# Quantum Mechanical Translations

## Discrete and Continuous Eigenvalues

### Discrete

$$A|\varphi_n\rangle = a_n|\varphi_n\rangle$$

$$\langle\varphi_n|\varphi_m\rangle = \delta^n_m$$

$$\sum_n|\varphi_n\rangle\langle\varphi_n|=\mathbb{I}$$

$$|\psi\rangle=\sum_n |\varphi_n\rangle\langle\varphi_n|\psi\rangle$$.

$$\sum_n|\langle\varphi_n|\psi\rangle|^2=1$$.

$$\langle\varphi|\psi\rangle = \sum_n\langle\varphi|\varphi_n\rangle\langle\varphi_n|\psi\rangle$$.

$$\langle|\varphi_n|A|\varphi_m\rangle = a_m\delta^n_m$$.

### Continuous

$$B|\xi'\rangle = b|\xi'\rangle$$

$$\langle\xi'|\xi''\rangle = \delta(\xi'-\xi'')$$

Where the delta function is the Dirac Delta Function.

$$\int d\xi'|\xi\rangle\langle\xi'|=\mathbb{I}$$.

$$|\chi\rangle = \sum_\mathbb{R}d\xi'|\xi'\rangle\langle\xi'|\chi\rangle$$.

$$\int_\mathbb{R}d\xi'|\langle\xi'|\chi\rangle|^2=1$$.

$$\langle\chi|\gamma\rangle = \int_\mathbb{R}d\xi' \chi^*(\xi')\gamma(\xi')$$.

$$\langle\xi''|B|\xi'\rangle = b\delta(\xi''-\xi')$$.

## Position

### Example

Gaussian Wave Packets $$X$$ in one dimension. $$X|x'\rangle = x'|x'\rangle$$. $$|\psi\rangle = \int_\mathbb{R}dx dx'|x'\rangle\langle x'|\psi\rangle$$.

Practically speaking, $$\mathcal{P}=|\langle x'|\psi\rangle|^2\cdot(\delta x)$$. Then, $$\int_\mathbb{R}|\langle x'|\psi\rangle|^2d x' = 1$$. $$|\psi'\rangle_{after}=\int_{x'-(\delta x)/2}^{x'+(\delta x)/2} dx |x\rangle\langle x|\psi\rangle$$.

$$\langle x'|\psi\rangle = \psi(x')$$.

### Spatial Translations

$$\hat{U}_{d\vec{x}} = \mathbb{I}-id\vec{x}\cdot \hat{G} = \mathbb{I}-id\vec{x}\cdot \frac{\hat{\vec{p}}}{\hbar}$$, $$\hat{U}_{d\vec{x}}|\vec{x}' \rangle = |\vec{x}'+d\vec{x}\rangle$$.

Successive translations: $$\hat{U}_{d\vec{x}'}\hat{U}_{d\vec{x}''} = \mathbb{I}-\frac{i}{\hbar}\hat{\vec{p}}(d\vec{x}'+d\vec{x}'') = \hat{U}_{d\vec{x}'+d\vec{x}''}$$.

### Momentum Eigenkets from Spatial Translations

$$\hat{U}_{d\vec{x}}|\vec{p}'\rangle = (1-i/\hbar\vec{p}'\cdot d\vec{x}')|\vec{p}'\rangle$$.

### Inserting Identity

$$\psi_\alpha(x')=\langle x'|\alpha\rangle$$, $$\langle\beta|\alpha\rangle = \int dx'\langle \beta|x'\rangle\langle x'|a\rangle = \int dx'\psi^*_\beta(x')\psi_\alpha(x')=\int dp'\langle\beta|p'\rangle\langle p'|\alpha\rangle = \int dp' \Psi^*_\beta(p')\Psi_\alpha(p')$$

$$\langle\beta|A|\alpha\rangle = \iint dx'dx'' \psi^*_\beta(x')\langle x'|A|x''\rangle \psi_\alpha(x'')$$.

#### Example

$$A=\hat{X}^2$$. Then, $$\langle x'|\hat{X}^2|x''\rangle = (x'')^2\delta(x''-x')$$.

$$\langle \beta|A|\alpha\rangle = \int dx'\psi^*_\beta(x')\psi_\alpha(x')x'^2$$.

### Passive and Active Transformations

#### Active

$$\vec{r}\to\vec{r}'$$. The coordinates remain the same. (Our negative sign indicates that this we are using an active translation)

#### Passive

$$\vec{r}\to\vec{r}'$$. The coordinate system changes.

