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formulas:canonical_commutation_relations [2018/03/29 14:40]
jakobadmin ↷ Page moved from advanced_notions:canonical_commutation_relations to equations:canonical_commutation_relations
formulas:canonical_commutation_relations [2018/05/13 09:18] (current)
jakobadmin ↷ Page moved from equations:canonical_commutation_relations to formulas:canonical_commutation_relations
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-<WRAP lag>​$[\hat{p}_i,​\hat{x}_j] =  -i \delta_{ij}$</​WRAP>​+<WRAP lag>​$ ​ ​\color{black}{[}\color{royalblue}{\hat{p}_i},\color{firebrick}{\hat{x}_j}\color{black}{]} \color{black}{ ​=  -i} \color{olive}{\hbar} \color{black}{\delta_{ij}}, \ \color{black}{[}\color{firebrick}{\hat{x}_i} , \color{firebrick}{\hat{x}_j}\color{black}{]} \color{black}{=} \color{darksalmon}{0},​ \  \color{black}{[} \color{royalblue}{\hat{p}_i} , \color{royalblue}{\hat{p}_j} \color{black}{]} \color{black}{=} \color{darksalmon}{0}$</​WRAP>​
  
 ====== Canonical Commutation Relations ====== ====== Canonical Commutation Relations ======
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 <tabbox Intuitive> ​ <tabbox Intuitive> ​
  
-The canonical commutation relations tell us that we can't measure the momentum and the location of a particle at the same time with arbitrary precision.+The canonical commutation relations tell us that we can't measure the <color royalblue>​momentum</​color> ​and the <color firebrick>​location</​color> ​of a particle at the same time with <color olive>arbitrary precision</​color>​. 
 + 
 +However, can measure the <color firebrick>​location</​color>​ on different axes - e.g. the location on the x-axis and the location on the y-axis - with <color darksalmon>​arbitrary precision</​color>​. Equally, we can measure the <color royalblue>​momentum</​color>​ in the direction of different axes with <color darksalmon>​arbitrary precision</​color>​. 
 + 
 + 
 +<​blockquote>​The action, in the form h/2π, also occurs in the Heisenberg Uncertainty Relationship,​ and here it sets the limit of precision to which the conjugate coordinates (momentum and position; or energy and time) can be determined simultaneously for one particle. It also occurs in the Dirac commutator relations (discovered by the British theoretical physicist, Paul Dirac (1902–84)) which show that the order in which measurements of these conjugate coordinates are carried out does make a difference - a tiny difference of magnitude h/2π . 
 + 
 +<​cite>​The Lazy Universe by Coopersmith</​cite></​blockquote>​  
 + 
 <tabbox Concrete> ​ <tabbox Concrete> ​
  
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 **Quantum Mechanics** **Quantum Mechanics**
  
-\begin{equation} ​ \label{eq:​commquantummech} [\hat{p}_i,​\hat{x}_j] =  -i \delta_{ij} .\end{equation}+\begin{equation} ​ \label{eq:​commquantummech} [\hat{p}_i,​\hat{x}_j] =  -i \hbar \delta_{ij} .\end{equation}
  
  
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 \begin{equation} \label{qftcomm} ​ [\Phi(x), \pi(y)]=\Phi(x) \pi(y) - \pi(y) \Phi(x) = i \delta(x-y) \end{equation} where $\delta(x-y)$ is the Dirac delta distribution and $\pi(y) = \frac{\partial \mathscr{L}}{\partial(\partial_0\Phi)}$ is the conjugate momentum. ​ \begin{equation} \label{qftcomm} ​ [\Phi(x), \pi(y)]=\Phi(x) \pi(y) - \pi(y) \Phi(x) = i \delta(x-y) \end{equation} where $\delta(x-y)$ is the Dirac delta distribution and $\pi(y) = \frac{\partial \mathscr{L}}{\partial(\partial_0\Phi)}$ is the conjugate momentum. ​
  
-It tells us that the fields in [[theories:​quantum_field_theory|quantum field theory]] $\Phi(x)$ can‘t be simply a function, but must be __operators__. In contrast, ordinary functions and numbers commute:+It tells us that the fields in [[theories:​quantum_field_theory:canonical|quantum field theory]] $\Phi(x)$ can‘t be simply a function, but must be __operators__. In contrast, ordinary functions and numbers commute:
  
