# I Weinberg Lectures in QM (2013 Ed.), Equation 7.10.15

1. Mar 20, 2016

### jouvelot

Hello everyone,
I don't get how the second-order derivative $\partial^2 S/\partial x_i \partial x_j$ of the phase S arrives here. If one performs a power series expansion of the Hamiltonian around $\nabla S$, then I do get where the first term $A$ comes from, but then adding higher-order derivatives doesn't seem to introduce this second-order derivative of the phase S (even taking the somewhat
fuzzy footnote into account).
Does anyone who has read this book have an idea?
Pierre

2. Mar 20, 2016

### jouvelot

To put my question in perspective, I put below an excerpt of the corresponding page (I hope small quotes are ok on the forum; if not, please feel fee to let me know) in Weinberg's book.

Schrodinger's equation 7.10.1 is $H(p,x)\psi(x) = E \psi(x)$. The idea is to assume a solution $\psi(x)= N(x)e^{-iS(x)\hbar}$ (Equation 7.10.2), in which case, after identifying as usual $p$ with $-i\hbar\nabla$, Schrodinger's equation is equivalent to Equation 7.10.13 below. Then, my question amounts to understanding the right part of Equation 7.10.15, and thus 7.10.14.

Thanks.

Last edited: Mar 20, 2016
3. Mar 22, 2016

### muppet

I'm not entirely sure I follow (7.10.13) (why p is identified with the expression containing grad S), but is it not simply the case that by acting on the product $\psi = N e^{ i S}$ with two derivatives it's inevitable that you'll pull down a term that looks like the second derivative of S, just because of the product rule and the derivatives of exponentials?

4. Mar 23, 2016

### vanhees71

One way to derive classical mechanics from quantum theory is to use singular perturbation theory, i.e., a formal expansion in powers of $\hbar$, starting with the term $\propto 1/\hbar$ around $\hbar=0$. In physics it's also known as the Wentzel-Kramers-Brillouin approximation. It's exactly the same technique how you derive ray optics from wave optics, i.e., the Maxwell equations. In leading order $\mathcal{O}(1/\hbar)$ you get the Hamilton-Jacoby partial differential equation for the "eikonal" $S$, which is nothing else than the classical action, and that's equivalent to good old Newtonian mechanics.

5. Mar 23, 2016

### jouvelot

Hi,

My understanding of Equation 7.10.13 is that it is the transcription of Schrodinger’s equation when applied to $\psi$ defined in terms of $N(x)$ and $S(x)$.

The application of the operator $p = -i\hbar\nabla$ to such $\psi$ yields $-i\hbar\nabla N(x) e^{iS(x)/\hbar}+N(x) \nabla S(x) e^{iS(x)/\hbar}$, which is equivalent to $((-i\hbar\nabla + \nabla S(x) )N(x)) e^{iS(x)/\hbar}$. So since the phase exponential part of $\psi$ is kept unchanged, it can be factored out of the whole Schrodinger’s equation on $\psi$, giving 7.10.13, which only applies to the $N(x)$ part of the wave function.

Now, if one performs a first-order power series expansion of the Hamiltonian at $\nabla S(x)$, one gets the first parts of Equations 7.10.14 and 7.10.15, but I don’t see how to obtain the second parts even with higher-order expansions.

Thanks.

Pierre

Last edited: Mar 23, 2016
6. Mar 23, 2016

### jouvelot

Hi vanhees71,

Thanks for your comments, and I do remember reading about doing expansions in powers of $\hbar$. But here, Weinberg speaks of orders of gradients, not of $\hbar$ (even though this is the same here, when dealing with the gradient of $N(x)$). Given the book self-containedness up to now, I assume I should be able to derive this formula and see where the second-order derivatives of $S$ come from. But I don't see how.

7. Jun 18, 2016

### jouvelot

Ok, going back to this after a small hiatus, I think I got it, and it's pretty simple in fact. Basically, from (7.10.13), one does a first-order expansion of H around $\nabla S(x)$, yielding
$$H(\nabla S)N + (-i\hbar\sum_i\nabla_i)[(\partial H/\partial p_i)_{p=\nabla S} N] = E N.$$ The zero-th order terms $H(\nabla S)N$ and $E N$ cancel, as per (7.10.3). The remaining $\nabla$ yields then
$$\sum_i(\partial H/\partial p_i)_{p=\nabla S} (\nabla_i N) + N\sum_{ij} (\partial p_j / \partial x_i)_{p=\nabla S} \partial/\partial p_j (\partial H/\partial p_i)_{p=\nabla S} = 0,$$
i.e., $$A.\nabla(N)+N {1\over 2}\sum_{ij} (\partial^2 S / \partial x_j\partial x_i) (\partial^2 H/\partial p_j\partial p_i)_{p=\nabla S} = A.\nabla(N)+NB = 0,$$
as in (7.10.14).