What is the commutator of x1 and the translation operator?

In summary, we discussed the translation operator F(\textbf{l}) and how to compute [x_i , F (\textbf {l} )]. We used the fundamental commutation relation [x_i, G(\textbf{p})] = i\hbar \frac{\partial G}{\partial p_i} and expanded F into a Taylor series. However, after applying the commutation relation, we found that the commutator [x_i , G(\textbf{p})] actually equals \frac{n i \hbar}{p_i} F(\textbf{l}). Finally, we looked at an example of this commutator with the exponential function and saw that it is not correct.
  • #1
indigojoker
246
0
Let the translation operator be:

[tex]F (\textbf {l} ) = exp \left( \frac{-i \textbf{p} \cdot \textbf{l}}{\hbar} \right) [/tex]

where p is the momentum operator and l is some finite spatial displacement

I need to find [tex] [x_i , F (\textbf {l} )] [/tex]

let me start with a fundamental commutation relation:

[tex] [x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i} [/tex]

also let me define:

[tex]\textbf{p} = (p_i,p_j,p_k) [/tex]
[tex]\textbf{l} = (\Delta x, \Delta y, \Delta z)[/tex]

we expand F in a taylor series:

[tex]F (\textbf {l} ) = \sum_{n=0}^{\infty} \frac{ \left( \frac{ -i \textbf{p } \cdot \textbf{ l}}{\hbar} \right) ^n}{n!} [/tex]

[tex]= \sum_{n=0}^{\infty} \frac{ \left( \frac{-i p_i \Delta x}{\hbar} \right) ^n}{n!} \frac{ \left( \frac{-i p_j \Delta y}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i p_k \Delta z}{\hbar} \right) ^n}{n!}[/tex]

[tex]= \sum_{n=0}^{\infty} \frac{ \left(\frac{ -i \Delta x}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i \Delta y}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i \Delta z}{\hbar} \right) ^n}{n!} p_i^n p_j^n p_k^n[/tex]

we now apply:

[tex] [x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i} [/tex]

which leaves us with:

[tex]= \sum_{n=0}^{\infty} \frac{ \left(\frac{ -i \Delta x}{\hbar} \right) ^n}{n!} \frac{\left(\frac{ -i \Delta y}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i \Delta z}{\hbar} \right) ^n}{n!} n i \hbar p_i^{n-1} p_j^n p_k^n[/tex]

[tex][x_i , G ( \textbf{p} ) ]=\frac{n i \hbar}{p_i} F (\textbf {l} ) [/tex]

wondering if this looks okay.
 
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  • #2
It's not okay. Compute this commutator

[tex] \left[x_{1},e^{\frac{p_{1}l_{1}+p_{2}l_{2}+p_{3}l_{3}}{i\hbar}} \right] [/tex]
 

1. What is commutation of operator?

Commutation of operator is a mathematical operation that involves changing the order of two operators in a given equation. It is also known as the commutator of operators or operator commutation.

2. Why is commutation of operator important in science?

Commutation of operator is important in science because it helps in understanding the relationship between different operators and how they affect the outcome of a mathematical equation. It is particularly useful in quantum mechanics and other branches of physics.

3. How is commutation of operator calculated?

Commutation of operator is calculated using the commutator, which is a mathematical expression that involves the two operators and their respective equations. The commutator is found by subtracting the product of the two operators in their original order from the product of the two operators in the reversed order.

4. What is the significance of the commutator in commutation of operator?

The commutator is significant because it determines whether two operators commute or not. If the commutator is equal to zero, then the operators commute and can be rearranged in any order. However, if the commutator is not equal to zero, then the operators do not commute and their order is important in determining the outcome of the equation.

5. Can commutation of operator be applied to all types of operators?

Commutation of operator can be applied to most types of operators, including linear and non-linear operators. However, it may not be applicable to all types of operators as some may not have well-defined commutators. It is important to consider the properties of the operators before applying commutation of operator.

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