# Approximations in deriving infinitesimal operator commutation relationship

I'm having trouble understanding some derivation from Sakurai's QM, chapter 1. To derive $$[x,J(dx)]=dx$$ he claimed that $$dx|x+dx\rangle \simeq dx|x\rangle$$ because the approximation is of second order in dx. I don't understand the meaning of this justification...Also, aren't $$|x+dx\rangle$$ and $$|x\rangle$$ orthogonal kets? So how come we can even do this kind of approximation?

• Near duplicate, and here, and also. Nov 3, 2020 at 20:36
• More duplicates. Nov 3, 2020 at 20:40

I'm not sure I see what the problem is, it's explained quite clearly in the section. The equation you have (Equation 1.6.24 in the Revised Edition) basically says:

$$\Big[\mathbf{x}, \mathcal{T}(\text{d}\mathbf{x}^{'})\Big]| \mathbf{x}'\rangle = \text{d}\mathbf{x}^{'} | \mathbf{x}^{'} + \text{d}\mathbf{x}^{'}\rangle,$$

and since you know that $$| \mathbf{x}^{'} + \text{d}\mathbf{x}^{'}\rangle = \mathcal{T}(\text{d}\mathbf{x}^{'})| \mathbf{x}^{'}\rangle.$$

Now, Sakurai chooses the form of the infinitesimal translation operator to be (in Equation 1.6.20) $$\mathcal{T}(\text{d}\mathbf{x}^{'}) = \mathbb{I} - i \mathbf{K}\cdot \text{d}\mathbf{x}^{'},$$

and so $$\text{d}\mathbf{x}^{'}\mathbb{I}| \mathbf{x}^{'} + \text{d}\mathbf{x}^{'}\rangle = \text{d}\mathbf{x}^{'}\mathcal{T}(\text{d}\mathbf{x}^{'})| \mathbf{x}^{'}\rangle = \text{d}\mathbf{x}^{'}\mathbb{I} \left(\mathbb{I} - i \mathbf{K}\cdot \text{d}\mathbf{x}^{'}\right)| \mathbf{x}^{'}\rangle = \text{d}\mathbf{x}^{'}|\mathbf{x}^{'}\rangle + \mathcal{O}(\text{d}\mathbf{x}^{'2})|\mathbf{x}^{'}\rangle,$$

as argued. Now, as to why one can write the Translation operator in that fashion, the chapter in Symmetries will probably help you understand that better. However, it's not too unreasonable. Clearly, as the infinitesimal translation gets smaller and smaller, the translation operator must approach the identity operation (translating something by nothing is equivalent to acting on it by the identity operator).

As to orthogonality, as the other answer points out, the states $$|x\rangle$$ and $$|x+\text{d} x\rangle$$ are indeed orthogonal. It can be shown that $$\langle x | x+\text{d} x\rangle = \delta(\text{d}x),$$ which is zero for all non-zero $$\text{d}x$$.

The claim being made is that the two states are equal only up to the leading order in $$\text{d}x$$.

EDIT: In case you are interested in how the above relation may be proved, here's a back-of-the-envelope calculation if you are willing to admit that the translation operator is given by $$\mathcal{T}(\text{d}x) = e^{-i \hat{P} dx},$$ where $$\hat{P}$$ is the momentum operator, and is just $$\hbar \mathbf{K}$$. (I'll be using units where $$\hbar = 1$$, and will probably miss a couple of pesky 2$$\pi$$s.) By definition,

$$\langle x | x + \text{d}x\rangle = \langle x | \mathcal{T}(\text{d}x) | x\rangle = \langle x | e^{- i P \text{d}x} | x\rangle = \langle x | \sum_{n}{\frac{(- i P \text{d}x)^n}{n!}} | x\rangle.$$

The states of definite position can be decomposed in the momentum basis as $$|x\rangle = \int_{-\infty}^\infty e^{i p x} |p\rangle,$$ and so

\begin{align} \langle x | \sum_{n}{\frac{(- i P \text{d}x)^n}{n!}} | x\rangle &= \sum_{n}\frac{(-i \text{d}x)^n}{n!}\int \int \text{d}p' \text{d}p\,\, e^{-i (p'-p) x} \langle p'|\hat{P}^n|p\rangle\\ &= \sum_{n}\frac{(-i \text{d}x)^n}{n!}\int \int \text{d}p' \text{d}p \,\, e^{-i (p'-p) x} p^n \delta(p-p')\\ &= \int \text{d}p\,\,\sum_{n}\frac{(-i \text{d}x)^n}{n!} p^n\\ &= \int \text{d}p\,\, e^{-i p \text{d}x}\\ &= \delta (\text{d}x). \end{align}

aren't $$|x+dx⟩$$ and $$|x⟩$$ orthogonal kets?

Yes of course they are. But they are actually not equal to each other but it's an approximation of order $$dx$$.

You can relate it in the following way: Consider a simple Taylor expansion of a function $$f(x)$$. $$f(x_0+\Delta x)=f(x_0)+\left |\frac{df}{dx} \right|_{x=x_0}\Delta x+\cdots$$ Now we know that n-tuples can be represented as a vector with n components and in the limit, we can represent a function of ,say $$x$$, with inifinite number of basis $$|x\rangle$$. So that the function $$f(x)$$ can be represented as vector $$|f\rangle$$ in $$|x\rangle$$ basis. Refer to Mathematical introduction for Quantum mechanics to any text (or see R.Shankar).

Now writing the taylor series in vector notation, We have $$|f(x_0+\Delta x)\rangle=|f(x_0)\rangle+\left |\frac{d|f\rangle}{dx} \right|_{x=x_0}|\Delta x\rangle+\cdots$$

For $$|x+dx\rangle$$ we can do the same

$$|x+dx\rangle=|x\rangle+\left |\frac{d|x\rangle}{dx} \right|_{x}| dx\rangle+\cdots$$ multiplying with $$dx$$ and taking the term with first order in $$dx$$ we have:

$$dx|x+dx\rangle\approx dx|x\rangle$$

• I suspect the OP is trying to play with $\langle x| x+dx\rangle=\delta(dx)\approx dx ~ \delta'(0)$... Nov 3, 2020 at 19:06
• @Philip I'm not sure what you mean by "actual"... It is a chimera, but in physics we often understand the δ-function as the limit of a Gaussian of width a as that goes to 0 . But a Taylor expansion of a vector w.r.t. a parameter does not work out as this above. Nov 3, 2020 at 20:16
• Apologies, I wasn't very clear. I was just surprised that my back-of-the-envelope calculation for $\langle x|x+\text{d}x\rangle = \delta(\text{d}x)$ was correct. Do you know of a reference that discusses this? It's the first time I'm seeing it. Nov 3, 2020 at 20:19
• Yeah, please I too never saw this relation. Nov 3, 2020 at 20:24
• I would not be surprised if even here on this site, but I do not recall particulars. It is obvious and general; the only thing you are asking is if one could Taylor expand around a different argument, to which the answer is "beware!". Needless to say, the above question has been asked and answered several times in this site, for the very same paragraph of the very same textbook. Nov 3, 2020 at 20:29