# Goldstein- CHAPTER 9 [SOLUTIONS]

Book: Classical Mechanics 3rd Edition
Author(s): Herbert Goldstein, Charles P. Poole, John L. Safko
So, I have tried solving some of the problems of the Chapter 9 of Goldstein Classical mechanics.
You can download the pdf version here: Solutions Goldstein Chapter 9
I have also embedded the pdf below as well as posted them in this blog post.

Solutions Goldstein Chapter 9

CHAPTER 9 – CANONICAL TRANSFORMATIONS
DERIVATIONS:

9.4. Show directly that the transformation
$Q=\log\left(\frac{1}{q}\sin{p} \right )$
$P=q\cot{p}$
is canonical.
9.4. Sol.
We are given a transformation as follows,
$Q=\log\left(\frac{1}{q}\sin{p} \right )$
$P=q\cot{p}$
We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, the Poisson Bracket of Q,P with respect to q & p should be equal to 1.

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q,P]_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}$
$\implies[Q,P]_{q,p}= \left(\frac{1}{\frac{1}{q}\sin{p}}\right)\frac{\partial \left(\frac{1}{q}\sin{p}\right)}{\partial q}(-q\csc^2{p})-\cot{p}\left(\frac{1}{\frac{1}{q}\sin{p}}\right)\frac{\partial \left(\frac{1}{q}\sin{p} \right )}{\partial p}$
$\implies[Q,P]_{q,p}= \left(\frac{q}{\sin{p}}\right)\left(\frac{-\sin{p}}{q^2}\right)\left(-q\csc^2{p}\right)-\cot{p}\left(\frac{q}{\sin{p}}\right)\left(\frac{1}{q}\cos{p}\right)$
$\implies[Q,P]_{q,p}= \csc^2{p}-\cot^2{p}$
$\implies[Q,P]_{q,p}= 1$
$\therefore [Q,P]_{q,p}= [Q,P]_{Q,P}$

Hence Proved.

9.5. Show directly that for a system of one degree of freedom, the transformation
$Q=\arctan{\frac{\alpha q}{p}}$
$P=\frac{\alpha q^2}{2}\left(1+\frac{p^2}{\alpha^2 q^2}\right)$
is canonical, where α is an arbitrary constant of suitable dimensions.
9.5. Sol.

We are given a transformation as follows,
$Q=\arctan{\frac{\alpha q}{p}}$
$P=\frac{\alpha q^2}{2}\left(1+\frac{p^2}{\alpha^2 q^2}\right)$

We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, the Poisson Bracket of Q,P with respect to q & p should be equal to 1.

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q,P]_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}$

$\implies [Q,P]_{q,p}= \left( \frac{1}{1+\frac{\alpha^2q^2}{p^2}}\right)\frac{\partial \left(\frac{\alpha q}{p} \right )}{\partial q}\left( \frac{p}{\alpha} \right )-2\left(\frac{\alpha q}{2} \right )\left(\frac{1}{1+\frac{\alpha^2 q^2}{p^2}} \right )\frac{\partial \left(\frac{\alpha q}{p} \right )}{\partial p}$
$\implies [Q,P]_{q,p}= \left( \frac{1}{1+\frac{\alpha^2q^2}{p^2}}\right)\left(\frac{\alpha}{p}\right )\left(\frac{p}{\alpha}\right)-\left(\alpha q \right )\left(\frac{1}{1+\frac{\alpha^2 q^2}{p^2}} \right )\left(\frac{-\alpha q}{p^2}\right)$
$\implies [Q,P]_{q,p}= \left(\frac{p^2}{p^2+\alpha^2q^2} \right)\left[1+\frac{\alpha^2q^2}{p^2} \right ]$
$\implies [Q,P]_{q,p}= \left(\frac{p^2}{p^2+\alpha^2q^2} \right)\left(\frac{p^2+\alpha^2q^2}{p^2} \right)$
$\implies [Q,P]_{q,p}= 1$
$\therefore [Q,P]_{q,p}= [Q,P]_{Q,P}$

Hence Proved.

