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peter_pan Group Title

Quantum Mechanics (the very beginning): I was studying the proof of the conservation of ∫|ψ|²d³r (ψ is the wave function) but I don't understand the following step, in which Gauss's theorem is used.

  • one year ago
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  1. peter_pan Group Title
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    The following integral is performed over the whole space and ψ* is the complex conjugate of ψ: \[\int\limits -i \hbar {\partial{| \psi|^2} \over \partial t} d^3r = - {\hbar^2 \over 2m} \int\limits \nabla \cdot [ \psi^* \nabla \psi- \psi \nabla \psi^*] d^3r\] Now, I don't understandthis step \[\int\limits \nabla \cdot [ \psi^* \nabla \psi- \psi \nabla \psi^*] d^3r =0\] How has the divergence theorem been used?

    • one year ago
  2. Carl_Pham Group Title
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    You can use the divergence theorem to convert that volume integral to an integral over the boundary surface of the quantity in brackets. However, you set the boundary surface quite far away from any interesting behaviour of your wavefunction, e.g. at infinity, so that either the wavefunction or its gradient, or both, are zero on that boundary. Hence the surface integral is zero.

    • one year ago
  3. peter_pan Group Title
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    Yes, because the wave function must be normalisable and goes to zero at infinity. Thank you for your answer!

    • one year ago
  4. Carl_Pham Group Title
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    Be careful. It only has to be normalizable in the sense that Y->0 at r->inf if it represents a bound state. If it represents a free particle, then it has extent over all space. However, in that case the gradient of the Y is zero. If the wavefunction represents a state with both free character and bound-state character, then fortunately either Y or its gradient goes to zero at infinity. I only mention these technicalities because you're involved in a proof, and a proof is only a proof if it dots every i and crosses every t. But clearly you get the idea.

    • one year ago
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