- #1

- 56

- 0

[tex]

\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

xi \hbar \pd {} {x} {}

[/tex]

I presume I can't just differentate the x as I need to preserver order, does this just sit like this till I can 'deal' with it?

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- Thread starter Lee
- Start date

- #1

- 56

- 0

[tex]

\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

xi \hbar \pd {} {x} {}

[/tex]

I presume I can't just differentate the x as I need to preserver order, does this just sit like this till I can 'deal' with it?

- #2

nrqed

Science Advisor

Homework Helper

Gold Member

- 3,764

- 295

[tex]

\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

xi \hbar \pd {} {x} {}

[/tex]

I presume I can't just differentate the x as I need to preserver order, does this just sit like this till I can 'deal' with it?

In calculating commutators of differntial operators, it is convenient to apply th commutator on a "test function", which is is just som arbitrary function of x, y and z that must be removed at the very end of the calculation.

So if you have two operators A and B (which are differential operators) and you want to compute their commutator, just consider

[tex] [A,B] f(x,y,z) = AB f(x,y,z) - BA f(x,y,z) [/tex]

Apply all the derivatives and at the very end, remove the test function.

- #3

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

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But remember that "x" is an operator, not a function. So

[tex]

\frac{\partial}{\partial x} x \neq 1

[/tex]

Instead, it's supposed to be the operator

[tex]

\psi(x, y, z, t) \rightarrow \frac{\partial (x \psi(x, y, z, t))}{\partial x}

[/tex]

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