From 'Modern Quantum Mechanics, revised edition' by J.J. Sakurai,(adsbygoogle = window.adsbygoogle || []).push({});

In page 44 the translation (spacial displacement) operator [itex]\mathcal{T}(d \boldsymbol{x'})[/itex] is introduced.

[tex]\mathcal{T}(d \boldsymbol{x'}) \mid \boldsymbol{x'} \rangle = \mid \boldsymbol{x'} + d \boldsymbol{x'} \rangle[/tex]

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.

[tex]\mathcal{T}^{\dagger} ( d \boldsymbol{x'} ) \mathcal{T} ( d \boldsymbol{x'} ) = 1[/tex]

[tex]\mathcal{T} ( d \boldsymbol{x''} ) \mathcal{T} ( d \boldsymbol{x'} ) = \mathcal{T} ( d \boldsymbol{x'} + d \boldsymbol{x''} )[/tex]

Two other properties are also stated, but they don't pertain to my question.

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

For example, in the proof that unitarity is satisfied,

[tex] \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} [/tex]

Here the second order terms of [itex]d \boldsymbol{x'}[/itex] are ignored. Thus the 0 in the second line.

In the proof for the second property,

[tex] \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} [/tex]

In the second line, the second order terms of [itex]d \boldsymbol{x}[/itex] 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Sakurai : Approximations in the construction of Quantum theory

**Physics Forums | Science Articles, Homework Help, Discussion**