# Berry Phase paper question

#### Dickfore

If I did the integrations correctly, my answer is
$$\langle n \vert \nabla_{a} \vert n \rangle = \int_{0}^{a}{dx \, \psi^{\ast}_{n}(x) \, \left( \frac{\partial}{\partial a} \psi_{n}(x) \right)} = -\frac{5}{2 a}$$

#### Dickfore

Your "complication" comes from the overuse of the term "vector" in two different contexts.

EDIT:
I guess he decided to erase his last post.

#### Syrus

The symbol $\nabla_{\mathbf{R}}$ 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.

#### Syrus

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

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!

#### Dickfore

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:
$$\frac{\partial}{\partial R_i} \, \vert n(\mathbf{R}) \rangle \equiv \lim_{t \rightarrow 0} \frac{\vert n(R_1, \ldots, R_i + t, \ldots, R_n) \rangle - \vert n(R_1, \ldots, R_i, \ldots, R_n) \rangle}{t}$$
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.

#### Jazzdude

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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