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, \(\langle x|p\rangle = \mathbf{C}\exp\left[\frac{i}{\hbar}px\right]\).