 For example $f(x)=3x$ and $g(y)= 7y^2 +3$ clearly commute $$ [f(x) , g(x)]= f(x)g(x) - g(x) f(x) = 3x (7y^2 +3) -(7y^2 +3) 3x =0. $$ For example $f(x)=3x$ and $g(y)= 7y^2 +3$ clearly commute $$ [f(x) , g(x)]= f(x)g(x) - g(x) f(x) = 3x (7y^2 +3) -(7y^2 +3) 3x =0. $$
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 Take note that, for brevity, it‘s conventional to write $\Phi(x)$ instead of $\Phi(x,t)$ , which means we include $t$ in $x$: $x_0=t$, $x_1=x$, $x_2=y$ and $x_3=z$. Take note that, for brevity, it‘s conventional to write $\Phi(x)$ instead of $\Phi(x,t)$ , which means we include $t$ in $x$: $x_0=t$, $x_1=x$, $x_2=y$ and $x_3=z$.
  
 +-->​Derivation#​
  
 +Analogous to what we do in quantum mechanics, we identify the conserved quantities like energy and momentum with the generators of the corresponding symmetry.
 +
 +The invariance under displacements of the field itself $\Phi \rightarrow \Phi - i \epsilon$ yields a new conserved quantity, called conjugate momentum $\Pi$. So completely analogous to what we discussed above for quantum mechanics, we now identify the conjugate momentum density with the corresponding generator ​
 +
 +
 +\[ \mathrm{conj. \ mom. \ density \ } \pi(x) \rightarrow {\mathrm{gen. \  of  \ displ. \ of  \ the  \ field \ itself: \ }} -i \frac{\partial}{\partial \Phi(x)} \]
 +
 +We can then calculate
 +
 +\[[\Phi(x), \pi(y)] \Psi = \left[\Phi(x),​ -i \frac{\partial}{\partial \Phi(y)}\right] \Psi
 +\]
 +\begin{equation} \label{eq:​field-conjmom-commutator} ​ \underbrace{=}_{\text{product rule}} -i \Phi(x) \frac{\partial \Psi}{\partial \Phi(y)} +   ​i ​ \Phi(x) \left(\frac{\partial\Psi}{\partial \Phi(y)} \right) + i \left( \frac{\partial \Phi(x)}{\partial \Phi(y)} \right) ​ \Psi =  i \delta(x-y) ​ \Psi \end{equation}
 +Again, the equations hold for arbitrary $\Psi$ and we can therefore write
 +
 +\begin{equation} [\Phi(x), \pi(y)] = i \delta(x-y) ​ \end{equation}
 +
 +<--
    
 <tabbox Abstract> ​ <tabbox Abstract> ​
 +The mathematically rigorous theory of these commutation relations can be found in 
  
-<note tip> +  * JørgensenP.E.T.; Moore, R.T.: Operator commutation relations. Dordrecht: Reidel, 1984 
-The motto in this section is: //the higher the level of abstractionthe better//+  * Schmüdgen, K.: Unbounded operator algebras and representation theory. Birkhäuser Verlag, Basel, 1990
-</​note>​+
  
 +However, take note that the canonical commutation relations are mathematically problematic nevertheless.
 +
 +One problem is that the operators involved are unbounded. If we want that they represent physical observables they have to be self-adjoint;​ but on their respective domains of self-adjointness the commutator on the left-hand side is undefined.
 +
 +Another problem is that the canonical commutation relations rely on the possibility of choosing global coordinates on $\mathbb{R}^3$. Since this is, in general not possible on arbitrary configuration spaces it is unclear how they can then be generally valid.
 +
 +The problem of finding an appropriate mathematical interpretation of the canonical commutation
 +relations is the subject of [[advanced_tools:​quantization|quantization theory]]. ​
 +
 +----
 +
 +  * A good discussion can be found [[https://​mathoverflow.net/​questions/​55988/​quantum-mechanics-formalism-and-c-algebras/​56003|here]].
 <tabbox Why is it interesting?> ​   <tabbox Why is it interesting?> ​  
  
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 In quantum field theory it tells us that our fields are operators. ​ In quantum field theory it tells us that our fields are operators. ​
 +
  
 </​tabbox>​ </​tabbox>​
  
  
formulas/canonical_commutation_relations.1522327232.txt.gz · Last modified: 2018/03/29 12:40 (external edit)