- #26

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[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]

- Thread starter Syrus
- Start date

- #26

- 2,967

- 5

[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]

- #27

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EDIT:

I guess he decided to erase his last post.

- #28

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This is what I was getting at. Thank you.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".

- #29

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- #30

tom.stoer

Science Advisor

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sic!

- #31

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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

- #32

- 239

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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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