Translations and momentum
Contents:
# Continuous spectrum
In a discrete, finite world: $\hat A|a'\rangle = a'|a'\rangle$. In a continuous one: $\hat\xi|\xi'\rangle = \xi'|\xi'\rangle$. They mean the same shit.
Everything we had carries over, with sums becoming integrals and the Kronecker delta becoming the Dirac delta:
$$ \begin{aligned} \langle a'|a''\rangle = \delta_{a'a''} &\quad\longrightarrow\quad \langle\xi'|\xi''\rangle = \delta(\xi' - \xi'') \\ \sum_{a'}|a'\rangle\langle a'| = 1 &\quad\longrightarrow\quad \int d\xi'|\xi'\rangle\langle\xi'| = 1 \\ |\alpha\rangle = \sum_{a'}|a'\rangle\langle a'|\alpha\rangle &\quad\longrightarrow\quad |\alpha\rangle = \int d\xi'|\xi'\rangle\langle\xi'|\alpha\rangle \\ \sum_{a'}|\langle a'|\alpha\rangle|^2 = 1 &\quad\longrightarrow\quad \int d\xi'|\langle\xi'|\alpha\rangle|^2 = 1 \\ \langle\beta|\alpha\rangle = \sum_{a'}\langle\beta|a'\rangle\langle a'|\alpha\rangle &\quad\longrightarrow\quad \langle\beta|\alpha\rangle = \int d\xi'\langle\beta|\xi'\rangle\langle\xi'|\alpha\rangle \\ \langle a''|\hat A|a'\rangle = a'\delta_{a'a''} &\quad\longrightarrow\quad \langle\xi''|\hat\xi|\xi'\rangle = \xi'\delta(\xi'' - \xi') \end{aligned} $$
(the last one assuming eigenkets).
# Position eigenkets
$$ \hat x|x'\rangle = x'|x'\rangle $$
The eigenkets $|x'\rangle$ of $\hat x$ form a complete set. The state ket of any state $|\alpha\rangle$ can be written as
$$ |\alpha\rangle = \int_{-\infty}^{\infty}dx'|x'\rangle\langle x'|\alpha\rangle $$
$\implies$ when we detect the position ($x'$), $|\alpha\rangle$ jumps into the $|x'\rangle$ state. Same way as an SG$\hat x$ measurement makes $|+\rangle \to |S_x;+\rangle$, for example.
In practice of course, we can’t detect “exactly” at $x'$, so we do it around a small band, $x' - \frac{\Delta}{2}$ to $x' + \frac{\Delta}{2}$:
$$ \underbrace{|\alpha\rangle = \int_{-\infty}^{\infty}dx''|x''\rangle\langle x''|\alpha\rangle}_{\text{before measurement: can be anywhere}} \xrightarrow{\text{measure}} \int_{x' - \Delta/2}^{x' + \Delta/2}dx''|x''\rangle\langle x''|\alpha\rangle $$
with $\Delta \to dx'$. The prob. for this detector to click is
$$ |\langle x'|\alpha\rangle|^2dx' $$
The prob. of the particle being “somewhere” is $\int dx'|\langle x'|\alpha\rangle|^2$, and if $|\alpha\rangle$ is normalized, this is 1. Duh.
$$ \int_{-\infty}^{\infty}dx'|\langle x'|\alpha\rangle|^2 = 1 $$
We of course know that $\langle x'|\alpha\rangle$ is the wave function (more on that in the next notes).
# In 3 dimensions
$$ |\alpha\rangle = \int d^3x'|\vec x'\rangle\langle\vec x'|\alpha\rangle $$
where $|\vec x'\rangle = |x', y', z'\rangle$ is a simultaneous eigenket:
$$ \hat x|\vec x'\rangle = x'|\vec x'\rangle, \qquad \hat y|\vec x'\rangle = y'|\vec x'\rangle, \qquad \hat z|\vec x'\rangle = z'|\vec x'\rangle $$
We know from the simultaneous eigenket definition
$$ \implies [x_i, x_j] = 0 $$
# Translations
Take a state $|\alpha\rangle$ from $\vec x'$ to $\vec x' + d\vec x'$, leaving everything else unchanged $\to$ an infinitesimal translation. We have an operator for this:
$$ \mathcal J(d\vec x')|\vec x'\rangle = |\vec x' + d\vec x'\rangle $$
(not an eigenvalue equation). Now, we can expand $|\alpha\rangle$ in position eigenkets:
$$ \begin{aligned} \mathcal J(d\vec x')|\alpha\rangle &= \mathcal J(d\vec x')\int d^3x'|\vec x'\rangle\langle\vec x'|\alpha\rangle \\ &= \int d^3x'|\vec x' + d\vec x'\rangle\langle\vec x'|\alpha\rangle \\ &= \int d^3x'|\vec x'\rangle\langle\vec x' - d\vec x'|\alpha\rangle \end{aligned} $$
(the last line by a change of variables).
