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Chapter 6 Lecture 1 Canonical Transformations

Akhlaq Hussain 1 6.1 Canonical Transformations

푁 Hamiltonian formulation 퐻 푞푖, 푝푖 = σ푖=1 푝푖 푞푖ሶ − 퐿 (Hamiltonian) 휕퐻 푝푖ሶ = − 휕푞푖 (푯풂풎풊풕풐풏′풔 푬풒풖풂풕풊풐풏풔) 휕퐻 푞푖ሶ = 휕푝푖 one can get the same differential equations to be solved as are provided by the Lagrangian procedure.

퐿 푞푖ሶ , 푞푖 = 푇 − 푉 (Lagrangian) 푑 휕퐿 휕퐿 − = 0 (Lagrange’s equation) 푑푡 휕푞푖ሶ 휕푞푖 Therefore, the Hamiltonian formulation does not decrease the difficulty of solving Problems. The advantages of Hamiltonian formulation is not its use as a calculation tool, but rather in deeper insight it offers into the formal structure of the mechanics. 2 6.1 Canonical Transformations

➢ In 퐿 푞푖ሶ , 푞푖 system is described by “푞푖” and ” 푞푖ሶ ” in configurational , ➢ The parameters that define the configuration of a system are called and the vector space defined by these coordinates is called configuration space.

➢ The position of a single particle moving in ordinary Euclidean Space (3D) is defined by the vector 푞 = 푞(푥, 푦, 푧) and therefore its configuration space is 푄 = ℝ3 ➢ For n disconnected, non-interacting particles, the configuration space is 3푛 ℝ . 3 6.1 Canonical Transformations

➢ In Hamiltonian 퐻 푞푖, 푝푖 we describe the state of the system in by generalized coordinates and momenta.

➢ In dynamical system theory, a Phase space is a space in which all possible states of a system are represented with each possible state corresponding to one unique point in the phase space.

➢ There exist different momenta for particles with same position and vice versa.

4 6.1 Canonical Transformations

➢ To understand the importance of Hamiltonian let us consider a problem for which solution of Hamilton’s equations are trivial (simple) and Hamiltonian is constant of .

➢ For this case all the coordinates “푞푖” of the problem will be cyclic and all conjugate “푝푖” momenta will be constant.

Since 푝푖 = 훼푖 = 퐶표푛푠푡푎푛푡 휕퐻 휕퐻 And 푞푖ሶ = = = 휔푖 휕푝푖 휕훼푖

푞푖 = 휔푖푡 + 훽푖

훽푖 is constant and can be find by the initial conditions. But in real problem it is not necessary that all the coordinates are cyclic.

5 6.1 Canonical Transformations

➢ Practically, it rarely happens that all the coordinates are cyclic. ➢ However a system can be described by more than one set of generalized coordinates. ➢ The motion of particle in plane is described by generalized coordinates either the cartesian coordinates. In cartesian coordinates

푞1 = 푥, & 푞1 = 푦 In polar coordinates

푞1 = 푟, & 푞1 = 휃 Both choices are equally valid, but one of the set may be more convenient for the problem under the consideration. Not that for the central neither x, nor y is cyclic while the second set does contain a cyclic coordinate 휃 6 6.1 Canonical Transformations

➢ The number of cyclic coordinates thus depend on choice of generalized coordinates, and for each problem there may be one choice for which all the coordinates are cyclic.

➢ Since the generalized coordinates suggested by the problem will not be cyclic normally, we must first derive a specific procedure for transforming from one set of variables to some other set that may be more suitable.

7 6.1 Canonical Transformations

➢ Let us consider transformation equations

푄푖 = 푄푖 푞푖, 푝푖, 푡 , & 푃푖 = 푃푖 푞푖, 푝푖, 푡 ➢ Such that the general dynamical theory is invariant under these transformations.

➢ Let us consider a function 퐾 푄푖, 푃푖, 푡 such that 휕퐾 휕퐾 푃푖ሶ = − & 푄ሶ푖 = 휕푄푖 휕푃푖

푄푖 & 푃푖 are called canonical coordinates and transformation 푞푖 → 푄푖 & 푝푖 → 푃푖

푄푖 = 푄푖 푞푖, 푝푖, 푡 , & 푃푖 = 푃푖 푞푖, 푝푖, 푡 are known as canonical transformations.

