Canonical transformations

In summary, the conversation discusses a mathematical problem involving the calculation of the symplectic unit matrix in a canonical transformation. The speaker initially tried to calculate the matrix, but ran into difficulties with its representation in latex. They then received help from another person who clarified that the transformation equations only involve two variables, making the matrix a 2x2 rather than a larger size. The speaker then goes on to present their revised calculation and explains that they now need to solve four equations to find the correct values for the variables.
  • #1
Lambda96
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Homework Statement
See Screenshot
Relevant Equations
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Hi,

unfortunately, I have problems with the following task

Bildschirmfoto 2023-01-21 um 17.53.00.png

I first tried to calculate ##JIJ^T##.

$$\left( \begin{array}{rrr}
\frac{\partial q'_i}{\partial q_j} & \frac{\partial q'_i}{\partial P_j} \\\frac{\partial P'_i}{\partial q_j} & \frac{\partial P'_i}{\partial P_j} \\
\end{array}\right)\left( \begin{array}{rrr}
0 & \textbf{1} \\\ -{\textbf{1}} & 0 \\
\end{array}\right)\left( \begin{array}{rrr}
\frac{\partial q'_i}{\partial q_j} & \frac{\partial P'_i}{\partial P_j} \\\frac{\partial P'_i}{\partial q_j} & \frac{\partial q'_i}{\partial P_j} \\
\end{array}\right)=\left( \begin{array}{rrr}
\frac{\partial q'_i}{\partial q_j} \textbf{1} \frac{\partial q'_i}{\partial P_j} -\frac{\partial q'_i}{\partial P_j} \textbf{1} \frac{\partial q'_i}{\partial q_j} & \frac{\partial q'_i}{\partial q_j} \textbf{1} \frac{\partial P_i}{\partial P_j} -\frac{\partial q'_i}{\partial P_j} \textbf{1} \frac{\partial P_i}{\partial q_j} \\
\frac{\partial P_i}{\partial q_j} \textbf{1} \frac{\partial q'_i}{\partial P_j} -\frac{\partial P_i}{\partial P_j} \textbf{1} \frac{\partial q'_i}{\partial q_j} & \frac{\partial P_i}{\partial q_j} \textbf{1} \frac{\partial P_i}{\partial P_j} -\frac{\partial P_i}{\partial P_j} \textbf{1} \frac{\partial P_i}{\partial q_j} \\
\end{array}\right)$$

##\textbf{1}## is supposed to be the unit matrix, but unfortunately I don't know how to write it with latex, which is why I have represented it in my calculation like this

Then I took the following relation from my professor, I hope you can read it well.

Bildschirmfoto 2023-01-21 um 18.38.51.png

and get the following
$$\left( \begin{array}{rrr}
\Bigl\{ q'_i,q'_i \Bigr\} & \Bigl\{ q'_i,P_i \Bigr\} \\
\Bigl\{ P_i,q'_i \Bigr\} & \Bigl\{ P_i,P_i \Bigr\} \\
\end{array}\right)$$

Unfortunately, I am not getting anywhere now, because in order to show which values ##\alpha## and ##\beta## must assume in order for it to be a canonical transformation, I would have to get the symplectic unit matrix again, but with my calculation I would only get numbers as entries in the matrix and not unit matrices, as it should be.

My professor's script contains the following formulation

Bildschirmfoto 2023-01-21 um 18.58.05.png
 
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  • #2
Hi @Lambda96. I’m no expert, but (since you haven’t any other replies and I may have seen where you are going wrong) here goes...

What you have written suggests that you are trying to apply a general format to construct equations for systems with 2 or more degrees of freedom. But the problem is simpler than that.

The given transformation equations (for ##p’## and ##q’##) show the system has only one degree of freedom (so the phase space is 2-dimensional).

That means ##\vec x## has only two components, ##x_1 = q## and ##x_2 = p##.

Similarly ##\vec x’## has only two components, ##x_1’= q’## and ##x_2’ = p’##.

The Jacobian tranformation matrix,##J##, is 2x2.

For example ##J_{11} = \frac {∂x_1’}{∂x_1} = \frac {∂q’}{∂q}##

Since ##q’ = \sqrt 2 p^α \sin q ## then ##J_{11} = \frac {∂q’}{∂q}= \sqrt 2 p^ α \cos q##

Similarly, you need to find ##J_{12}, J_{21}## and ##J_{22}##.