What properties should $\mathcal J$ have?
- It’s reasonable to assume that if $|\alpha\rangle$ is normalized, then a shifted $|\alpha\rangle$ must also be. $\langle\alpha|\alpha\rangle = \langle\alpha|\mathcal J^\dagger\mathcal J|\alpha\rangle$, so $\mathcal J^\dagger\mathcal J = 1$ $\to$ unitary, from the prob. $= 1$ thing.
- Two translations add like vectors: $\mathcal J(d\vec x'')\mathcal J(d\vec x') = \mathcal J(d\vec x' + d\vec x'')$
- $\mathcal J(-d\vec x') = \mathcal J^{-1}(d\vec x')$
- $\lim_{d\vec x' \to 0}\mathcal J(d\vec x') = 1$ $\to$ no translation $\implies$ unchanged state.
These come from simply reasoning about $d\vec x$ as a vector.
$$ \implies \boxed{\mathcal J(d\vec x') = 1 - i\vec K\cdot d\vec x'} $$
$\to$ this fits, where $\vec K$ has $K_x, K_y, K_z$, and all 3 are Hermitian. Then all the properties are satisfied to first order. Eg. $\mathcal J^\dagger\mathcal J = (1 + i\vec K\cdot d\vec x')(1 - i\vec K\cdot d\vec x') = 1 + O(d\vec x'^2)$.
# Position and translations don’t commute
Note that,
$$ \begin{aligned} \hat{\vec x}\mathcal J(d\vec x')|\vec x'\rangle &= \hat{\vec x}|\vec x' + d\vec x'\rangle = (\vec x' + d\vec x')|\vec x' + d\vec x'\rangle \\ \mathcal J(d\vec x')\hat{\vec x}|\vec x'\rangle &= \vec x'\mathcal J(d\vec x')|\vec x'\rangle = \vec x'|\vec x' + d\vec x'\rangle \end{aligned} $$
Hence,
$$ \boxed{[\hat{\vec x}, \mathcal J(d\vec x')]|\vec x'\rangle = d\vec x'|\vec x' + d\vec x'\rangle \approx d\vec x'|\vec x'\rangle} $$
(to first order). Now $|\vec x'\rangle$ can be any position eigenket, and they form a complete set.
$\therefore$ we must have
$$ [\hat{\vec x}, \mathcal J(d\vec x')] = d\vec x' $$
$\to$ just that with the ket removed, since it holds always.
This basically says that “measure and move” vs “move and measure” are off by a factor that’s always how much you moved, which makes sense.
Substituting $\mathcal J = 1 - i\vec K\cdot d\vec x'$, the $i$-th component is
$$ -i\hat x_i(\vec K\cdot d\vec x') + i(\vec K\cdot d\vec x')\hat x_i = dx_i' \implies -i\sum_j[\hat x_i, K_j]dx_j' = dx_i' $$
$$ \implies \boxed{[x_i, K_j] = i\delta_{ij}} $$
$\to$ an imaginary no.
# Momentum as a generator of translations
We have $[x_i, K_j] = i\delta_{ij}$ $\to$ the commutation relation b/w the position operators $x, y, z$ and the $K$ operators $K_x, K_y, K_z$.
What’s the physical meaning of $\vec K$?
$\to$ it is related to the momentum operator in QM.
$\hat K \neq \hat p$, because the dimensions are all wrong ($\vec K\cdot d\vec x$ has to be dimensionless, so $K \sim 1/\text{length}$). We set
$$ \hat K = \frac{\hat p}{\text{constant}} $$
$\to$ just to set the dimensions right, but we have no way of knowing the exact value. We let it be $\hbar$, because it fixes the dimensionality problem. Recall that
$$ \frac{p}{\hbar} = \frac{2\pi}{\lambda} $$
from the de Broglie wavelength relation.
$\implies \hat K$ corresponds to the wave no.
$$ \implies \boxed{\mathcal J(d\vec x') = 1 - \frac{i\vec p\cdot d\vec x'}{\hbar}} $$
where $\vec p$ is the momentum operator. And from $[x_i, K_j] = i\delta_{ij}$ we get
$$ \boxed{[x_i, p_j] = i\hbar\delta_{ij}} $$
$\implies x$ and $p_x$ are incompatible observables with no simultaneous eigenkets, since $[x, p_x] = i\hbar \neq 0$ $\implies$ they cannot be measured simultaneously. We can even obtain the relation (from the general uncertainty relation, $\frac{1}{4}|i\hbar|^2$):
$$ \langle(\Delta x)^2\rangle\langle(\Delta p_x)^2\rangle \geq \frac{\hbar^2}{4} $$
# Finite translations
Note that $\mathcal J(d\vec x')$ is an operator for infinitesimal translations. Now we want to concern ourselves with finite translations $\implies$ compounding $N$ infinitesimal ones.
