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In this Physics.SE post, there is a transformation: $$Q = q,$$ $$P = \sqrt{p} - \sqrt{q}.$$

for Hamiltonian $H = \frac{p^2}{2}$. The post discusses the validity of this transformation as a canonical transformation.

But here I'm want to ask if the transformation is valid, because the dimension of $\sqrt{p}$ is not always same with that of $\sqrt{q}$. If these two dimensions are different, how can they be connected through a minus sign, and what is the dimension of the result?

Community
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Negelis
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1 Answers1

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Suppose that instead, one wrote \begin{align} P = a\sqrt{p} - b\sqrt{q} \end{align} such that $a$ and $b$ had dimensions engineered to make the dimensions match in both terms, then there clearly would be no issue.

Now imagine that the values of $a$ and $b$ in a given system of units (say SI units) are $1$; then we obtain the canonical transformation written in the question. As long as we work in this system of units where $a$ and $b$ both have value $1$, then everything will be fine.

Assuming that this has been done is useful because then we don't have to carry around the parameters $a$ and $b$ that would make computations more cumbersome. This is similar to how particle physicists tend to work in units where the value of the speed of light is unity; $c=1$.

joshphysics
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  • However, the dimensions of $p$ (or $q$) may be different for different problems. Then you have to choose different dimensions of a and b in different problems too. But this is not what canonical transformation is. Canonical transformation should not be problem-dependent. In other words, the transformation should apply for all kinds of problems with Hamiltonian $H = \frac{1}{2}p^2$. – Negelis Jul 24 '14 at 16:32
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    @Negelis Right. For a given system, one simply notes that there exist $a$ and $b$ that make the dimensions work out, and then one chooses a system of units in which their values are unity. In this way, one can interpret that transformation for any given system. Given this interpretation, the canonical transformation isn't problem dependent (it's the same mathematical transformation), but how one interprets the dimensions is problem-dependent. – joshphysics Jul 24 '14 at 18:42