The corresponding symplectic identity matrix, ##I##, is also a 2x2 matrix (the elements are simple integers).

You are dealing only with 2x2 matrices (but still probably have some messy working ahead!). Multiply-out ##JIJ^T## and then find what values of ##\alpha## and ##\beta## are needed to make ##JIJ^T=I##.
 
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  • #3
The matrix notation is a bit clumsy when it comes to concrete calculations in my opinion. It's nice for formal calculations. It just says that the "canonical Poisson brackets" should hold for the new phase-space coordinates, i.e., you have to find values for ##\alpha## and ##\beta## such that
$$\{x',x' \}=\{p',p'\}=0, \quad \{x',p'\}=1.$$
 
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  • #4
Thanks for your help Steve4Physics, I had completely misunderstood the meaning of the matrix and its entries then, also thanks to vanhees71 for your help.

I have now done the calculation again:

$$\left( \begin{array}{rrr}
\frac{\partial q'}{\partial q} & \frac{\partial q'}{\partial p} \\
\frac{\partial p'}{\partial q} & \frac{\partial p'}{\partial p} \\
\end{array}\right)\left( \begin{array}{rrr}
0 & 1 \\
-1 & 0 \\
\end{array}\right)\left( \begin{array}{rrr}
\frac{\partial q'}{\partial q} & \frac{\partial p'}{\partial q} \\
\frac{\partial q'}{\partial p}& \frac{\partial p'}{\partial p} \\
\end{array}\right)$$

Inserting values then gives:

$$\left( \begin{array}{rrr}

\frac{\partial q'}{\partial q} & \frac{\partial q'}{\partial p} \\

\frac{\partial p'}{\partial q} & \frac{\partial p'}{\partial p} \\

\end{array}\right)\left( \begin{array}{rrr}

0 & 1 \\

-1 & 0 \\

\end{array}\right)\left( \begin{array}{rrr}

\frac{\partial q'}{\partial q} & \frac{\partial p'}{\partial q} \\

\frac{\partial q'}{\partial p}& \frac{\partial p'}{\partial p} \\

\end{array}\right)=\left( \begin{array}{rrr}

\sqrt{2}p^{\alpha}\cos(q) & -\sqrt{2}p^{\beta}\sin(q) \\

\sqrt{2} \alpha p^{\alpha-1}\sin(q) & \sqrt{2} \beta p^{\beta-1}\cos(q) \

\end{array}\right)
\left( \begin{array}{rrr}

\sqrt{2} \alpha p^{\alpha-1} \sin(q) & -\sqrt{2}p^{\beta} \sin(q) \\

-\sqrt{2}p^{\alpha} \cos(q) & \sqrt{2} \beta p^{\beta-1} \cos(q) \

\end{array}\right)$$$$=\left( \begin{array}{rrr}

2 \alpha p^{2\alpha-1} \cos(q) \sin(q) -2 \alpha p^{2\alpha -1} \cos(q) \sin(q) \ & 2 \beta p^{\alpha} \cos^2(1)p^{\beta -1} +2 \alpha p^{\beta} \sin^2(q)p^{\alpha -1} \\

2 \alpha p^{\beta} \cos^2(1)p^{\alpha -1} -2 \beta p^{\alpha} \sin^2(q)p^{\beta -1} & -2 \beta p^{2\beta-1} \cos(q) \sin(q) +2 \beta p^{2\beta -1} \cos(q) \sin(q) \

\end{array}\right)$$

Now the following holds:

$$a_{11}: 2 \alpha p^{2\alpha-1} \cos(q) \sin(q) -2 \alpha p^{2\alpha -1} \cos(q) \sin(q) =0$$
$$a_{12}: 2 \beta p^{\alpha} \cos^2(1)p^{\beta -1} +2 \alpha p^{\beta} \sin^2(q)p^{\alpha -1}=1$$
$$a_{21}: 2 \alpha p^{\beta} \cos^2(1)p^{\alpha -1} -2 \beta p^{\alpha} \sin^2(q)p^{\beta -1} =-1$$
$$a_{22}: -2 \beta p^{2\beta-1} \cos(q) \sin(q) +2 \beta p^{2\beta -1} \cos(q) \sin(q)=0$$

So now I just have to solve the 4 equations for ##\alpha## and ##\beta##.
 
