# Sakurai : Approximations in the construction of Quantum theory

1. Apr 21, 2013

### omoplata

From 'Modern Quantum Mechanics, revised edition' by J.J. Sakurai,

In page 44 the translation (spacial displacement) operator $\mathcal{T}(d \boldsymbol{x'})$ is introduced.
$$\mathcal{T}(d \boldsymbol{x'}) \mid \boldsymbol{x'} \rangle = \mid \boldsymbol{x'} + d \boldsymbol{x'} \rangle$$
It is shown that the translation operator must be unitary and that two successive translations must be the same as one single translation to the same effect.
$$\mathcal{T}^{\dagger} ( d \boldsymbol{x'} ) \mathcal{T} ( d \boldsymbol{x'} ) = 1$$
$$\mathcal{T} ( d \boldsymbol{x''} ) \mathcal{T} ( d \boldsymbol{x'} ) = \mathcal{T} ( d \boldsymbol{x'} + d \boldsymbol{x''} )$$
Two other properties are also stated, but they don't pertain to my question.

Then it is demonstrated that if we assume,$$\mathcal{T} ( d \boldsymbol{x'} ) = 1 - i \mathbf{K} \cdot d \boldsymbol{x'}$$, these properties are satisfied. But when showing proof, second order and above terms of $d \boldsymbol{x'}$ are ignored.

For example, in the proof that unitarity is satisfied,
$$\begin{eqnarray} \mathcal{T}^{\dagger} (d \boldsymbol{x'} ) \mathcal{T} ( d \boldsymbol{x'} ) & = & (1 + i \mathbf{K}^{\dagger} \cdot d \boldsymbol{x'} ) ( 1 - i \mathbf{K} \cdot d \boldsymbol{x'} ) \\ & = & 1 - i ( \mathbf{K} - \mathbf{K}^{\dagger} ) \cdot d \boldsymbol{x'} + 0 \left[ ( d \boldsymbol{x'} )^{2} \right] \\ & \approx & 1 \end{eqnarray}$$
Here the second order terms of $d \boldsymbol{x'}$ are ignored. Thus the 0 in the second line.

In the proof for the second property,
$$\begin{eqnarray} \mathcal{T} (d \boldsymbol{x''} ) \mathcal{T} ( d \boldsymbol{x'} ) & = & (1 - i \mathbf{K} \cdot d \boldsymbol{x''} ) ( 1 - i \mathbf{K} \cdot d \boldsymbol{x'} ) \\ & \approx & 1 - i \mathbf{K} \cdot ( d \boldsymbol{x''} + d \boldsymbol{x'} ) \\ & = & \mathcal{T} ( d \boldsymbol{x'} + d \boldsymbol{x''} ) \end{eqnarray}$$
In the second line, the second order terms of $d \boldsymbol{x}$ are ignored.

This is only the beginning. In numerous times later in the book, when constructing infinitesimal operators, second and higher order terms are ignored after a Taylor expansion.

Since such methods are used in the construction of Quantum theory, does this mean that the whole of Quantum Mechanics is an approximation?

2. Apr 21, 2013

### fzero

No, it just means that the proofs given above are not complete. In order to complete the proofs we could instead consider finite transformations instead of the infinitesimal transformations. It turns out to be much easier to consider the infinitesimal transformations, which at least illustrate how things will work for the 2nd and higher-order terms.

For example, at 2nd order, we must take

$$\mathcal{T} ( d \boldsymbol{x'} ) = 1 - i \mathbf{K} \cdot d \boldsymbol{x'} -\frac{1}{2} (\mathbf{K} \cdot d \boldsymbol{x'})^2.$$

Then, to 2nd order,
$$\begin{eqnarray} \mathcal{T}^{\dagger} (d \boldsymbol{x'} ) \mathcal{T} ( d \boldsymbol{x'} ) & = & 1 - \frac{1}{2} (\mathbf{K}^{\dagger} \cdot d \boldsymbol{x'} )^2 -\frac{1}{2} ( \mathbf{K} \cdot d \boldsymbol{x'} )^2 + (\mathbf{K}^{\dagger} \cdot d \boldsymbol{x'} ) ( \mathbf{K} \cdot d \boldsymbol{x'} ) \\ & = & 1 - \frac{1}{2} \left[( \mathbf{K} - \mathbf{K}^{\dagger} ) \cdot d \boldsymbol{x'} \right]^2 .\end{eqnarray}$$

The 2nd-order term vanishes if $K$ is Hermitian, which was the same condition needed at 1st-order.

The claim is that this will persist to all orders. It is not a rigorous mathematical proof, but it is clear that such a proof could be made. I have not personally studied the text, but many people here seem to recommend the book by Ballentine whenever people want a more mathematically rigorous treatment of QM. I do not know if he addresses similar issues there.

3. Apr 21, 2013

### micromass

Staff Emeritus
The rigorous version of this is the following: http://en.wikipedia.org/wiki/Stone's_theorem_on_one-parameter_unitary_groups

The problem from pure mathematical point-of-view is that $dx$ is not well-defined. Sakurai treats it as a number such that $(dx)^2 = 0$, this seems to be nonrigorous. The right argument involves tangent spaces and differential geometry. This is of course too difficult to present in an introductory books that doesn't involve much math.

So, the argument of sakurai can be made rigorous, but it doesn't seem easy to do so.