8 6.1 Canonical Transformations

Here “퐾 푄푖, 푃푖, 푡 ” play role of Hamiltonian and 푄푖 & 푃푖 must satisfy Hamilton’s principle.

푡2 (훿 ׬ σ 푃푖 푄ሶ푖 − 퐾 푄푖, 푃푖, 푡 푑푡 (1 푡1

푡2 (훿 ׬ σ 푝푖 푞푖ሶ − 퐻 푞푖, 푝푖, 푡 푑푡 = 0 (2 푡1 Equation (1) and Equation (2) may not be qual, therefore we can find a function “F” such that

푡2 푑퐹 ׬ 푑푡 = 퐹 푡2 − 퐹 푡1 푡1 푑푡

푡2 푑퐹 and δ ׬ 푑푡 = 0 where δ퐹 푡2 = δ퐹 푡1 푡1 푑푡 푑퐹 and σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + (3) 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 9 Function “퐹” is called generating function. 6.1 Canonical Transformations

➢ There are four different possibilities for “퐹”

1) 퐹1 푞푖, 푄푖, 푡 Provided that 푞푖, 푄푖 are treated as independent

2) 퐹2 푞푖, 푃푖, 푡 Provided that 푞푖, 푃푖 are treated as independent

3) 퐹3 푝푖, 푄푖, 푡 Provided that 푝푖, 푄푖 are treated as independent

4) 퐹4 푝푖, 푃푖, 푡 Provided that 푝푖, 푃푖 are treated as independent

10 6.1 Canonical Transformations

Case I: Eq. 3 can be written as 푑퐹 σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 (4) 푖 푖 푖 푖 푖 푖 푖 푖 푑푡

푑퐹1 푞푖,푄푖,푡 휕퐹1 휕퐹1 휕퐹1 Since = σ 푞푖ሶ + σ 푄ሶ푖 + Therefore Eq. (4)… 푑푡 휕푞푖 휕푄푖 휕푡

휕퐹1 휕퐹1 휕퐹1 σ 푝푖 푞푖ሶ − 퐻 푞푖, 푝푖, 푡 = σ 푃푖 푄ሶ푖 − 퐾 푄푖, 푃푖, 푡 + σ 푞푖ሶ + σ 푄ሶ푖 + 휕푞푖 푑푄푖 휕푡

Comparing coefficient of “푞푖ሶ ” & “푄ሶ푖” on both sides

휕퐹1 푝푖 = (5)a 휕푞푖

휕퐹1 푃푖 = − (5)b 휕푄푖

휕퐹1 And 퐾 푄푖, 푃푖, 푡 = 퐻 푞푖, 푝푖, 푡 + (5)c 휕푡 11 6.1 Canonical Transformations

➢ From Eq. (5a) we can determine “푝푖” in terms of 푞푖, 푄푖 and t and the inverse transformation 푄푖 in terms of 푞푖, 푝푖 푎푛푑 푡

Eq. (5)a ⇒ 푄푖 = 푄푖 푞푖, 푝푖, 푡

Eq. (5)b ⇒ 푃푖 = 푃푖 푞푖, 푝푖, 푡 & Eq. (5)c provide connection between new and old Hamiltonian

12 6.1 Canonical Transformations

Case II: For 퐹2 푞푖, 푃푖, 푡 generating function

휕퐹1 Since 푃푖 = − 휕푄푖

Therefore, 퐹1 푞푖, 푄푖, 푡 + σ 푃푖 푄푖 = 퐹2 푞푖, 푃푖, 푡 푑퐹 Since σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푑 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 퐹 푞 , 푃 , 푡 − σ 푃 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 2 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 2 − σ 푃 푄ሶ − σ 푃ሶ 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = −퐾 푄 , 푃 , 푡 + 2 − σ 푃ሶ 푄 푖 푖 푖 푖 푖 푖 푑푡 푖 푖

푑 휕퐹2 휕퐹2 휕퐹2 Since 퐹2 푞푖, 푃푖, 푡 = σ 푞푖ሶ + σ 푃푖ሶ + 푑푡 휕푞푖 휕푃푖 휕푡

13 6.1 Canonical Transformations

Putting in previous equation

휕퐹2 휕퐹2 휕퐹2 ෍ 푝푖 푞푖ሶ − 퐻 푞푖, 푝푖, 푡 = −퐾 푄푖, 푃푖, 푡 + ෍ 푞푖ሶ + ෍ 푃푖ሶ + − ෍ 푃푖ሶ 푄푖 휕푞푖 휕푃푖 휕푡

