# Berry Phase paper question

1. Oct 15, 2012

### Syrus

I am examining Berry's original 1984 paper of geometrical phases. I am having trouble understanding a portion of it, however (which is attached as a photo). In particular, equation (6) appears to have a bra-ket expression which includes the bra- complex conjugate of a state, <n(R)|, along with a ket |∇n(R)>. My question is, how is the resulting expression (the integral over space? of the product of these functions) another vector (as it must be since it is included in a line integral)?

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2. Oct 16, 2012

### tom.stoer

I think this is a sloppy notation and instead of |n(R)> it should read ψn(R)

Last edited: Oct 16, 2012
3. Oct 16, 2012

### Jazzdude

$\nabla _\mathbf{R}$ is a vector operator, so it's really a 3-vector containing the derivative operators. Your result is therefore a 3-vector too. I personally don't like this notation, as it mixes basis dependent operators with basis independent formalism. I would even call it broken, but some people seem to like it and find the intended meaning to be "obvious enough".

4. Oct 16, 2012

### Syrus

Yes, but how is a 3-vector (the result of del oeprating on the ket vector) integrated (since it is included within a bra-ket) to obtain yet another vector (as it is dotted with the line differential)?

Is this a non-standard bra-ket integration?

5. Oct 16, 2012

### Jazzdude

Not sure what you mean, but it seems you misunderstand something fundamental there. The result of <n_R|\nabla_R n_R> is a 3-vector of complex numbers. That 3-vector is then line integrated with dR.

6. Oct 16, 2012

### Syrus

This is my reasoning:

-Bra-ket expression denote integration over all space (or so I read).
-The notion of integrating nablaR acting on (what must be a scalar) ket produces a vector quantity, which is then multiplied by the bra- expression, and then integrated over all space.
-However, space integrals produce scalar (not vector) quantities. But a vector is needed since a dot product exists.

I hope this helps you understand where my mistake arises.

In particular, how does the bra-ket expression result in a complex 3-vector?

7. Oct 16, 2012

### Jazzdude

A combination of a bra and a ket in this order describes an inner product on the vector space of quantum states. Integration only comes in if you expand the inner product in a continuous basis.

Not integration, just the inner product. Since nabla is a vector really you get 3 distinct inner products with three scalar results that form a vector with 3 components again.

Like I just said, you get 3 scalar products because nabla is a vector operator. Like this: <a |*(d/dx,d/dy,d/dz) |b> = (<a|d/dx|b>,<a|d/dy|b>,<a|d/dz|b>).

This vector is then dotted with dR and integrated along the closed loop.

8. Oct 16, 2012

### Syrus

Ah, thank you Jazzdude. It seems I need to better familiarize myself with the notion of inner product. The phys chem 2 notion of bra-ket was probably simplified to idiocracy (as is the case with most concepts treated in undergrad cirriculae)- hence my misleading.

9. Oct 16, 2012

### tom.stoer

Now I remember.

|n(R)> is not a ket with which depends on coordinates R; R is the set of parameters. Sorry for the confusion.

Let's make an example.

Consider a Hamiltonian

$$H[\vec{B},\Phi_0] = \frac{1}{2m}\left(\vec{p}-q\vec{A}\right)^2 +qV$$

for a particle in a uniform magnetic field B derived from a vector potential

$$\vec{A} = \frac{1}{2} \vec{B} \times \vec{r}$$

and an electric potential

$$V(r) = \Phi_0\,u(r)$$

Then the parameters for A and V define a 4-dim. parameter vector

$$\vec{R} = \{B_i, \Phi_0 \}$$

For each set of parameters we define the eigenstates

$$(H[\vec{B},\Phi_0] - E[\vec{B},\Phi_0])\,|\psi[\vec{B},\Phi_0]\rangle = 0$$

In Berry's phase what matters is how the eigenstates depend on the parameters and how this affects the time evolution of a state when the parameters vary slowly with time (adiabatic approximation). To study this effect one has to calculate the integral

$$\gamma_\psi[\mathcal{C}] = \int_{\mathcal{C}} dR\,\langle |\psi[\vec{R}]| i\nabla_R |\psi[\vec{R}]\rangle$$

where C is a curve in the parameter space

$$\vec{R} = \{B_i, \Phi_0 \}$$

Last edited: Oct 16, 2012
10. Oct 16, 2012

### Syrus

Tom.stoer, thank you for the helpful information. Can you expand on the 4 dimensional parameter vector R you define above? Is this supposed to be regarded as (some kind of) function of these parameters?