9.6. The transformation equations between two sets of coordinates are
$Q=\log(1+\sqrt{q}\cos{p})$
$P=2(1+\sqrt{q}\cos{p})\sqrt{q}\sin{p}$
(a) Show directly from these transformation equations that Q,P are canonical variables if q and p are.
(b) Show that the function that generates this transformation is

$F_3=-(e^Q-1)^2 \tan{p}$
9.6 Sol. (a)
We are given a transformation as follows,
$Q=\log(1+\sqrt{q}\cos{p})$
$P=2(1+\sqrt{q}\cos{p})\sqrt{q}\sin{p}$

We can re-write the second equation as:
$\implies P=2\sqrt{q}\sin{p}+2q\cos{p}\sin{p}$

We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, the Poisson Bracket of Q,P with respect to q & p should be equal to 1.

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q,P]_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}$
$\implies [Q,P]_{q,p}= \left(\frac{1}{1+\sqrt{q}cosp} \right )\frac{\partial \left(1+\sqrt{q}cosp \right ) }{\partial q}\left[2\sqrt{q}cosp+2q \frac{\partial (cospsinp)}{\partial p} \right ]-\left[ 2sinp\left(\frac{1}{2\sqrt{q}} \right )+2cospsinp\right]\left(\frac{1}{1+\sqrt{q}cosp} \right )\frac{\partial \left(1+\sqrt{q}cosp \right ) }{\partial p}$
$\implies [Q,P]_{q,p}= \left(\frac{1}{1+\sqrt{q}cosp} \right )\left(\frac{cosp}{2\sqrt{q}} \right )\left[2\sqrt{q}cosp+2qcos^2p-2qsin^2p \right ]-\left[ 2sinp\left(\frac{1}{2\sqrt{q}} \right )+2cospsinp\right]\left(\frac{1}{1+\sqrt{q}cosp} \right )\left(-\sqrt{q}sinp\right )$
$\implies [Q,P]_{q,p}= \left(\frac{1}{1+\sqrt{q}cosp} \right )\left[cos^2p+\sqrt{q}cos^3p-\sqrt{q}cos psin^2p \right ]+\left(\frac{1}{1+\sqrt{q}cosp} \right )\left[sin^2p+2\sqrt{q}cospsin^2p\right]$
$\implies [Q,P]_{q,p}= \left(\frac{1}{1+\sqrt{q}cosp} \right )\left[cos^2p+sin^2p+\sqrt{q}cos^3p-\sqrt{q}cospsin^2p+2\sqrt{q}cospsin^2p\right]$
$\implies [Q,P]_{q,p}= \left(\frac{1}{1+\sqrt{q}cosp} \right )\left[1+\sqrt{q}cosp\left(cos^2p-sin^2p+2sin^2p\right)\right]$
$\implies [Q,P]_{q,p}= \left(\frac{1}{1+\sqrt{q}cosp} \right )\left[1+\sqrt{q}cosp\left(cos^2p+sin^2p\right)\right]$
$\implies [Q,P]_{q,p}= \left(\frac{1}{1+\sqrt{q}cosp} \right )\left(1+\sqrt{q}cosp\right)$
$\implies [Q,P]_{q,p}= 1$
$\therefore [Q,P]_{q,p}= [Q,P]_{Q,P}$

Hence Proved.
(b)
We are given the following generating function of the F3 type:
$F_3=-(e^Q-1)^2\tan p$
For a generating function of F3 type, q is given as:
$q=-\frac{\partial F_3}{\partial p}$
$\implies q=(e^Q-1)^2\sec^2p$
$\implies \sqrt{q}\cos p=e^Q-1$
$\implies \boxed{Q=\log(1+\sqrt{q}\cos p)} \hspace{1cm} \textup{(i)}$

and P is given as:
$P=-\frac{\partial F_3}{\partial Q}$
$P=2\tan p(e^Q-1)e^Q$
Plugging the value of Q from eq(i) in the above equation,
$\implies P=2\tan p(1+\sqrt{q}\cos p-1)(1+\sqrt{q}\cos p)$
$\implies P=2\frac{\sin p}{\cos p}\sqrt{q}\cos p(1+\sqrt{q}\cos p)$
$\implies \boxed {P=2(1+\sqrt{q}\cos p)\sqrt{q}\sin p}$
Therefore, the given generating does in fact generate the given transformation.

