-1
$\begingroup$

I'm fairly new to QM and was reading about the Taylor series expansion of $V'(x)$ from Chapter 6 (The Classical Limit) of Shankar's Principles of Quantum Mechanics, pages 182-183.

The terms in the expansion have a multiple in powers of $(x-x_0)$. Where $x_0 = \langle x\rangle$.

I can see how $\langle x - x_0\rangle$ is zero. And also why $\langle(x - x_0)^2\rangle$ is greater than zero.

I think $\langle(x - x_0)^3\rangle$ is also zero and so are all odd powered terms. Can you please confirm if this is correct? I'm confused since the book doesn't state this and seems to suggest that all succeeding terms beyond first power make a contribution to this Taylor Series.

$\endgroup$
4
  • $\begingroup$ Shankar means Ramamurti Shankar? $\endgroup$ Sep 30, 2017 at 21:47
  • 1
    $\begingroup$ Many of us may not know that book here. Could you add more context? $\endgroup$
    – FGSUZ
    Sep 30, 2017 at 22:33
  • 1
    $\begingroup$ What if $x$ takes the values $1,2,6$ all with probability $1/3$? $\endgroup$
    – WillO
    Sep 30, 2017 at 22:39
  • $\begingroup$ @WrichikBasu Yes, Ramamurti Shankar Quantum Mechanics book. $\endgroup$
    – frank_010
    Oct 1, 2017 at 9:59

2 Answers 2

2
$\begingroup$

So this is a question somewhat more general than quantum mechanics and it applies to all random distributions. Just to review, we write that the probability for a random variable $X$ to be between two values $x$ and $x+dx$ for small $dx$ is given by its probability density function $f(x)$ as $$\operatorname{Pr}(x < X < x+dx) = f(x)~dx.$$ In your particular case you may recognize $f(x)$ as $\Psi^*(x)\Psi(x)$ but this is not terribly important for your present issue.

We can then define the moments of the distribution according to $$\langle X^n\rangle = \int_{\mathcal S}dx~f(x)~x^n,$$where $\mathcal S$ is the set of possible values that the random variable $X$ can take on, generally called its "support." This works for so-called continuous random variables but we can extend it to discrete random variables (and mixes thereof) by introducing the so-called Dirac $\delta$-function, which is not really a function (it does not have a graph) but can still have a meaning inside one of these integrals, where it is defined by its property that $\int_a^b dx~\delta(x - c) = \{1\text{ if } a < c < b\text{ else }0\}.$ It can be approximated as a smooth extremely-thin extremely-tall Gaussian such that the area under the curve is 1, but there are other approximations and this is not unique by any means.

Worked out in an example

Here is an example of a nontrivial distribution: consider $\exp(-x/\lambda)$ on the support $\mathcal S = [0, \infty).$ Integrating over the whole space gives $\lambda$ so we find that the normalized probability distribution is, $$f(x) = \frac1\lambda~e^{-x/\lambda}.$$ This is called the exponential distribution. If $X$ is exponentially distributed then its moments are: $$\langle X^n\rangle = \int_0^\infty dx~\frac1\lambda~e^{-x/\lambda}~x^n.$$Substituting $u=x/\lambda$ we can find a well-known integral for the factorial function, $$\langle X^n\rangle = \lambda^n~n!.$$Now suppose that we want to consider the random variable $Y=X-\lambda.$ This has mean 0, now. What do all of the rest of the moments do? Well we can just expand out this exponential as a binomial series and use the linearity of the expectation value over that sum to say,$$\langle Y^n\rangle = \langle (X-\lambda)^n \rangle = \sum_{k=0}^n \frac{n!}{k!~(n-k)!} \langle X^{n-k} \rangle ~(-1)^k~\lambda^k.$$We can then use our previous result to get $$\langle Y^n\rangle = \lambda^n~n!~\sum_{k=0}^n \frac{(-1)^k}{k!} = \lambda^n ~D_n,$$ where $D_n$ are the derangement numbers $\{1,0,1,2,9,\dots\}$ defined by $$D_0 = 1,\\ D_1 = 0,\\ D_{n+1} = n~(D_n + D_{n-1}).$$ So these are clearly not all $0$ nor are the odd ones zero nor anything like that; the only zero term is the $n=1$ term and that is by design.