Comparing coefficient of “푞푖ሶ ” & “푃푖ሶ ” on both sides

휕퐹2 푝푖 = (6)a 휕푞푖

휕퐹1 푄푖 = (6)b 휕푃푖 휕퐹 And 퐾 푄 , 푃 , 푡 = 퐻 푞 , 푝 , 푡 + 2 (6)c 푖 푖 푖 푖 휕푡

14 6.1 Canonical Transformations

Case III: For 퐹3 푝푖, 푄푖, 푡 generating function

휕퐹1 Since 푝푖 = 휕푞푖

Therefore, we can write 퐹1 푞푖, 푄푖, 푡 − σ 푝푖 푞푖 = 퐹3 푝푖, 푄푖, 푡 푑퐹 Since σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푑 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 퐹 푝 , 푄 , 푡 + σ 푝 푞 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 3 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 3 + σ 푝 푞ሶ + σ 푝ሶ 푞 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푖 푖 푑퐹 ⇨ −퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 3 + σ 푝ሶ 푞 푖 푖 푖 푖 푖 푖 푑푡 푖 푖

푑 휕퐹3 휕퐹3 휕퐹3 Since 퐹3 푝푖, 푄푖, 푡 = σ 푝푖ሶ + σ 푄ሶ푖 + , putting in above equation. 푑푡 휕푝푖 휕푄푖 휕푡

15 6.1 Canonical Transformations

Putting in previous equation

휕퐹3 휕퐹3 휕퐹3 −퐻 푞푖, 푝푖, 푡 = ෍ 푃푖 푄ሶ푖 − 퐾 푄푖, 푃푖, 푡 + ෍ 푝푖ሶ + ෍ 푄ሶ푖 + + ෍ 푝푖ሶ 푞푖 휕푝푖 휕푄푖 휕푡

Comparing coefficient of “푝푖ሶ ” & “푄ሶ푖” on both sides

휕퐹3 푞푖 = − (7)a 휕푝푖

휕퐹3 푃푖 = − (7)b 휕푄푖 휕퐹 And 퐾 푄 , 푃 , 푡 = 퐻 푞 , 푝 , 푡 + 3 (7)c 푖 푖 푖 푖 휕푡

16 6.1 Canonical Transformations

Case IV: For 퐹4 푝푖, 푃푖, 푡 generating function

휕퐹1 휕퐹1 Since 푝푖 = & 푃푖 = − 휕푞푖 휕푄푖

Therefore, we can write 퐹1 푞푖, 푄푖, 푡 − σ 푝푖 푞푖 + σ 푃푖 푄푖 = 퐹4 푝푖, 푄푖, 푡 푑퐹 Since σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푑 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 퐹 푝 , 푄 , 푡 + σ 푝 푞 − σ 푃 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 4 푖 푖 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 4 + σ 푝 푞ሶ + σ 푝ሶ 푞 − σ 푃 푄ሶ − σ 푃ሶ 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푖 푖 푖 푖 푖 푖 푑퐹 −퐻 푞 , 푝 , 푡 = −퐾 푄 , 푃 , 푡 + 4 + σ 푝ሶ 푞 − σ 푃ሶ 푄 푖 푖 푖 푖 푑푡 푖 푖 푖 푖

푑 휕퐹4 휕퐹4 휕퐹4 Since 퐹4 푝푖, 푃푖, 푡 = σ 푝푖ሶ + σ 푃푖ሶ + , putting in above equation. 푑푡 휕푝푖 휕푃푖 휕푡 17 6.1 Canonical Transformations

Putting in previous equation

휕퐹4 휕퐹4 휕퐹4 −퐻 푞푖, 푝푖, 푡 = −퐾 푄푖, 푃푖, 푡 + ෍ 푝푖ሶ + ෍ 푃푖ሶ + + ෍ 푝푖ሶ 푞푖 − ෍ 푃푖ሶ 푄푖 휕푝푖 휕푃푖 휕푡

