"XVI.
THE HAMILTON EQUATIONS OF MOTION
The Hamiltonian is the Legendre transformation of the Lagrange function: H(q, p, t) =
i
qi pi − L(q, q, t) ˙ ˙
Note: The Lagrangian is the function of q and q. The Hamiltonian is the function of the ˙ canonically conjugate variables q and p. The differential of H(q, p, t) is given by (this is the general form for a function depending on p, q, t) dH =
i
∂H dqi + ∂qi
i
∂H ∂H dpi + dt ∂pi ∂t
(1)
The differential of H(q, p, t) using its definition is (this is the particular form for a H given in the definition) dH =
i
qi dpi + ˙
i
pi dqi − ˙ ∂L = pi ˙ ∂qi
i
∂L dqi − ∂qi ∂L = pi ∂ qi ˙ pi dqi − ˙
i
∂L ∂L dqi − ˙ dt ∂ qi ˙ ∂t
Using the expressions
we have dH =
i
qi dpi − ˙
i
∂L dt. ∂t
Comparing this to eq. ( 1) we have the following 2n + 1 equations ∂H ∂pi ∂H −pi = ˙ ∂qi ∂L ∂H − = ∂t ∂t qi = ˙ called “canonical equations of Hamilton”. Note that these are first order equations. In the Lagrangian mechanics we had n generalized ccordinates and we had to solve a second order equation (the Euler-Lagrange) equation (with initial conditions for q and q to characterize ˙ the system. Now we have 2n independent variables q and p and 2n first order equations to be solved.
1
The total time derivative of the Hamiltonian dH = dt =
i
i
∂H qi + ˙ ∂qi
i
∂H ∂H pi + ˙ ∂pi ∂t ∂H ∂H ∂H + ∂pi ∂qi ∂t
∂H ∂H − ∂qi ∂pi
i
∂L ∂H =− = ∂t ∂t So if L is not explicitly depend on time then H is constant of motion. In particular, if the potential is velocity independent and the transformation which defines the generalized coordinates is time independent then H is the total energy.
A.
Example: Central potential
1 ˙ ˙ L(r, φ, r, φ) = T − V (r) = m(r 2 + r 2 φ2 ) + V (r) ˙˙ 2 The generalized momenta ∂L = mr ˙ ∂r ˙ ∂L ˙ = mr 2 φ. pφ = ˙ ∂φ pr = The Hamiltonian, by definition: ˙ H(r, φ, pr , pφ ) = rpr + φpφ − L(r, φ, r, φ) = T + V (r) ˙ ˙˙
B. Example: Particle in EM field
1 ˙ L(r, r) = m˙ 2 − eφ(r) + e/cA˙ r r 2 The generalized momenta p = m˙ + e/cA r the Hamiltonian 1 1 ˙ ˙ r (p − e/cA)2 + eφ(r) H(r, p) = rp − L(r, r) = m˙ 2 + eφ(r) = 2 2m Note: H is constant of motion for time independent fields only. 2
XVII.
THE MODIFIED HAMILTON’S PRINCIPLE
t2
t2
δS = δ
t1
Ldt = δ
t1 i
pi qi − H dt ˙
In derivation of the Lagrange equations we have assumed that q and q are related (δ q = ˙ ˙ dδq/dt). Here we assume that q and p are independent variables: qi → qi + ηi (t) pi → pi + χi (t) with fixed endpoints: ηi (t1 ) = ηi (t2 ) = 0 The variation of the action δS = δ where H(p + χ, q + η) = H(p, q) +
i t2 t1
χi (t1 ) = χi (t2 ) = 0
[
i (pi
+ χi )(qi + ηi ) − H(p + χ, q + η)] dt − ˙ ˙
t2 t1
[
i
pi qi − H(p, q)] dt ˙
=0
∂H dqi + ∂qi d
i
∂H dpi = H(p, q) + ∂pi d
i
∂H ηi + ∂qi
i
∂H χi . ∂pi
Due to the fixed endpoint:
t2 t1
ηi pi = [ηi pi ]t2 − ˙ t1
t2
pi ηi = ˙
t1
Substituting these expressions into the variation of action δS = δ
t2 t1 i
qi − ˙
∂H ∂pi
ηi − pi + ˙
∂H ∂qi
χi = 0
and because η and χ arbitrary, we get the Hamilton equations ∂H ∂pi ∂H −pi = ˙ ∂qi qi = ˙
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XVIII.