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  • #5
Are all these equations independent from each other? Are all of them give any restrictions on ##\alpha## and ##\beta##? Finally note that the right-hand sides of the equations are independent of ##q## and ##p##. What does this imply for ##\alpha## and ##\beta##?
 
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  • #6
Lambda96 said:
So now I just have to solve the 4 equations for ##\alpha## and ##\beta##.
Well, only 2 equations really. On inspection, the values of ##a_{11}## and ##a_{22}## are necessarily both zero.

I haven't checked you working but you appear to have some typo's: ##a_{12}## and ##a_{21}## contain ##cos^2(1)## which doesn't look right.

More generally, if you think it would be useful, the video below provides a wonderful (IMHO) treatment of material relevant to your Post #1 question.

However, the video is 1h 18m long (excluding the YouTube ad’s). If you watch it, be prepared to regularly hit the ‘Skip ad’ button! And don’t be put-off by the quite-basic first 5 minutes.
 
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  • #7
Thanks Steve4Physics and vanhees71 for your help👍, thanks also for the videos 👍

@Steve4Physics I have now checked my calculation several times regarding ##a_{12}## and ##a_{21}## but I always get ##cos^2## and ##sin^2## or do you mean that the 1 in ##cos^2## and ##sin^2## is wrong? If so, that's right it should actually be ##cos^2(q)## and ##sin^2(q)##, I don't know how the 1 got in their :smile:
 
  • #8
Lambda96 said:
If so, that's right it should actually be ##cos^2(q)## and ##sin^2(q)##, I don't know how the 1 got in their :smile:
Yes. You are given ##q’ = \sqrt 2 p^α \sin (q) ## and ##p’ = \sqrt 2 p^ β \cos (q) ##.

When you differentiate, you can only get expressions containing ##\cos (q)## and ##\sin (q) ##. Terms containing ##\sin (1) ## and/or ##\cos (1) ## can't arise. Looks like some sort of transcription error that turned ‘##q##’ into ‘##1##’.

If that’s the only mistake, I’d guess that you should have
##a_{12}: 2 \beta p^{\alpha} \cos^2(q)p^{\beta -1} +2 \alpha p^{\beta} \sin^2(q)p^{\alpha -1}=1##
##a_{21}: 2 \alpha p^{\beta} \cos^2(q)p^{\alpha -1} -2 \beta p^{\alpha} \sin^2(q)p^{\beta -1} =-1##

These can be simplified/combined in different ways. But I'll admit I don't see how to find ##\alpha## and ##\beta## from them.
 
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  • #9
Some hints: First of all there's only one non-trivial Poisson bracket here, which is ##a_{12}=-a_{21}## the other are ##0##, because ##\{Q,Q\}=\{P,P\}## for all phase-space functions ##Q(q,p)## and ##P(q,p)##.

Finally you need two equations to solve for ##\alpha## and ##\beta##. After simplifying your equations for ##a_{12}## a bit, you'll see that one equation follows from the demand that the right-hand side must not depend on ##p##. After solving for this, ##a_{12}=1## gives the 2nd equation.
 
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  • #10
Steve4Physics said:
But I'll admit I don't see how to find ##\alpha## and ##\beta## from them.
After eliminating the dependence on ##p##, rather than trying to solve for ##\alpha## or ##\beta## individually, think about what pair of values will work to get rid of the dependence on ##q##.
 