## Momentum

### Operator in Position Basis

$$\langle\beta|A|\alpha\rangle = \iint dx'dx'' \langle \beta|x'\rangle\langle x'|A|x''\rangle\langle x''|\alpha\rangle = \iint dx'dx'' \psi^*_\beta(x')\psi_\alpha(x'')\langle x'|A|x''\rangle$$

If $$A=F(X)$$ then this is easy to compute, $$\langle x'|A|x''\rangle =F(x'')\delta(x'-x'')$$. So, $$\langle\beta|A|\alpha\rangle =\int dx' \psi^*_\beta(x')\psi_\alpha(x')F(x')$$.

If $$A=f(\hat{P})$$? Then, $$\langle x'|f(\hat{P})|x''\rangle$$. $$\langle p'|A|p''\rangle = f(p'')\delta(p'-p'')$$.

#### Recall

$$\hat{U}_{\Delta x} = \mathbb{I}-\frac{i}{\hbar}\hat{P}_x \Delta x$$, $$\hat{U}_{\Delta x}|x\rangle = |x+\Delta x\rangle$$.

$$\hat{U}_{\Delta x}|\alpha\rangle = \hat{U}_{\Delta x}\int dx |x\rangle\langle x|\alpha\rangle = \int dx |x+\Delta x\rangle\langle x|\alpha\rangle = \int dx' |x'\rangle \langle x'-\Delta x|\alpha\rangle = \int dx'|x'\rangle \left[\langle x'|\alpha - \Delta x\frac{\partial}{\partial x'}\langle x'|\alpha\rangle\right] = |\alpha\rangle - \Delta x\int dx'\frac{\partial}{\partial x'}\langle x'|\alpha\rangle = (\mathbb{I}-\frac{i}{\hbar}\hat{P}_x\Delta x)|\alpha\rangle$$. So, $$\hat{P_x}|\alpha\rangle = -i\hbar\int dx' |x'\rangle \frac{\partial}{\partial x'}\langle x'|\alpha\rangle$$. $$\langle x''|\hat{P}_x|\alpha\rangle = -i\hbar \frac{\partial}{\partial x''}\langle x''|\alpha\rangle= -i\hbar \frac{\partial}{\partial x''}\psi_\alpha(x'')$$.

If $$|\alpha\rangle = |x'\rangle$$ then $$\langle x''|\hat{P}_x|x'\rangle = -i\hbar \frac{\partial}{\partial x''}\delta(x''-x')$$.

Thus, if $$A=f(\hat{P})$$ then $$\langle x'|f(\hat{P})|x''\rangle = \iint dx'dx'' \psi^*_\beta(x')f\left(-i\hbar\frac{\partial}{\partial x''}\delta(x'-x'')\right)\psi_\alpha(x'')= \int dx' \psi^*_\beta(x')f\left(-i\hbar\frac{\partial}{\partial x'}\right)\psi_\alpha(x')$$

### Momentum Space Wavefunction

$$\hat{P}|p'\rangle = p'|p'\rangle$$, $$\langle p'|p''\rangle = \delta(p'-p'')$$, $$|\alpha\rangle = \int dp' |p'\rangle\langle p'|\alpha\rangle$$.

$$\hat{P}_x \doteq -i\hbar \frac{\partial}{\partial x}$$, $$\hat{X}\doteq i\hbar \frac{\partial}{\partial p}$$, $$\mathcal{P}(p'-\Delta p/2,p'+\Delta p/2) = |\langle p'|\alpha\rangle|^2\Delta p = |\Phi_\alpha(p')|^2\Delta p$$.

$$\langle x''|\hat{P}|\alpha = -i\hbar \frac{\partial}{\partial x''}\langle x''|\alpha\rangle$$.

$$|\alpha\rangle = |p'\rangle$$. Then, $$\langle x''|\hat{P}|\alpha\rangle = -i\hbar \frac{\partial}{\partial x''}\langle x''|p'\rangle=p'\langle x''|p'\rangle$$. So, $$p'\langle x''|p'\rangle = -i\hbar\frac{\partial}{\partial x''}\langle x''|p'\rangle$$. So, $$\langle x|p\rangle = \mathbf{C}\exp\left[\frac{i}{\hbar}px\right]$$.

Created: 2023-06-25 Sun 02:32

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