POISSON BRACKETS
The time derivative of function f (q, p) depending on q and p only df = dt ∂f ∂qi + ∂qi ∂t ∂f ∂pi = ∂pi ∂t ∂f ∂H − ∂qi ∂pi ∂f ∂H = [H, f ] ∂pi ∂qi
i
i
i
i
where we have defined the Poisson-Brackets: [f, g] =
i
∂g ∂f − ∂pi ∂qi
i
∂g ∂f = ∂qi ∂pi
This operation is an example of a Lie algebra. Properties of Lie algebras (including Poisson brackets) (1) Anticommutation [g, f ] = −[f, g] (2) Linearity [g1 + g2 , f ] = [g1 , f ] + [g2 , f ] (3) Jacobi’s identity [g, [h, f ]] + [f, [g, h]] + [h, [f, g]] = 0 (4) Derivatives ∂ ∂g ∂f [g, f ] = , f + g, ∂t ∂t ∂t (5) Multiplication [g1 g2 , f ] = g1 [g2 , f ] + g2 [g1 , f ] Important examples of Lie algebras: Vector product, product of antisymmetric matrices, commutators in QM, etc. f is conserved if, and only if, [H, f ] = 0. The Poisson-bracket of qi and pi [pi , pj ] = 0 [qi , qj ] = 0 [pi , qj ] = δij
Cyclic variables: If H is independent of qi then, according to the Hamilton equations, the corresponding canonically conjugate momentum is conserved.
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XIX.
CANONICAL TRANSFORMATIONS
The Lagrangian for a harmonic oscillator is 1 1 ˙ L(q, q) = mq 2 − kq 2 ˙ 2 2 the Hamiltonian ˙ ˙ H(q, p) = pq − L(q, q) = The Hamilton equations are ∂H p = ∂p m ∂H = kq −p = ˙ ∂q q= ˙ The Poisson-bracket of q and p [p, p] = 0 [q, q] = 0 [p, q] = 1 12 12 p + kq 2m 2
We can introduce a new set of variables Q and P by defining P = eiπ/4 (p + imωq) Q = eiπ/4 (p − imωq) (ω 2 = k/m)
Using these variables, the Hamiltonian becomes 1 H(Q, P ) = ω(P 2 + Q2 ) = ωP Q 2 The Hamilton equations are ∂H ˙ = ωP Q= ∂P ∂H ˙ −P = = ωQ ∂Q The Poisson-bracket of q and p [P, P ] = 0 [Q, Q] = 0 [P, Q] = 1
This is an example of a canonical transformation. In general, the change of variable form p, q to P, Q in such a way that the Hamiltonian mechanics preserves its canonical form are called canonical transformations. There are equivalent definitions of canonical transformations: (1) The Hamilton equations remain valin in the new coordinates 5
(2) Poisson bracket based criterion [Pi , Pj ] = 0 [Qi , Qj ] = 0 [Pi , Qj ] = δij
The Hamilton equations can be derived from the modified Hamilton principle.
t2
δS = δ
t1 i
pi qi − H(p, q) dt = 0 ˙
The canonically transformed quantity should also satisfy the same condition:
t2
δS = δ
t1 i
˙ Pi Qi − H (P, Q) dt = 0
These two form are equivalent if the two integrands apart from a total time derivative are equal. The time integral of total time derivative of a function F of two of the variables choosing from p, q, Q and P is
t2 t1
dF dt = F (t2 ) − F (t1 ) dt dF ˙ Pi Qi − H (P, Q) + dt
and this is gives no contribution when we take the variation. We have pi qi − H(p, q) = ˙
i
i
The function F is called generating function. Let’s first assume that F = F (q, Q), then dF =
i
∂F dqi + ∂qi
i
∂F ∂F dQi + dt ∂Qi ∂t
(2)
From the above two equations: ∂F ∂qi ∂F Pi = − ∂Qi ∂F H −H = ∂t pi = Using Legendre transformations, we can choose different generating functions: FB (q, P ) = FA (q, Q) − FC (p, Q) = FA (q, Q) − FD (p, P ) = FA (q, Q) +
i
Pi Q i
i
pi =
∂FB ∂qi ∂FC ∂pi
Qi =
∂FB ∂Pi ∂FC ∂Qi Qi = − ∂FD ∂Pi (3)
p i qi
i
qi = −
Pi = − qi = −
q i Pi −
Q i Pi
i
∂FD ∂pi
6
These canonical transformations give us a great flexibility to choose the most appropriate variables to solve the Hamilton equations.