A finite translation in $\hat x$ by $\Delta x'$:
$$ \mathcal J(\Delta x'\hat x)|\vec x'\rangle = |\vec x' + \Delta x'\hat x\rangle $$
($\hat x$ just to specify the direction).
$$ \implies \mathcal J(\Delta x'\hat x) = \lim_{N\to\infty}\left(1 - \frac{ip_x\Delta x'}{N\hbar}\right)^N $$
$$ \boxed{\mathcal J(\Delta x'\hat x) = \exp\left(-\frac{ip_x\Delta x'}{\hbar}\right)} $$
$\to$ the operator of finite translations.
# Translations commute
Finite translations along $\hat x$ and $\hat y$ commute, that is, going to a point D can be done via A $\to$ B $\to$ D or A $\to$ C $\to$ D.
C ─────────────► D
▲ ▲
│ Δy │ Δy
│ │
A ─────────────► B
Δx
$$ \begin{aligned} \mathcal J(\Delta y'\hat y)\mathcal J(\Delta x'\hat x) &= \mathcal J(\Delta x'\hat x + \Delta y'\hat y) \\ \mathcal J(\Delta x'\hat x)\mathcal J(\Delta y'\hat y) &= \mathcal J(\Delta x'\hat x + \Delta y'\hat y) \end{aligned} $$
$$ \implies [\mathcal J(\Delta x'\hat x), \mathcal J(\Delta y'\hat y)] = 0 $$
Recall that $\exp(x) = 1 + x + \frac{x^2}{2!} + \dots$
$$ \begin{aligned} \exp\left(-\frac{ip_x\Delta x'}{\hbar}\right) &= 1 - \frac{ip_x\Delta x'}{\hbar} - \frac{p_x^2\Delta x'^2}{2\hbar^2} + \dots \approx \mathcal J(\Delta x'\hat x) \\ \exp\left(-\frac{ip_y\Delta y'}{\hbar}\right) &= 1 - \frac{ip_y\Delta y'}{\hbar} - \frac{p_y^2\Delta y'^2}{2\hbar^2} + \dots \approx \mathcal J(\Delta y'\hat y) \end{aligned} $$
Plug these into the commutator. The $1$s commute with everything, and the squared terms only show up at higher orders, so to 2nd order
$$ \left[1 - \frac{ip_x\Delta x'}{\hbar} - \frac{p_x^2\Delta x'^2}{2\hbar^2},\ 1 - \frac{ip_y\Delta y'}{\hbar} - \frac{p_y^2\Delta y'^2}{2\hbar^2}\right] \approx -\frac{\Delta x'\Delta y'}{\hbar^2}[p_x, p_y] $$
Since we know that the commutator is 0,
$$ \implies [p_x, p_y] = 0 $$
and more generally,
$$ \boxed{[p_i, p_j] = 0} $$
This relation is due to the fact that translations in diff. directions commute. (Note how rotations won’t commute!)
So, $[p_i, p_j] = 0$ $\implies$ we have a simultaneous eigenket
$$ |p_x', p_y', p_z'\rangle \equiv |\vec p'\rangle, \qquad p_x|\vec p'\rangle = p_x'|\vec p'\rangle, \dots $$
for momentum. And clearly the infinitesimal translation operator also gives an eigenvalue equation here:
$$ \mathcal J(d\vec x')|\vec p'\rangle = \left(1 - \frac{i\vec p\cdot d\vec x'}{\hbar}\right)|\vec p'\rangle = \left(1 - \frac{i\vec p'\cdot d\vec x'}{\hbar}\right)|\vec p'\rangle $$
So even though $|\vec x'\rangle$ is not an eigenket of $\mathcal J(d\vec x')$ (as $\mathcal J(d\vec x')|\vec x'\rangle = |\vec x' + d\vec x'\rangle$), $|\vec p'\rangle$ is an eigenket of $\mathcal J(d\vec x')$.
$$ [\vec p, \mathcal J(d\vec x')] = 0 $$
# The fundamental commutation relations
So, we have the following commutation relations:
$$ \boxed{[x_i, x_j] = 0, \qquad [p_i, p_j] = 0, \qquad [x_i, p_j] = i\hbar\delta_{ij}} $$
These are the “fundamental commutation relations”.
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