When are the odd central moments zero?

We have seen that it is not a universal truth that odd moments $\langle X^n\rangle$ are zero, nor that odd central moments $\langle (X-\bar X)^n\rangle$ are zero (where $\bar X = \langle X \rangle$).

In fact if $f(x)$ has a Fourier transform $f[k]$ then we can speak of its characteristic function $$\phi_f[k] = \ln~f[k] = \ln\left[\int_{-\infty}^{\infty} dx~e^{-ikx}~f(x)\right].$$On the theory-side this perspective is extremely helpful because if $Z = X + Y$ and $X$ and $Y$ are independent random variables and $X$ is distributed with $f(x)$ and $Y$ is distributed with $g(y)$ and $Z$ is distributed with $h(z)$, the expression for $h$ is very complicated because it is a convolution of $f$ and $g$, but the expression for $\phi_h$ is very simple, it is $\phi_h[k] = \phi_f[k] + \phi_g[k],$ because convolutions become multiplications in Fourier space and the logarithm turns this into a summation. Similarly when we shift to $Y = X - \langle X \rangle$ this in Fourier space looks like multiplying by $e^{ik\langle X \rangle}$ and therefore this just adds $ik\langle X\rangle$ to the characteristic function.

Now the claim is that $\phi[k]$ has a Taylor expansion related to the moments you've calculated. For example $\phi'[k] = f'[k]/f[k]$ and at $k=0$ we find $\phi'[0] = -i\langle X\rangle.$ Similarly $$\phi''[0] = (-i)^2\frac{f[0]~f''[0] - f'[0]~f'[0]}{f[0]^2} = (-i)^2\big(\langle X^2 \rangle - \langle X \rangle^2\big).$$These terms (that is, $\phi^{(n)}[0]/(-i)^n$) are called the cumulants of the random variable and they always lead with $\langle X^n \rangle$. However a straightforward dimensional-analysis argument shows that every term which makes up an odd cumulant must have at least one term of an odd moment, so the condition that all of the odd moments vanish necessarily implies that all of the odd cumulants vanish, which implies that $\phi[k]$ is purely real-valued and even, $\phi[-k] = \phi[k].$ In fact this directly implies that $f[k] = e^{\phi[k]}$ is real-and-even and, with some argumentation we can then find that $f(x)$ must be real-and-even, too.

So for these central-moments $\langle (X - \langle X\rangle)^n\rangle$ to vanish for odd $n,$ generally (a) they must exist in the first place, and (b) the probability distribution should be symmetric about its mean.

$\endgroup$
2
$\begingroup$

The first term is zero because \begin{equation} \langle x - x_0 \rangle = \langle x \rangle - x_0 = x_0 - x_0 = 0 \end{equation}

But for terms of other orders

\begin{equation} \langle( x - x_0)^n \rangle \neq \langle x^n\rangle - x_0^n, \end{equation} so there is no reason why they should be zero.

For third order in particular \begin{equation} \langle( x - x_0)^3 \rangle =\langle x^3\rangle - 3\langle x^2 \rangle x_0 + 3 x_0^2 \langle x\rangle + x_0^3 \neq 0. \end{equation}

So the book is correct, every term above first order should contribute.

$\endgroup$
2
  • $\begingroup$ Thanks for this explanation. Also, can you comment when is it permissible to express expectation of products as product of expectations. E.g. You have expressed $<(3x^2x_0>$ as 3$<x^2>x_0$. But clearly this expansion is not possible with $<(x - x_0)^2>$ = $<(x - x_0)(x - x_0)>$ $\endgroup$
    – frank_010
    Oct 1, 2017 at 10:08
  • $\begingroup$ @andreas_jacob the rules are that $\langle X Y\rangle = \langle X\rangle\langle Y\rangle$ whenever $X$ and $Y$ are independent random variables, and, as sort of a special case of this, $\langle k X\rangle = k \langle X \rangle$ whenever $k$ is a constant (constants are random variables which are independent of everything else because they take on their value with probability 1). Note that $x_0 = \langle X \rangle$ is a constant. It may come from a random variable but it is a property of the whole random distribution, not of the value you observe in a particular experiment in passing. $\endgroup$
    – CR Drost
    Oct 2, 2017 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.