A.
Example: Harmonic oscillator
The Hamiltonian H(q, p) = Generating function F (q, Q) = then ∂F = mωqcotQ ∂q mωq 2 ∂F = P =− ∂Q 2sin2 Q We can express the old variables p and q with the new ones: p= q= p= √ 2P sinQ mω 2mωP cosQ m2 ωq cotQ 2 12 1 p + mω 2 q 2 2m 2
Now we have the new Hamiltonian as (substituting p and q into the old Hamiltonian) H(P, Q) = 12 1 p + mω 2 q 2 = ωP ˙ 2m 2
The new Hamiltonian is cyclic in Q, the canonically conjugate momentum is therefore constant! As the energy is constant of motion in this case, H = E and P= The equation of motion for Q is ∂H ˙ Q= =ω ∂P therefore Q = ωt + α By sustituting this back into q q= that is we got the familiar solution. 7 2E sin(ωt + α) mω 2 E ω
B.
Infinitesimal canonical transformations
First let’s construct a generating function for an identity transformation Q=q we can use FB (q, P ) for this by choosing FB (q, P ) = qP as generating function. We then have Q= ∂FB ∂P p= ∂FB ∂q P =p
next let’s construct a transformation which is infinitesimally different form the identity transformation FB (q, P ) = qP + G(q, P ) where G(q, P ) is called “infinitesimal generator”. In this case Q= or by rearranging: ∂G ∂G P =p− ∂P ∂q Now if there is a function W (p, q) the change of this function due to the infinitesimal Q=q+ transformation is (obtained with Taylor expansion) δW = W (P, Q) − W (p, q) = In particular, if W = H δH = [W, G] so if H is invariant to the infinitesimal transformation generated by G then the generator, G is a constant of motion, and vica versa. Let us try this on an infinitesimal rotation with angle δθ X = xcosδθ + ysinδθ = x + yδθ Y = −xsinδθ + ycosδθ = y − xδθ Z=z (4) 8 ∂W ∂G ∂W ∂G − = [W, G] ∂q ∂p ∂p ∂q ∂FB ∂G ∂FB ∂G =q+ p= =p+ ∂P ∂P ∂q ∂q
that is ( = δθ) ∂G =y ∂px which shows that G = p x y − p y x = Lz the generator of the infinitesimal rotation around z is Lz . If the Hamiltonian is invariant to a rotation around the z axis then the generator Lz is a constant of motion. Next, if G = H and = δt then Q=q+ ∂H δt = q + qδt = q(t + δt) ˙ ∂p P =p− ∂H δt = p + pδt = p(t + δt) ˙ ∂q ∂G = −x ∂py
XX.
HAMILTON-JACOBI EQUATION
What is the most appropriate transformation? The one for which the transformed Hamiltonian H = 0. In this case the generating function (and to emphasize the special choice we call it S) should satisfy (see above the equation for H ) ∂S + H(q, p, t) = 0 ∂t this is called the Hamilton-Jacobi equation. The suitable generating function is in this case S(q, P, t) = FA (q, Q) − and ∂S ∂qi ∂S Qi = ∂Pi pi = As H is zero, Q and P are trivially cyclic variables: Pi = α i Qi = βi Pi Q i
i
where αi and βi can be calculated from the initial conditions. S(q, P, t) = S(q, α, t) ∂S(q, α, t) ∂S + H(q, , t) = 0 ∂t ∂q 9 ( ∂S = βi ∂αi
A.