11. Oct 16, 2012

### tom.stoer

what do you mean by "expand"?

12. Oct 16, 2012

### Syrus

I'm sorry, can you elaborate. I also edited my previous post a bit. Which four parameters are these? Are they B, r, r and phi?

13. Oct 16, 2012

### tom.stoer

As I said

with i=1,2,3. Three constants B1, B2 and B3 for the magnetic field and one for the electric potential. These are simply four parameters in the definition of the Hamiltonian. They are constants. Perhaps it's confusing to use the letter R which seems to indicate that it has something to do with a position; it hasn't.

Let's make another example, the d-dim. anisotropic harmonic oscillator

$$H[\vec{\omega}] = \frac{\vec{p}^2}{2m} + \frac{m}{2}\sum_{i=1}^d \omega_i^2 x_i^2$$

with i=1,2,...d. Now the d-dim. parameter vector is simply

$$\vec{R} = \{\omega_1,\omega_2,\ldots,\omega_d\}$$

The eigenfunctions and eigenvalues depend on these parameters, that means for different frequencies we get different wave functions and different energies.

Again: the parameter vector R and a position space vector x have nothing to do with each other

14. Oct 16, 2012

### Syrus

Ah, I see. So if the parameter vector is a function of time (as is the case in Berry's paper), then its components must also be functions of time. Is this true?

15. Oct 16, 2012

### tom.stoer

Yes, the parameters become time dependent in Berry's approach

16. Oct 17, 2012

### Syrus

Tom.Stoer, I understand the concept of R(t). However, how is the nabla operator enacted on a vector (or vector-based quantity) such as, for instance, n(R(t))?

I am familiar with the del operator acting on scalar functions.

For example, the inner product in (6) yields a complex scalar (by the definition of inner product of a complex vector space). but how can this be dotted with the line differential?

Last edited: Oct 17, 2012
17. Oct 17, 2012

### tom.stoer

Think about a simple, finite-dimensional example where e.g. a fermion (with fixed location i.e. with vanishing momentum) is coupled to a uniform magnetic field with components Bi, i=1,2,3 using Pauli matrices σi:

$$H[B_i] = \mu\sum_i\sigma_iB_i$$

The Schrödinger wave function is replaced by an algebraic equation

$$(H[B_i]-E[B_i])u{[B_i]} = 0$$

where u is a 2-component spinor (w/o dependency on x,y,z) representing the fermion spin.

There are two solutions, i.e. eigenvectors u1,2 and eigenvalues E1,2 for a 2*2-matrix eigenvector problem (with a=1,2 numbering the solution - just another index ;-).

Each solution ua can be written as 2-component spinor

$$u[B_i] = \begin{pmatrix} u_+[B_i] \\ u_-[B_i] \end{pmatrix}$$

where each component depends on the parameters Bi. That means that for different magnetic field strengths you get different solutions to the 2*2-matrix eigenvector problem.

Solutions to this problem can be found in many QM textbooks, e.g. Sakurai.

In order to follow Berry's approach you have to calculate

$$a_k[B_i] =\begin{pmatrix} u^\ast_+[B_i] \\ u^\ast_-[B_i] \end{pmatrix} i \frac{\partial}{\partial B_k}\begin{pmatrix} u_+[B_i] \\ u_-[B_i] \end{pmatrix}$$

where i=1,2,3 indicates that the three functions ak with k=1,2,3 still depend on the magnetic field Bi.

The next step would then be to calculate the integral for a specific curve C in the 3-dim. B-space.

Last edited: Oct 17, 2012
18. Oct 17, 2012

### Syrus

Yes, but you've simply taken the partial derivate (in your last expression), not the gradient? Does the spinor not represent |n(R(t))> here?

