Berry Phase paper question

  • Thread starter Syrus
  • Start date
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If I did the integrations correctly, my answer is
[tex]
\langle n \vert \nabla_{a} \vert n \rangle = \int_{0}^{a}{dx \, \psi^{\ast}_{n}(x) \, \left( \frac{\partial}{\partial a} \psi_{n}(x) \right)} = -\frac{5}{2 a}
[/tex]
 
2,956
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Your "complication" comes from the overuse of the term "vector" in two different contexts.

EDIT:
I guess he decided to erase his last post.
 
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The symbol [itex]\nabla_{\mathbf{R}}[/itex] refers to differentiating the state ket w.r.t. each parameter, i.e. it becomes "a vector of kets".
This is what I was getting at. Thank you.
 
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It seems the notion I was lacking was that nabla acting on a ket vector in Hilbert space yields another vector in hilbert space (I hope).
 

tom.stoer

Science Advisor
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It seems the notion I was lacking was that nabla acting on a ket vector in Hilbert space yields another vector in hilbert space (I hope).
sic!
 
2,956
5
It seems the notion I was lacking was that nabla acting on a ket vector in Hilbert space yields another vector in hilbert space (I hope).
Yes it does. Think about the partial derivative:
[tex]
\frac{\partial}{\partial R_i} \, \vert n(\mathbf{R}) \rangle \equiv \lim_{t \rightarrow 0} \frac{\vert n(R_1, \ldots, R_i + t, \ldots, R_n) \rangle - \vert n(R_1, \ldots, R_i, \ldots, R_n) \rangle}{t}
[/tex]
Since the expression in the limit is always a linear combination of two kets, it must be a ket (that is what is meant by the Hilbert space being a linear space). Thus, every partial derivative is a ket. You may treat these partial derivatives as COVARIANT components of a gradient in some n dimensional metric space. This metric space may be curved, depending on the nature of the parameters.
 
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It seems the notion I was lacking was that nabla acting on a ket vector in Hilbert space yields another vector in hilbert space (I hope).
No, it yields a vector of hilbert space vectors. You have to understand that there are two vector spaces here, one for states, one for parameters.
 

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