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  • #11
Lambda96 said:
$$\left( \begin{array}{rrr}
\frac{\partial q'}{\partial q} & \frac{\partial q'}{\partial p} \\
\frac{\partial p'}{\partial q} & \frac{\partial p'}{\partial p} \\
\end{array}\right)\left( \begin{array}{rrr}
0 & 1 \\
-1 & 0 \\
\end{array}\right)\left( \begin{array}{rrr}
\frac{\partial q'}{\partial q} & \frac{\partial p'}{\partial q} \\
\frac{\partial q'}{\partial p}& \frac{\partial p'}{\partial p} \\
\end{array}\right)$$
Instead of substituting the explicit expressions for ##q'## and ##p'## in terms of ##q## and ##p## at this point, I think it's much nicer to carry out the matrix multiplication above and show that it reduces to what your professor indicated:
$$\left( \begin{array}{rrr}
\{q', q'\} &\{q', p'\} \\
\{p', q'\} & \{p', p'\} \\
\end{array}\right)$$ It should be easy to see that two of the entries here are zero and the other two differ only in sign. Since the matrix is required to equal the matrix
$$\left( \begin{array}{rrr}
0 & 1 \\
-1 & 0 \\
\end{array}\right)$$ you end up with only one independent equation to be satisfied. This equation is your ##a_{12}## equation given below

Lambda96 said:
$$a_{12}: 2 \beta p^{\alpha} \cos^2(1)p^{\beta -1} +2 \alpha p^{\beta} \sin^2(q)p^{\alpha -1}=1$$
Your ##a_{21}## equation has a sign error. The ##a_{21}## equation should be equivalent to the ##a_{12}## equation.

So, you just need to find values of ##\alpha## and ##\beta## such that the ##a_{12}## equation is satisfied.
 
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  • #12
vela said:
After eliminating the dependence on ##p##, rather than trying to solve for ##\alpha## or ##\beta## individually, think about what pair of values will work to get rid of the dependence on ##q##.
Thanks. I realised how to do it after reading @vanhees71's Post #9.
 
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  • #13
Thank you Steve4Physics, vanhees71, vela and TSny for your help👍 👍 👍👍

I have now started to rewrite ##a_{12}## as follows:

$$2p^{\alpha} \cos^2(q) \beta p^{\beta -1}+2 \alpha p^{\alpha -1} \sin^2(q) p^{\beta}$$

$$2 \cos^2(q) \beta p^{\beta +\alpha -1}+ 2 \alpha p^{\alpha +\beta -1} \sin^2(q)$$

After all, I want p to vanish or rather become 1, which is the case when ##\alpha + \beta -1=0## simultaneously, I also want the 2 to vanish and become a 1. I would achieve all this if ##\alpha= \frac{1}{2}## and ##\beta=\frac{1}{2}##.

Regarding task b, I still have a question, if I have now understood correctly, the system has two degrees of freedom, must the Poisson bracket then be calculated as follows?

$$\Bigl\{ q'_j, p'_k \Bigr\}=\sum\limits_{i=1}^{2}\frac{\partial q'_j}{\partial q_i}\frac{\partial p'_k}{\partial p_i}- \frac{\partial q'_j}{\partial p_i}\frac{\partial p'_k}{\partial q_i}=\frac{\partial q'_j}{\partial q_1}\frac{\partial p'_k}{\partial p_1}- \frac{\partial q'_j}{\partial p_1}\frac{\partial p'_k}{\partial q_1}+\frac{\partial q'_j}{\partial q_2}\frac{\partial p'_k}{\partial p_2}-\frac{\partial q'_j}{\partial p_2}\frac{\partial p'_k}{\partial q_2}$$

I have tried to determine ##\alpha, \beta, \delta, \gamma## by calculating the following

$$\Bigl\{ q'_1,p'_1 \Bigr\}=1$$
$$\Bigl\{ q'_2,p'_2 \Bigr\}=1$$
$$\Bigl\{ q'_1,q'_2 \Bigr\}=0$$
$$\Bigl\{ p'_1,p'_2 \Bigr\}=0$$

With this I found out that ##\gamma=1##, ##\alpha=0##, ##\beta=0## and ##\delta=0## but since I got so many zeros I don't know if I defined the Poisson brackets correctly?
 
  • #14
Lambda96 said:
Regarding task b, I still have a question, if I have now understood correctly, the system has two degrees of freedom, must the Poisson bracket then be calculated as follows?

$$\Bigl\{ q'_j, p'_k \Bigr\}=\sum\limits_{i=1}^{2}\frac{\partial q'_j}{\partial q_i}\frac{\partial p'_k}{\partial p_i}- \frac{\partial q'_j}{\partial p_i}\frac{\partial p'_k}{\partial q_i}=\frac{\partial q'_j}{\partial q_1}\frac{\partial p'_k}{\partial p_1}- \frac{\partial q'_j}{\partial p_1}\frac{\partial p'_k}{\partial q_1}+\frac{\partial q'_j}{\partial q_2}\frac{\partial p'_k}{\partial p_2}-\frac{\partial q'_j}{\partial p_2}\frac{\partial p'_k}{\partial q_2}$$
That looks ok to me.