9.8. Prove directly that the transformation
$Q_1=q_1$
$P_1=p_1-2p_2$
$Q_2=p_2$
$P_2=-2q_1-q_2$
is canonical and find a generating function.
9.8. Sol.
We are given a transformation as follows,
$Q_1=q_1$
$P_1=p_1-2p_2$
$Q_2=p_2$
$P_2=-2q_1-q_2$
We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, Q,P should satisfy the following condition:
$[Q_j,P_k]_{q,p}=\delta_{jk}$
i.e. we need to prove, $[Q_{1},P_{1}]_{q,p}=1$
$[Q_{2},P_{2}]_{q,p}=1$
and
$[Q_1,Q_2]_{q,p}=[P_1,P_2]_{q,p}=[Q_1,P_2]_{q,p}=[Q_2,P_1]_{q,p}=0$

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q_{1},P_{1}]_{q,p}=\frac{\partial Q_1}{\partial q_i}\frac{\partial P_1}{\partial p_i}-\frac{\partial P_1}{\partial q_i}\frac{\partial Q_1}{\partial p_i}$
$\implies [Q_{1},P_{1}]_{q,p}=\frac{\partial Q_1}{\partial q_1}\frac{\partial P_1}{\partial p_1}-\frac{\partial P_1}{\partial q_1}\frac{\partial Q_1}{\partial p_1}+\frac{\partial Q_1}{\partial q_2}\frac{\partial P_1}{\partial p_2}-\frac{\partial P_1}{\partial q_2}\frac{\partial Q_1}{\partial p_2}$
$\implies [Q_1,P_1]_{q,p}=1-0+0-0$
$\implies [Q_1,P_1]_{q,p}=1$

Now,

$[Q_{2},P_{2}]_{q,p}=\frac{\partial Q_2}{\partial q_i}\frac{\partial P_2}{\partial p_i}-\frac{\partial P_2}{\partial q_i}\frac{\partial Q_2}{\partial p_i}$
$\implies [Q_{2},P_{2}]_{q,p}=\frac{\partial Q_2}{\partial q_1}\frac{\partial P_2}{\partial p_1}-\frac{\partial P_2}{\partial q_1}\frac{\partial Q_2}{\partial p_1}+\frac{\partial Q_2}{\partial q_2}\frac{\partial P_2}{\partial p_2}-\frac{\partial P_2}{\partial q_2}\frac{\partial Q_2}{\partial p_2}$
$\implies [Q_2,P_2]_{q,p}=0-0+0-(-1)1$
$\implies [Q_2,P_2]_{q,p}=1$

Hence the given transformation is canonical.

9.10. Find under what conditions
$Q=\frac {\alpha p}{x}$
$P=\beta x^2$
where α and β are constants, represents a canonical transformation for a system of one degree of freedom, and obtain a suitable generating function. Apply the transformation to the solution of the linear harmonic oscillator.
9.10.Sol.
We are given a transformation as follows,
$Q=\frac {\alpha p}{x}$
$P=\beta x^2$

We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, the Poisson Bracket of Q,P with respect to q & p should be equal to 1.

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q,P]_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}$
$\implies [Q,P]_{q,p}=\left(\frac{-\alpha p}{x^2}\right)0-2\beta x \frac{\alpha}{x}$
$\implies [Q,P]_{q,p}=-2\beta \alpha$

But $[Q,P]_{q,p}=1$ for canonical transformation.
$\therefore -2\beta \alpha=1$
$\implies \beta =-\frac{1}{2\alpha}$
which is the condition for the given transformation to be canonical.

