How can I determine the values of α and β for a canonical transformation?

[Lambda96]

Homework Statement

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Relevant Equations

none

Hi,

unfortunately, I have problems with the following task

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.

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

 

[Steve4Physics]

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 $JIJ^T=I$.

 

[vanhees71]

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
$$\{x',x' \}=\{p',p'\}=0, \quad \{x',p'\}=1.$$

 

[Lambda96]

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 $\beta$.

 

[vanhees71]

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 $\beta$?

 

[Steve4Physics]

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.

 

[Lambda96]

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

 

[Steve4Physics]

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

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.

 

[vanhees71]

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.

 

[vela]

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 $q$.

 

[TSny]

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

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.

 

[Steve4Physics]

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.

 

[Lambda96]

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?

 

[Steve4Physics]

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 $\delta$.

 

[TSny]

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 $\gamma$.

 

[Lambda96]

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

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

 

[TSny]

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 $\Bigl\{ q'_2,p'_1 \Bigr\}$?

 

[Lambda96]

Thank you TSny, for your help and for looking over my calculations

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

$$\Bigl\{ q_2' , p_1' \Bigr\}=-\frac{p_1 \beta}{2q_1^2}-\frac{1}{2q_1}*0+0*0-0* \gamma=-\frac{p_1 \beta}{2q_1^2}=0 \rightarrow \beta=0$$

 

[TSny]

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.

 

[Lambda96]

Thanks for your help TSny and for looking over my calculation