19. Oct 17, 2012

### tom.stoer

There are three partial derivative k=1,2,3 resulting in three functions ak. This is the gradient!

$$\nabla_{{}_B} = \left( \frac{\partial}{\partial B_k} \right)$$

You are right, the spinor u represents |n(R)>; I simply wanted to provide an example where the gradient does not act on an abstact ket |n(R)> but on something that is rather familiar to everybody and that can be calculated directly w/o taking gradients in infinite dimensional Hilbert spaces ;-)

R represents the parameters for B.

Note that up to now I have not introduced any time-dependency, therefore I omitted the 't'

20. Oct 17, 2012

### Syrus

Ah, but then shouldn't the resulting quantity be, in a sense, a (3 component) vector including the differentiated spinors?

21. Oct 17, 2012

### tom.stoer

There is a scalar product for the spinors.

Let's look at k=1 for B1

$$a_1[B_i] =\begin{pmatrix} u^\ast_+[B_i] \\ u^\ast_-[B_i] \end{pmatrix} i \frac{\partial}{\partial B_1}\begin{pmatrix} u_+[B_i] \\ u_-[B_i] \end{pmatrix} = u^\ast_+[B_i] i \frac{\partial}{\partial B_1} u_+[B_i] + u^\ast_-[B_i] i \frac{\partial}{\partial B_1} u_-[B_i]$$

The same applies to k=2,3. You have three equations for k=1,2,3 and therefore you have three functions ak

22. Oct 17, 2012

### Syrus

So does ∇R refer to simply one partial derivative? Namely, that with respect to R? Or is this the gradient with partial derivative with respect to the parameters contained in R?

23. Oct 17, 2012

### Syrus

I'm interested in seeing what the expression∇R|n(R(t))> looks like in your example above, and how the direct product with <n(R(t)| is taken.

24. Oct 17, 2012

### tom.stoer

Are you familiar with the gradient? There are three partial derivatives ∂k w.r.t. the three components Bk. These three partial derivatives define the gradient in B-space (in my example) or in R-space (in the general case). What is unclear? What is missing?

This is shown explicitly in post #21. Instead of the symbol |n(R)> I use u[Bi] just to indicate that:
- we are not using an abstract ket |n> but a two-dim. object u
- we are not discussing the general R-space but a rather specific example with parameters Bi

You apply the ∂k to each component u±; each u± is a function of the parameters Bi; then you calculate the scalar product. Again: What is unclear? What is missing?

The formal replacement is just

$$\langle n(R)|\nabla_R|n(R)\rangle \;\to\;a_k[B_i];\,k=1,2,3$$

Do you have problems with the algebra? It seems that you miss a rather trivial point how to differentiate vector-valued functions or something like that.

Last edited: Oct 17, 2012
25. Oct 17, 2012

### Dickfore

Jesus, you are obtuse. $\vert n(\mathbf{R}(t) \rangle$ is a state ket, a vector in a Hilbert space, that depends on a set of time dependent (but slowly varying in the adiabatic approximation) set of parameters, that are grouped as a k-dimensional vector (with k being the number of parameters) $\mathbf{R}$. The symbol $\nabla_{\mathbf{R}}$ refers to differentiating the state ket w.r.t. each parameter, i.e. it becomes "a vector of kets". If there is one parameter, this vector has only one ket component. For example, look at the stationary wavefunctions for a particle in a box:
$$\psi_{n}(x) = \langle x \vert n \rangle = \sqrt{\frac{2}{a}} \, \sin \left( \frac{n \, \pi \, x}{a} \right)$$
Here, a is a parameter. Different w.r.t. it, we get
$$\langle x \vert \nabla_{a} \vert n \rangle = -\frac{1}{a \, \sqrt{2 \, a}} \, \sin \left( \frac{n \, \pi \, x}{a} \right) - \frac{n \, \pi \, x}{a^2} \, \sqrt{\frac{2}{a}} \, \cos\left( \frac{n \, \pi \, x}{a} \right)$$
Now, if you want to take a scalar product with another bra, you have to integrate w.r.t. x from 0 to a.

Last edited: Oct 17, 2012