9.11. Determine whether the transformation
$Q_1=q_1q_2$
$P_1=\frac{p_1-p_2}{q_2-q_1}+1$
$Q_2=q_1+q_2$
$P_2=-\frac{q_2p_2-q_1p_1}{q_2-q_1}-(q_2+q_1)$
is canonical.
Sol.9.11

We are given a transformation as follows,
$Q_1=q_1q_2$
$P_1=\frac{p_1-p_2}{q_2-q_1}+1$
$Q_2=q_1+q_2$
$P_2=-\frac{q_2p_2-q_1p_1}{q_2-q_1}-(q_2+q_1)$
We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, the Poisson Bracket of Qi,Pi with respect to q & p should be equal to 1.
i.e. we need to prove, $[Q_{1},P_{1}]_{q,p}=1$
$[Q_{2},P_{2}]_{q,p}=1$

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q_{1},P_{1}]_{q,p}=\frac{\partial Q_1}{\partial q_i}\frac{\partial P_1}{\partial p_i}-\frac{\partial P_1}{\partial q_i}\frac{\partial Q_1}{\partial p_i}$
$\implies [Q_{1},P_{1}]_{q,p}=\frac{\partial Q_1}{\partial q_1}\frac{\partial P_1}{\partial p_1}-\frac{\partial P_1}{\partial q_1}\frac{\partial Q_1}{\partial p_1}+\frac{\partial Q_1}{\partial q_2}\frac{\partial P_1}{\partial p_2}-\frac{\partial P_1}{\partial q_2}\frac{\partial Q_1}{\partial p_2}$
$\implies [Q_{1},P_{1}]_{q,p}=q_2\left(\frac{1}{q_2-q_1}\right)-0+q_1\left(\frac{-1}{q_2-q_1} \right )$
$\implies [Q_1,P_1]_{q,p}=\frac {q_2-q_1}{q_2-q_1}$
$\implies [Q_1,P_1]_{q,p}=1$

Now,

$[Q_{2},P_{2}]_{q,p}=\frac{\partial Q_2}{\partial q_i}\frac{\partial P_2}{\partial p_i}-\frac{\partial P_2}{\partial q_i}\frac{\partial Q_2}{\partial p_i}$
$\implies [Q_{2},P_{2}]_{q,p}=\frac{\partial Q_2}{\partial q_1}\frac{\partial P_2}{\partial p_1}-\frac{\partial P_2}{\partial q_1}\frac{\partial Q_2}{\partial p_1}+\frac{\partial Q_2}{\partial q_2}\frac{\partial P_2}{\partial p_2}-\frac{\partial P_2}{\partial q_2}\frac{\partial Q_2}{\partial p_2}$
$\implies [Q_2,P_2]_{q,p}=\frac{-q_1}{q_2-q_1}-0+\frac{q_2}{q_2-q_1}-0$
$\implies [Q_2,P_2]_{q,p}=\frac {q_2-q_1}{q_2-q_1}$
$\implies [Q_2,P_2]_{q,p}=1$

Hence, the given transformation is canonical.

9.14. Prove that the transformation
$Q_1=q_1^2$
$P_1=\frac{p_1 \cos p_2-2q_2}{2q_1 \cos p_2}$
$Q_2=q_2\sec p_2$
$P_2=-\frac{q_2p_2-q_1p_1}{q_2-q_1}-(q_2+q_1)$
is canonical, by any method you choose. Find a suitable generating function that will lead to this transformation.
Sol.9.14.

We are given a transformation as follows,
$Q_1=q_1^2$
$P_1=\frac{p_1 \cos p_2-2q_2}{2q_1 \cos p_2}$
$Q_2=q_2\sec p_2$
$P_2=-\frac{q_2p_2-q_1p_1}{q_2-q_1}-(q_2+q_1)$
We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, the Poisson Bracket of Qi,Pi with respect to q & p should be equal to 1.
i.e. we need to prove, $[Q_{1},P_{1}]_{q,p}=1$
$[Q_{2},P_{2}]_{q,p}=1$

