !t i' −i!t = e 2 cos θ j0i + e sin θe j1i Now look at the time dependence of the expectation value of the position operator, which we write in terms q ^ ~ y of raising and lowering operators as X = 2m! a^ +a ^ . Remembering that the states are orthonormal, −i' i!t i !t − i !t i' −i!t h ; tj X^ j ; ti = cos θ h0j + e sin θe h1j e 2 Xe^ 2 cos θ j0i + e sin θe j1i r = ~ cos θ h0j + e−i' sin θei!t h1j a^ +a ^y cos θ j0i + ei' sin θe−i!t j1i 2m! r p = ~ cos θ h0j + e−i' sin θei!t h1j cos θ j1i + ei' sin θe−i!t j0i + 2 j2i 2m! r = ~ cos θ sin θe−i(!t−') + sin θ cos θei(!t−') 2m! r = ~ sin 2θ cos (!t − ') 2m! p where we have used a^ j0i = 0; a^ j1i = j0i ; a^y j0i = j1i and a^y j1i = 2 j2i. We see that the expected position q ~ oscillates back and forth between ± 2m! sin 2θ with frequency !. Superpositions involving higher excited states will bring in harmonics, n!, and will then allow for varied traveling waveforms. p ^ 1 y Q9: Find the expectation value of the momentum, P = 2i 2~m! a^ − a^ D E2 D E D E2 Q10: Find the uncertainty in position, (∆X)2 = X^ − X^ = X^ 2 − X^ 7 Coherent states and the correspondence principle Any alternative physical theory must approach, in some suitable limit, the previously tested and accepted standard model. Such a limit was stated concretely for quantum mechanics by Niels Bohr in 1920.