Skip to main content
Physics

Return to Question

edited body; edited title; edited tags
Source Link
Qmechanic ♦
Qmechanic ♦
  • 213.1k
  • 48
  • 590
  • 2.3k

Is the momentum operator in position basis a first order-order approximation?

In my course we have just derived the momentum operator in the position basis to be the first order-order derivative of the position. This was achieved by using a first order-order taylor expansion:

$$ \psi(\boldsymbol r-d\boldsymbol r)\approx\psi(\boldsymbol r)⁻\nabla\psi(\boldsymbol r)\cdot d\boldsymbol r $$$$ \psi(\boldsymbol r-d\boldsymbol r)\approx\psi(\boldsymbol r)-\nabla\psi(\boldsymbol r)\cdot d\boldsymbol r $$

The same derivation can also be found on wikipedia.

Now my question is whether the final result, $\boldsymbol p = -i\hbar\nabla$, is an approximation? What if we were to use a second order term in the taylor expansion, would that change anything?

Is the momentum operator in position basis a first order approximation?

In my course we have just derived the momentum operator in the position basis to be the first order derivative of the position. This was achieved by using a first order taylor expansion:

$$ \psi(\boldsymbol r-d\boldsymbol r)\approx\psi(\boldsymbol r)⁻\nabla\psi(\boldsymbol r)\cdot d\boldsymbol r $$

The same derivation can also be found on wikipedia.

Now my question is whether the final result, $\boldsymbol p = -i\hbar\nabla$, is an approximation? What if we were to use a second order term in the taylor expansion, would that change anything?

Is the momentum operator in position basis a first-order approximation?

In my course we have just derived the momentum operator in the position basis to be the first-order derivative of the position. This was achieved by using a first-order taylor expansion:

$$ \psi(\boldsymbol r-d\boldsymbol r)\approx\psi(\boldsymbol r)-\nabla\psi(\boldsymbol r)\cdot d\boldsymbol r $$

The same derivation can also be found on wikipedia.

Now my question is whether the final result, $\boldsymbol p = -i\hbar\nabla$, is an approximation? What if we were to use a second order term in the taylor expansion, would that change anything?

Source Link
catmousedog
catmousedog
  • 728
  • 6
  • 10

Is the momentum operator in position basis a first order approximation?

In my course we have just derived the momentum operator in the position basis to be the first order derivative of the position. This was achieved by using a first order taylor expansion:

$$ \psi(\boldsymbol r-d\boldsymbol r)\approx\psi(\boldsymbol r)⁻\nabla\psi(\boldsymbol r)\cdot d\boldsymbol r $$

The same derivation can also be found on wikipedia.

Now my question is whether the final result, $\boldsymbol p = -i\hbar\nabla$, is an approximation? What if we were to use a second order term in the taylor expansion, would that change anything?