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q_{1},P_{1}]_{q,p}=\frac{\partial Q_1}{\partial q_i}\frac{\partial P_1}{\partial p_i}-\frac{\partial P_1}{\partial q_i}\frac{\partial Q_1}{\partial p_i}$
$\implies [Q_{1},P_{1}]_{q,p}=\frac{\partial Q_1}{\partial q_1}\frac{\partial P_1}{\partial p_1}-\frac{\partial P_1}{\partial q_1}\frac{\partial Q_1}{\partial p_1}+\frac{\partial Q_1}{\partial q_2}\frac{\partial P_1}{\partial p_2}-\frac{\partial P_1}{\partial q_2}\frac{\partial Q_1}{\partial p_2}$
$\implies [Q_1,P_1]_{q,p}= 2q_1\frac{\cos{p_2}}{2q_1\cos{p_2}}-0+0-0$
$\implies [Q_1,P_1]_{q,p}=1$
Now,

$[Q_{2},P_{2}]_{q,p}=\frac{\partial Q_2}{\partial q_i}\frac{\partial P_2}{\partial p_i}-\frac{\partial P_2}{\partial q_i}\frac{\partial Q_2}{\partial p_i}$
$\implies [Q_{2},P_{2}]_{q,p}=\frac{\partial Q_2}{\partial q_1}\frac{\partial P_2}{\partial p_1}-\frac{\partial P_2}{\partial q_1}\frac{\partial Q_2}{\partial p_1}+\frac{\partial Q_2}{\partial q_2}\frac{\partial P_2}{\partial p_2}-\frac{\partial P_2}{\partial q_2}\frac{\partial Q_2}{\partial p_2}$
$\implies [Q_2,P_2]_{q,p}= 0-0+\sec{p_2}\cos{p_2}-0$
$\implies [Q_2,P_2]_{q,p}=1$

Hence the given transformation is canonical.

9.15. (a) Using the fundamental Poisson Brackets find the values of α and β for which the equation
$Q=q^\alpha \cos{\beta p}$
$P=q^\alpha \sin{\beta p}$
represents a canonical transformation.
(b) For what values of α and β do these equations represent an extended canonical transformation? Find a generating function of the F3 form for the transformation.
(c) On the basis of part (b), can the transformation equations be modified so that they describe a canonical transformation for all values of β?

Sol.9.15.(a)
We are given a transformation as follows,
$Q=q^\alpha \cos{\beta p}$
$P=q^\alpha \sin{\beta p}$

We know that the fundamental Poisson Brackets of the transformed variables have the same value when evaluated with respect to any canonical coordinate set. In other words the fundamental Poisson Brackets are invariant under canonical transformation.

Therefore, in order for the given transformation to be canonical, the Poisson Bracket of Q,P with respect to q & p should be equal to 1.

Using the formula for Poisson Bracket,

$[u,v]_{q,p}=\frac{\partial u}{\partial q_{i}}\frac{\partial v}{\partial p_{i}}-\frac{\partial v}{\partial q_{i}}\frac{\partial u}{\partial p_{i}}$

$\therefore [Q,P]_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}$
$\implies [Q,P]_{q,p}=\left(\alpha q^{\alpha-1} \cos{\beta p}\right)\left(q^\alpha \beta \cos{\beta p}\right)-\alpha q^{\alpha-1}\sin{\beta p}q^\alpha \beta(-\sin{\beta p})$
$\implies [Q,P]_{q,p}=\alpha q^{2\alpha-1}\beta\cos^2{\beta p}+\alpha q^{2\alpha -1}\beta\sin^2\beta p$
$\implies [Q,P]_{q,p}=\alpha \beta q^{2\alpha-1}$

9.17. Show that the Jacobi identity is satisfied if the Poisson Bracket sign stands for the commutator of two square matrices.
$[A,B] = AB-BA$
Show also that for the same representation of the Poisson bracket that
$[A,BC]=[A,B]C+B[A,C].$
Sol.9.17.
We are given that the Poisson bracket sign stands for the commutator of two square matrices:
$[A,B]=AB - BA$