Sorry, but I don't have the time to check your values for ##\alpha, \beta, \gamma## and ##\delta##.
 
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  • #15
Lambda96 said:
Thank you Steve4Physics, vanhees71, vela and TSny for your help👍 👍 👍👍

I have now started to rewrite ##a_{12}## as follows:

$$2p^{\alpha} \cos^2(q) \beta p^{\beta -1}+2 \alpha p^{\alpha -1} \sin^2(q) p^{\beta}$$

$$2 \cos^2(q) \beta p^{\beta +\alpha -1}+ 2 \alpha p^{\alpha +\beta -1} \sin^2(q)$$

After all, I want p to vanish or rather become 1, which is the case when ##\alpha + \beta -1=0## simultaneously, I also want the 2 to vanish and become a 1. I would achieve all this if ##\alpha= \frac{1}{2}## and ##\beta=\frac{1}{2}##.
Looks good to me.

Lambda96 said:
Regarding task b, I still have a question, if I have now understood correctly, the system has two degrees of freedom, must the Poisson bracket then be calculated as follows?

$$\Bigl\{ q'_j, p'_k \Bigr\}=\sum\limits_{i=1}^{2}\frac{\partial q'_j}{\partial q_i}\frac{\partial p'_k}{\partial p_i}- \frac{\partial q'_j}{\partial p_i}\frac{\partial p'_k}{\partial q_i}=\frac{\partial q'_j}{\partial q_1}\frac{\partial p'_k}{\partial p_1}- \frac{\partial q'_j}{\partial p_1}\frac{\partial p'_k}{\partial q_1}+\frac{\partial q'_j}{\partial q_2}\frac{\partial p'_k}{\partial p_2}-\frac{\partial q'_j}{\partial p_2}\frac{\partial p'_k}{\partial q_2}$$

I have tried to determine ##\alpha, \beta, \delta, \gamma## by calculating the following

$$\Bigl\{ q'_1,p'_1 \Bigr\}=1$$
$$\Bigl\{ q'_2,p'_2 \Bigr\}=1$$
$$\Bigl\{ q'_1,q'_2 \Bigr\}=0$$
$$\Bigl\{ p'_1,p'_2 \Bigr\}=0$$
There are some more brackets that you need to evaluate, such as ##\Bigl\{ q'_1,p'_2 \Bigr\}##.

Lambda96 said:
With this I found out that ##\gamma=1##, ##\alpha=0##, ##\beta=0## and ##\delta=0## but since I got so many zeros I don't know if I defined the Poisson brackets correctly?
Check your result for ##\gamma##.
 
Last edited:
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  • #16
Steve4Physics said:
That looks ok to me.

Sorry, but I don't have the time to check your values for ##\alpha, \beta, \gamma## and ##\delta##.
No problem, you have already helped me a lot with task 1 👍

Thank you TSny for your help 👍👍👍

I have checked my calculation for ##\gamma## again and unfortunately overlooked the minus sign, so the following applies ##\gamma=-1##

unfortunately I did not have so much time yesterday to add my complete calculation, my calculations are as follows

$$\Bigl\{ q'_1,p'_1 \Bigr\}=0 * \beta - \alpha * 0 + 0*0-1* \gamma=-\gamma \rightarrow \gamma=-1$$
$$\Bigl\{ q'_2,p'_2 \Bigr\}=-\frac{p_1}{2q_1^2} \delta + 1 + 0*0-0*0=-\frac{p_1}{2q_1^2} \delta + 1 \rightarrow \delta=0$$
$$\Bigl\{ q'_1,q'_2 \Bigr\}=-0 * \frac{p_1}{2q_1^2} + \alpha \frac{p_1}{2q_1^2} + 0*0-1*0=\alpha \frac{p_1}{2q_1^2} \rightarrow \alpha=0$$
$$\Bigl\{ p'_1,p'_2 \Bigr\}=0 * \delta + 2\beta q_1 + \gamma*0-0*0=2\beta q_1 \rightarrow \beta=0$$
$$\Bigl\{ q'_1,p'_2 \Bigr\}=0 * \delta +2 \alpha q_1 + 0*0-1*0=2 \alpha q_1 \rightarrow \alpha=0$$
 