Example: Harmonic oscillator
The Hamiltonian 12 1 p + mω 2 q 2 2m 2 We replace p by ∂S/∂q in H and substitute it into to Hamilton-Jacobi equation: H(q, p) = 1 ∂S + ∂t 2m ∂S ∂q
2
1 + mω 2 q 2 = 0 2
This can be solved by separation of variables: S(q, E, t) = W (q, E) − Et where E is a constant of integration. By substituting this into the Hamilton-Jacobi equation 1 2m which can be solved: 2E W = mω dq − q2 = mω 2 mωq E mωq arcsin √ = +√ ω 2mE 2mE ∂W ∂q
2
1 + mω 2 q 2 = E 2
1−
mωq √ 2mE
2
(5)
By differentiating the equation for S with respect to E we have ∂S 1 mωq √ = arcsin ∂E ω −t = β = constant 2mE q= and the momentum ∂s = 2mE − m2 ω 2 q 2 . ∂q The reason that this special generating function is labeled by S is that it is actually equal p= to the action! The time derivative of S, using the Hamilton-Jacobi equation: dS = dt therefore S= 10 Ldt ∂S ∂S qi + ˙ = ∂qi ∂t p i qi − H = L ˙ 2E/msin(ωt + ω0 )
That is
i
i
B.
Hamilton-Jacobi equation for central potential
The Lagrangian for central potential using spherical coordinates: L= The canonical momenta pr = mr ˙ and the Hamiltonian is H= p2 p2 p2 φ r + V (r) + θ2+ 2 sinθ 2m 2mr 2mr ˙ pθ = mr 2 θ ˙ pφ = mr 2 sin2 θ φ m2 ˙ ˙ (r + r 2 θ 2 + r 2 sin2 θ φ2 ) + V (r) ˙ 2
The Hamilton-Jacobi equation is then ∂S 1 + ∂t 2m ∂S ∂r
2
+
1 2mr 2
∂S ∂θ
2
+
1 2mr 2 sinθ
∂S ∂φ
2
+ V (r) = 0
By separating the variables S(r, θ, φ, t) = St (t) + Sr (r) + Sθ (θ) + Sφ (φ) the Hamilton-Jacobi equation becomes 1 dSt + dt 2m dSr dr
2
1 + 2mr 2
dSθ dθ
2
1 + 2 sin2 θ 2mr
dSφ dφ
2
+ V (r) = 0
Because there is not explicit t and φ dependence, we conclude that dSt /dt and dSφ /dφ must be constant: dSt =E dt and S(r, θ, φ, t) = Sr (r) + Sθ (θ) + Et − Lz φ The Hamilton-Jacobi equation now is: 1 2m dSr dr
2
dSφ = −Lz dφ
1 + 2mr 2 dSθ dθ
dSθ dθ
2
2
+
2 Lz + V (r) = E 2mr 2 sin2 θ
When only θ is changed, + Lz 2 2 =L sin θ
11
should be constant. L2 is the angular momentum. Then Sθ = From this we can express φ by φ= dS ∂ =− dLz ∂Lz dθ L2 − L2 z = −arctan sin2 θ Lz cosθ L2 sin2 θ − L2 z + φ0 dθ L2 − L2 z sin2 θ
Now the Hamilton-Jacobi equation is for the remaining radial piece: 1 dSr 2 L2 + + V (r) = E 2m dr 2mr 2 and from this Sr = and t= dS = dE dr m 2mE − 2mV (r) − L2 /r 2 dr 2mE − 2mV (r) − L2 /r 2
C.
Bohr-Sommerfeld Quantization
Bohr-Sommerfeld hypotesis: Assume that the action integrals (for periodic motions) are integer multiples of h = 2π . First: z component of the angular momentum dSφ dφ = 2πLz = 2πm dφ Second: the angular momentum Sθ = dθ L2 − L2 z = −2π(L − Lz ) = 2π (l − m) sin2 θ
Third: the energy momentum (little more tedious), we assume V (r) = −GmM/r Sr = dr 2mE − 2mV (r) − L2 /r 2 = 2 −Lπ + Gm3/2 M π 2|E| = 2π n
using L = l this gives the energy levels of hydrogen-like atoms (replace GmM by Ze2 ): E=− G 2 m3 M 2 2(n + l)2 2
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XXI.
PHASE SPACE
In analogy to the configuration space in Lagrange formalism, we can define a phase space formed by coordinates q1 , q2 , . . ., qn and momenta p1 , p2 , . . ., pn . Each path represent a solution of the Hamilton equations. For periodic motion the path is closed it returns to its original starting point. The motion is deterministic, different paths can never cross.
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