We need to show that the Jacobi identity is satisfied by the new(above) definition of Poisson Brackets.
Therefore, we need to prove:
$\left[A,[B,C] \right ]+\left[B,[C,A] \right ]+\left[C,[A,B] \right ]=0$

Proof:
$\left[A,[B,C]\right]=[A,BC-CB]$ using the new definition of P.B.
$\implies \left[A,[B,C]\right]= A(BC-CB)-(BC-CB)A$
$\implies \left[A,[B,C]\right]= ABC-ACB-BCA+CBA \hspace{1cm}\textup{(i)}$
Note: We can’t simply cancel ABC by ACB as these are matrices and hence their product isn’t commutative.

Similarly,
$\left[B,[C,A]\right]=[B,CA-AC]$
$\implies \left[B,[C,A]\right]= B(CA-AC)-(CA-AC)B$
$\implies \left[B,[C,A]\right]= BCA-BAC-CAB+ACB \hspace{1cm}\textup{(ii)}$

and,
$\left[C,[A,B]\right]=[C,AB-BA]$
$\implies \left[C,[A,B]\right]= C(AB-BA)-(AB-BA)C$
$\implies \left[C,[A,B]\right]= CAB-CBA-ABC+BAC \hspace{1cm}\textup{(iii)}$

Adding (i), (ii) & (iii), we get
$\left[A,[B,C]\right]+\left[B,[C,A]\right]+\left[C,[A,B]\right]=ABC-ACB-BCA+CBA+BCA-BAC-CAB+ACB+CAB-CBA-ABC+BAC$
$\implies \boxed{\left[A,[B,C]\right]+\left[B,[C,A]\right]+\left[C,[A,B]\right]=0}$

Hence, Jacobi Identity is satisfied by the commutator of two square matrices.

Next, we are asked to prove that,
$[A,BC]=[A,B]C+B[A,C]$

Proof:
$[A,BC]=ABC-BCA$ from the given definition of P.B.
$\implies [A,BC]=ABC-BAC-BCA+BAC$ adding and subtracting $BAC$
$\implies [A,BC]=(AB-BA)C-B(CA-AC)$
$\implies [A,BC]=[A,B]C+B(AC-CA)$
$\implies \boxed{[A,BC]=[A,B]C+B[A,C]}$
Hence Proved.

9.21. (a) For a one-dimensional system with the Hamiltonian
$H=\frac{p^2}{2}-\frac{1}{2q^2}$
show that there is a constant of motion
$D=\frac{pq}{2}-Ht$
(b) As a generalization of part (a), for motion in a plane with the Hamiltonian
$H=|\mathbf{p}|^n-ar^{-n},$
where p is the vector of the momenta conjugate to the Cartesian coordinates, show that there is a constant of motion
$D=\frac{\mathbf{p.r}}{n}-Ht$
(c) The transformation Q=λq, p=λ is obviously canonical. However, the same transformation with with t time dilation, Q=λq, p=λP, t’=λ2t, is not. Show that, however, the equaitons of motion for q and p for the Hamiltonian in part(a) are invariant under this transformation. The constant of motion D is said to be associated with this invariance.
Sol.9.21. a)

$H=\frac{p^2}{2}-\frac{1}{2q^2}$
$D=\frac{pq}{2}-Ht$
$\frac{\mathrm{d} D}{\mathrm{d} t}=[D,H]+\frac{\partial D}{\partial t}$
$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = \left[ \frac{pq}{2}-Ht,H \right] -H$
$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = \frac{1}{2}[pq,H]-t[H,H]-H$
$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = \frac{1}{2}\left[ pq,\frac{p^2}{2}-\frac{1}{2q^2} \right]-0-H$
$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = \frac{1}{4}[pq,p^2]-\frac{1}{4}\left[ pq,\frac{1}{q^2} \right]-H$
$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = \frac{1}{4}\left\{ p[q,p^2]+[p,p^2]q-p\left[ q,\frac{1}{q^2} \right] -\left[p,\frac{1}{q^2}\right]q \right\}-H$
$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = \frac{1}{4}\left\{ p^2[q,p]+p[q,p]p+0-0-\frac{1}{q}\left[p,\frac{1}{q}\right]q-\left[p,\frac{1}{q}\right]\frac{q}{q}\right\}-H$

