Canonical transformations

  • Thread starter Lambda96
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  • #1
Lambda96
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Homework Statement:
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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
 

Answers and Replies

  • #2
Steve4Physics
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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
vanhees71
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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
Lambda96
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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
vanhees71
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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##?
 
  • #6
Steve4Physics
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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.
 
  • #7
Lambda96
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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
Steve4Physics
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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.
 
  • #9
vanhees71
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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
vela
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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
TSny
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$$\left( \begin{array}{rrr}
\frac{\partial q'}{\partial q} & \frac{\partial q'}{\partial p} \\
\frac{\partial p'}{\partial q} & \frac{\partial p'}{\partial p} \\
\end{array}\right)\left( \begin{array}{rrr}
0 & 1 \\
-1 & 0 \\
\end{array}\right)\left( \begin{array}{rrr}
\frac{\partial q'}{\partial q} & \frac{\partial p'}{\partial q} \\
\frac{\partial q'}{\partial p}& \frac{\partial p'}{\partial p} \\
\end{array}\right)$$
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

$$a_{12}: 2 \beta p^{\alpha} \cos^2(1)p^{\beta -1} +2 \alpha p^{\beta} \sin^2(q)p^{\alpha -1}=1$$
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
Steve4Physics
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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.
 
  • #13
Lambda96
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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
Steve4Physics
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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##.
 
  • #15
TSny
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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.

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\}##.

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:
  • #16
Lambda96
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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
TSny
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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\}##?
 
  • #18
Lambda96
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Thank you TSny, for your help and for looking over my calculations 👍👍👍

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

$$\Bigl\{ q_2' , p_1' \Bigr\}=-\frac{p_1 \beta}{2q_1^2}-\frac{1}{2q_1}*0+0*0-0* \gamma=-\frac{p_1 \beta}{2q_1^2}=0 \rightarrow \beta=0$$
 
  • #19
TSny
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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.
 
  • #20
Lambda96
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Thanks for your help TSny and for looking over my calculation 👍👍👍
 

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