Derivation in bohm's Quantum Theory

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SUMMARY

Bohm's Quantum Theory involves the derivation of the action variable J, expressed as J = 2∫_{a(E)}^{b(E)} dq √{2m[E-V(q)]}. The discussion highlights the calculation of the partial derivative of J with respect to energy E, utilizing the theorem of differentiation under the integral sign. This theorem allows for the evaluation of the derivative of an integral with variable limits, which is crucial for understanding the relationship between action and energy in quantum mechanics.

PREREQUISITES
  • Understanding of classical mechanics and action variables
  • Familiarity with quantum mechanics concepts, particularly Bohm's interpretation
  • Knowledge of calculus, specifically differentiation under the integral sign
  • Basic understanding of potential energy functions V(q)
NEXT STEPS
  • Study the theorem of differentiation under the integral sign in calculus
  • Explore the implications of action variables in quantum mechanics
  • Investigate Bohm's interpretation of quantum mechanics in detail
  • Review examples of energy-dependent potential functions V(q) in quantum systems
USEFUL FOR

Students and researchers in physics, particularly those focusing on quantum mechanics and action principles, as well as anyone interested in the mathematical foundations of Bohm's Quantum Theory.

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halfway page 41 Bohm obtains for the action variable
[tex] J = 2\int_{a(E)}^{b(E)} dq \sqrt{2m[E-V(q)]}[/tex]
Then he obtians the partial derivative to E "by a well-known theorem of the calculus":

[tex] \frac{\partial J}{\partial E} = 2\left\{ \sqrt{2m\left[E-V(q)\right]} \right\}_{q=b} \frac{\partial b}{\partial E} - 2\left\{ \sqrt{2m[E-V(q)]} \right\}_{q=a} \frac{\partial a}{\partial E} + 2 \int_a^b \sqrt{\frac{m}{2[E-V(q)]}} dq.[/tex]

What is this "well-known theorem"?

Thanks,
A_B
 
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thanks! ...Never came across that one before...

A_B
 

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