$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = \frac{1}{4}\left\{p^2+p^2-\frac{1}{q^2}-\frac{1}{q^2}\right\}-H$

where we have used,
$\left[p,\frac{1}{q}\right]=\frac{1}{q^2}$

$\therefore \frac{\mathrm{d} D}{\mathrm{d} t} = \frac{p^2}{2}-\frac{1}{2q^2}-H$
$\implies \frac{\mathrm{d} D}{\mathrm{d} t} = 0$
Therefore, D is a constant of motion.

9.22. Show that the following transformation is canonical by using Poisson Bracket:
$Q=\sqrt{2q}e^\alpha \cos p$
$P=\sqrt{2q}e^{-\alpha} \sin p$
Sol.9.22.
$Q=\sqrt{2q}e^\alpha \cos p$
$P=\sqrt{2q}e^{-\alpha} \sin p$

$[Q,P]_{q,p}= \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}$
$\implies [Q,P]_{q,p}= \frac{e^\alpha \cos p}{\sqrt{2q}}\sqrt{2q}e^-\alpha \cos p-\frac{e^{-\alpha}\sin p}{\sqrt{2q}}\sqrt{2q}e^\alpha(-\sin p)$
$\implies [Q,P]_{q,p}= \cos^2 p + \sin^2 p$
$\implies [Q,P]_{q,p}= 1$
$\therefore [Q,P]_{q,p}= [Q,P]_{Q,P}$

9.23. Prove that the following transformation from (q,p) to (Q,P) basis is canonical using Poisson Bracket
$Q=\sqrt{2}q\tan p$
$P=\sqrt{2}log(\sin p)$
Sol.9.23. NOTE: There seems to be some typo in the question as the Poisson Bracket of (Q,P) with respect to (q,p) doesn’t come out to be 1.

9.24. Prove that the transformation defined by Q=1/p and P=qp2 is canonical using Poisson Bracket.
Sol.9.24.

$Q=\frac{1}{p}$
$P=qp^2$

$[Q,P]_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}$
$\implies [Q,P]_{q,p}=0-p^2\left(-\frac{1}{p^2}\right)$
$\implies [Q,P]_{q,p}=1$
$\therefore [Q,P]_{q,p}= [Q,P]_{Q,P}$

9.30.(a) Prove that the Poisson Bracket of two constants os motion is itself a constant of motion even when when the constants depend on time explicitly.
(b) Show that if the Hamiltonian and a quantity F are constants of motion, then the nth partial derivative F with respect to t must also be a constant of motion.
(c) As an illustration of this result, consider the uniform motion of a free particle of mass m. The Hamiltonian is certainly conseerved, and there exists a constant of motion

$F=x-\frac{pt}{m}$
Show by direct computation od the Po
Sol.9.30.
(a)

Let the two constants of motion be A(t) & B(t).
Then since A(t) and B(t) are given to be constants of motion, therefore
$\frac{\mathrm{d} A(t)}{\mathrm{d} t}=0$
$\implies [A,H]+\frac{\partial A}{\partial t}=0 \hspace{1cm}\textup{(i)}$
and
$\frac{\mathrm{d} B(t)}{\mathrm{d} t}=0$
$\implies [B,H]+\frac{\partial B}{\partial t}=0 \hspace{1cm}\textup{(ii)}$

$\frac{\mathrm{d} [A,B]}{\mathrm{d} t} = \left[[A,B],H\right]+\frac{\partial [A,B]}{\partial t} \hspace{1cm}\textup{(iv)}$

Let’s evaluate the second term in the R.H.S.:
$\frac{\partial [A,B]}{\partial t} = \left[ \frac{\partial A}{\partial t},B \right] + \left[ A,\frac{\partial B}{\partial t} \right] \hspace{1cm}\textup{(iv)}$
from (i) and (ii) we have
$\frac{\partial A}{\partial t}=-[A,H]$
$\frac{\partial B}{\partial t}=-[B,H]$

