- Translation Operator

1. Translational Symmetry in Space #

The inner product between two states remains invariant under translation:

$\langle\phi|\psi\rangle = \langle\phi|T^{\dagger}(\Delta x’)T(\Delta x)|\psi\rangle$

where $T(\Delta x’)|x’\rangle = |x’+\Delta x’\rangle$

Therefore, $T^{\dagger}(\Delta x’)T(\Delta x) = 1$

This means that $T(\Delta x’)$ is a unitary operator.

2. General Form of Unitary Operators #

The general form of a unitary operator $U$ is:

$U = e^{ik}$, where $k^{\dagger} = k$

3. Representation of Translation Operator #

The translation operator can be represented as:

$T(\Delta x’) = e^{i\Delta x’k}$

4. Composition of Translation Operators #

Let’s verify if $T(\Delta x’_2)T(\Delta x’_1) = T(\Delta x’_1 + \Delta x’_2)$ holds:

$$T(\Delta x’_2)T(\Delta x’_1) = e^{i\Delta x’_2k}e^{i\Delta x’_1k} = e^{i(\Delta x’_1+\Delta x’_2)k}$$

This confirms that the composition property holds.

5. Approximation of Translation Operator #

For small translations, we can approximate:

$T(\Delta x’) \approx 1 + ik\Delta x'$

6. Unitary Equivalence Transformation #

The transformation $|x’\rangle \to |x’+ \Delta x’\rangle$ (which equals $T(\Delta x’)|x’\rangle$) corresponds to:

$x|x’\rangle = x T^{\dagger}(\Delta x’)T(\Delta x’)|x’\rangle = x’|x’\rangle$

From $x T^{\dagger}(\Delta x’)T(\Delta x’)|x’\rangle = x’|x’\rangle$, multiplying both sides by $T(\Delta x’)$:

$T(\Delta x’)x T^{\dagger}(\Delta x’)T(\Delta x’)|x’\rangle = x’T(\Delta x’)|x’\rangle$

Since $T(\Delta x’)|x’\rangle = |x’+ \Delta x’\rangle$:

$T(\Delta x’)x T^{\dagger}(\Delta x’)|x’+ \Delta x’\rangle = x’|x’+ \Delta x’\rangle$

Thus, $x \to T(\Delta x’)x T^{\dagger}(\Delta x’)$

7. Approximation Calculation for Position Operator #

Using the approximation $T(\Delta x’) = 1 + ik \Delta x’$:

Applying this to the state $|x’+ \Delta x’\rangle$:

Therefore, $\Delta x’- i \Delta x’[x,k]=0$, which implies $p=-\hbar k$, or $k = -\frac{p}{\hbar}$

8. Exponential Representation of Translation Operator #

Given that $T(\Delta x’) = e^{ik \Delta x’}$ and $k = -\frac{p}{\hbar}$:

$$T(\Delta x’) = e^{-\frac{i}{\hbar}p \Delta x’}$$

9. Bra representation of Translation Operator #

$$ \langle x’| T(\Delta x’) = \langle x’-\Delta x’|$$

$$\leftrightarrow$$

$$ T^{-1}(\Delta x’)|x’\rangle=T(-\Delta x’)|x’\rangle = |x’ - \Delta x’\rangle$$

10. Translation of a function #

$$\langle x|T(\Delta x)|\psi\rangle = \langle x-\Delta x|\psi\rangle = \psi(x-\Delta x)$$