How to Derive This Four-Component Expression?

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SUMMARY

The discussion focuses on deriving the expression \(\partial_{\mu}Ae^{(-i/\hbar)x\bullet p}u(p)\). The user successfully applies the product rule of differentiation, resulting in the expression \(Ae^{(-i/\hbar)x\bullet p}u(p)[-i\hbar\partial_{\mu}(x^{\nu}p_{\nu})]\). The key challenge lies in recombining the components after differentiation, particularly the momentum four-vector \(p\). The solution confirms that the derivative can be explicitly calculated using established rules of calculus.

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  • Understanding of quantum mechanics notation and concepts
  • Familiarity with four-vector calculus
  • Knowledge of the product rule in differentiation
  • Basic principles of wave functions in quantum physics
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Students and professionals in physics, particularly those studying quantum mechanics and advanced calculus, will benefit from this discussion. It is especially relevant for anyone working with wave functions and four-vector calculus.

ercagpince
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Homework Statement


I need to take the derivative of this expression explicitly

Homework Equations


[tex]\partial_{\mu}Ae^{(-i/\hbar)x\bullet p}u(p)[/tex]

The Attempt at a Solution


I tried to take the derivative of it by taking the derivatives of components of the momentum four vector "p".
However, I got stuck while trying to recombine the expression left from derivation.
 
Last edited:
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ercagpince said:

Homework Statement


I need to take the derivative of this expression explicitly

Homework Equations


[tex]\partial_{\mu}Ae^{(-i/\hbar)x\bullet p}u(p)[/tex]

The Attempt at a Solution


I tried to take the derivative of it by taking the derivatives of components of the momentum four vector "p".
However, I got stuck while trying to recombine the expression left from derivation.
[tex]\partial_{\mu}Ae^{(-i/\hbar)x\bullet p}u(p)<br /> =Ae^{(-i/\hbar)x\bullet p}u(p)[-i\hbar\partial_{\mu}(x^{\nu}p_{\nu})][/tex],
which should be easy.
 
Last edited:
thanks a lot!
 

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