Plugging these back in eq(iv)
$\frac{\partial [A,B]}{\partial t}=\left[ [H,A],B \right]+\left[ A,[H,B] \right]$
Plugging the above back in eq(iii)
$\frac{\mathrm{d} [A,B]}{\mathrm{d} t} = \left[[A,B],H\right]+\left[ [H,A],B \right]+\left[ A,[H,B] \right]$
$\implies \frac{\mathrm{d} [A,B]}{\mathrm{d} t} =\left[H,[B,A]\right]+\left[ B,[A,H] \right]+\left[ A,[H,B] \right]$
Now using Jacobi’s Identity,
$\implies \frac{\mathrm{d} [A,B]}{\mathrm{d} t} = 0$

(b)
Given:
$\frac{\mathrm{d}H}{\mathrm{d}t}=0$
and,
$\because \frac{\mathrm{d}H}{\mathrm{d}t}=[H,H]+\frac{\partial H}{\partial t}$
$\implies \frac{\partial H}{\partial t}=0 \hspace{1cm}\textup{(i)}$
Therefore, H is not a function of time.

We are also given that,
$\frac{\mathrm{d}F}{\mathrm{d}t}=0$
$\implies [F,H]=-\frac{\partial F}{\partial t}$
$\implies \frac{\partial F}{\partial t}=[H,F] \hspace{1cm\textup{(ii)}}$

Now, let’s have a look at the equation of motion of the nth partial derivative of F:
$\frac{\mathrm{d}\frac{\partial^n F}{\partial t^n}}{\mathrm{d}t}=\left[\frac{\partial^n F}{\partial t^n},H\right]+\frac{\partial }{\partial t}\left(\frac{\partial^n F}{\partial t^n}\right)$
$\implies \frac{\mathrm{d}\frac{\partial^n F}{\partial t^n}}{\mathrm{d}t}=\left[\frac{\partial^n F}{\partial t^n},H\right]+\frac{\partial^n }{\partial t^n}[H,F]$
$\implies \frac{\mathrm{d}\frac{\partial^n F}{\partial t^n}}{\mathrm{d}t}=\left[\frac{\partial^n F}{\partial t^n},H\right]+\left[ \frac{\partial^n H}{\partial t^n},F\right]+\left[H, \frac{\partial^n F}{\partial t^n}\right]$
the second term on the R.H.S. is clearly zero as H is independent of time as shown earlier already.
$\implies \frac{\mathrm{d}\frac{\partial^n F}{\partial t^n}}{\mathrm{d}t}=\left[\frac{\partial^n F}{\partial t^n},H\right]-\left[ \frac{\partial^n F}{\partial t^n},H\right]$
$\implies \frac{\mathrm{d}\frac{\partial^n F}{\partial t^n}}{\mathrm{d}t}=0$
Therefore, the nth partial derivative of F is a constant of motion.

(c)
$F=x-\frac{pt}{m}$
$\implies \frac{\partial F}{\partial t}=-\frac{p}{m}$
For a free particle of mass m:
$H=\frac{p^2}{2m}$
$\therefore [H,F]=\left[\frac{p^2}{2m},x-\frac{pt}{m} \right]$
$\implies [H,F]=\left[\frac{p^2}{2m}\right]-\left[\frac{p^2}{2m},\frac{pt}{m}\right]$
$\implies [H,F]= \frac{1}{2m}[p^2,x]-0$
$\implies [H,F]=\frac{1}{2m}\left\{p[p,x]+[p,x]p\right\}$
$\implies [H,F]=\frac{1}{2m}(-p-p)$
$\implies [H,F]=\frac{-p}{m}$
$\therefore \frac{\partial F}{\partial t}=[H,F]$
Hence Proved.

## One thought on “Goldstein- CHAPTER 9 [SOLUTIONS]”

1. please provide the solution of chapter 9th of classical mechanics by goldstein up to problem 38