Comparing coefficient of “푝푖ሶ ” & “푃푖ሶ ” on both sides

휕퐹4 푞푖 = − (8)a 휕푝푖

휕퐹4 푄푖 = (8)b 휕푃푖 휕퐹 And 퐾 푄 , 푃 , 푡 = 퐻 푞 , 푝 , 푡 + 4 (8)c 푖 푖 푖 푖 휕푡

18 6.1 Canonical Transformations

Properties of the Four basic canonical transformations Generating Derivatives of Trivial Transformation function generating function special cases 퐹 푞 , 푄 , 푡 휕퐹1 휕퐹1 퐹 = 푞 푄 푝 = 푄 , 1 푖 푖 푝푖 = , 푃푖 = − 1 푖 푖 푖 푖 휕푞푖 휕푄푖 푃푖 = −푞푖 퐹 푞 , 푃 , 푡 휕퐹2 휕퐹1 퐹 = 푞 푃 푝 = 푃 2 푖 푖 푝푖 = , 푄푖 = 2 푖 푖 푖 푖 휕푞푖 휕푃푖 푄푖 = 푞푖 퐹 푝 , 푄 , 푡 휕퐹3 휕퐹3 퐹 = 푝 푄 푞 = −푄 3 푖 푖 푞푖 = − , 푃푖 = − 3 푖 푖 푖 푖 휕푝푖 휕푄푖 푃푖 = −푝푖 퐹 푝 , 푃 , 푡 휕퐹4 휕퐹4 퐹 = 푝 푃 푞 = −푃 4 푖 푖 푞푖 = − , 푄푖 = 4 푖 푖 푖 푖 휕푝푖 휕푃푖 푄푖 = 푝푖

19 6.2 Conditions for the transformation to be canonical

Conditions for the transformation to be canonical

For 푭ퟏ 풒풊, 푸풊, 풕 ⇨ 풅푭ퟏ = σ 풑풊풅풒풊 − σ 푷풊풅푸풊

For 푭ퟐ 풒풊, 푷풊, 풕 ⇨ 풅푭ퟐ = σ 풑풊풅풒풊 + σ 푸풊풅푷풊

For 푭ퟑ 풑풊, 푸풊, 풕 ⇨ 풅푭ퟑ = − σ 풒풊풅풑풊 − σ 푷풊풅푸풊

For 푭ퟒ 풑풊, 푷풊, 풕 ⇨ 풅푭ퟒ = − σ 풒풊풅풑풊 + σ 푷풊풅푸풊

20 6.2 Conditions for the transformation to be canonical

The transformation from 푞푖, 푝푖 to 푄푖, 푃푖 will be canonical if

෍ 풑풊풅풒풊 − ෍ 푷풊풅푸풊 is an exact differential

Solution: Consider the generating function 퐹1 푞푖, 푄푖 휕퐹1 휕퐹1 푑퐹1 = ෍ 푑푞푖 + ෍ 푑푄푖 휕푞푖 휕푄푖 휕퐹1 휕퐹1 Since 푝푖 = and 푃푖 = − 휕푞푖 휕푄푖 Therefore,

푑퐹1 = ෍ 푝푖푑푞푖 − ෍ 푃푖푑푄푖

which is an exact differential equation. 21 6.2 Conditions for the transformation to be canonical

Similarly, considering generating function 퐹4 푝푖, 푃푖, 푡 휕퐹4 휕퐹4 푑퐹4 푝푖, 푃푖, 푡 = σ 푑푝푖 + σ 푑푃푖 휕푝푖 휕푃푖 휕퐹4 휕퐹4 Since 푞푖 = − and 푄푖 = 휕푝푖 휕푃푖

Therefore, 푑퐹4 푝푖, 푃푖, 푡 = − σ 푞푖푑푝푖 + σ 푄푖푑푃푖 which is an exact differential Now subtracting 푑퐹4 from 푑퐹1

푑퐹1 − 푑퐹4 = σ 푝푖푑푞푖 − σ 푃푖푑푄푖 + σ 푞푖푑푝푖 − σ 푄푖푑푃푖 ⇒ 푑퐹1 − 푑퐹4 = σ 푞푖푑푝푖 + σ 푝푖푑푞푖 − σ 푄푖푑푃푖 + σ 푃푖푑푄푖 ⇒ 푑퐹1 − 푑퐹4 = 푑 푞푖푝푖 − 푑 푄푖푃푖 ⇒ 푑 퐹1 − 퐹4 = 푑 푞푖푝푖 − 푄푖푃푖 Which is exact differential. Therefore the transformation is canonical. And ⇒ 퐹1 = 퐹4 + 푞푖푝푖 − 푄푖푃푖 22