  • #17
Lambda96 said:
No problem, you have already helped me a lot with task 1 👍

Thank you TSny for your help 👍👍👍

I have checked my calculation for ##\gamma## again and unfortunately overlooked the minus sign, so the following applies ##\gamma=-1##

unfortunately I did not have so much time yesterday to add my complete calculation, my calculations are as follows

$$\Bigl\{ q'_1,p'_1 \Bigr\}=0 * \beta - \alpha * 0 + 0*0-1* \gamma=-\gamma \rightarrow \gamma=-1$$
$$\Bigl\{ q'_2,p'_2 \Bigr\}=-\frac{p_1}{2q_1^2} \delta + 1 + 0*0-0*0=-\frac{p_1}{2q_1^2} \delta + 1 \rightarrow \delta=0$$
$$\Bigl\{ q'_1,q'_2 \Bigr\}=-0 * \frac{p_1}{2q_1^2} + \alpha \frac{p_1}{2q_1^2} + 0*0-1*0=\alpha \frac{p_1}{2q_1^2} \rightarrow \alpha=0$$
$$\Bigl\{ p'_1,p'_2 \Bigr\}=0 * \delta + 2\beta q_1 + \gamma*0-0*0=2\beta q_1 \rightarrow \beta=0$$
$$\Bigl\{ q'_1,p'_2 \Bigr\}=0 * \delta +2 \alpha q_1 + 0*0-1*0=2 \alpha q_1 \rightarrow \alpha=0$$
That all looks good.

What about ##\Bigl\{ q'_2,p'_1 \Bigr\}##?
 
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  • #18
Thank you TSny, for your help and for looking over my calculations 👍👍👍

For ##\Bigl\{ q_2' , p_1' \Bigr\}## I have obtained the following:

$$\Bigl\{ q_2' , p_1' \Bigr\}=-\frac{p_1 \beta}{2q_1^2}-\frac{1}{2q_1}*0+0*0-0* \gamma=-\frac{p_1 \beta}{2q_1^2}=0 \rightarrow \beta=0$$
 
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  • #19
Lambda96 said:
Thank you TSny, for your help and for looking over my calculations 👍👍👍

For ##\Bigl\{ q_2' , p_1' \Bigr\}## I have obtained the following:

$$\Bigl\{ q_2' , p_1' \Bigr\}=-\frac{p_1 \beta}{2q_1^2}-\frac{1}{2q_1}*0+0*0-0* \gamma=-\frac{p_1 \beta}{2q_1^2}=0 \rightarrow \beta=0$$
Yes. Nice work.
 
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  • #20
Thanks for your help TSny and for looking over my calculation 👍👍👍
 
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1. What are canonical transformations?

Canonical transformations are mathematical transformations that preserve the canonical form of Hamilton's equations in classical mechanics. They are used to transform the coordinates and momenta of a system while maintaining the same physical properties and equations of motion.

2. Why are canonical transformations important?

Canonical transformations are important because they allow us to simplify the equations of motion for a system by transforming to a new set of coordinates and momenta that may be more convenient or intuitive. They also help us to identify conserved quantities, such as energy and angular momentum, which are crucial in understanding the behavior of a system.

3. How are canonical transformations different from other transformations?

Canonical transformations are different from other transformations because they preserve the symplectic structure of a system, which is a fundamental property in classical mechanics. This means that the transformation does not change the Poisson brackets between the coordinates and momenta, and therefore does not alter the equations of motion.

4. What are the two types of canonical transformations?

The two types of canonical transformations are point transformations and generating function transformations. Point transformations involve a direct transformation of the coordinates and momenta, while generating function transformations use a generating function to determine the new coordinates and momenta.

5. How are canonical transformations related to the principle of least action?

Canonical transformations are related to the principle of least action because they preserve the equations of motion, which are derived from the principle of least action. This means that the transformed coordinates and momenta still satisfy the principle of least action, and the action remains unchanged under a canonical transformation.

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