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[Reference: Modern Quantum Mechanics, J.J. Sakurai, Chapter 5]

The Berry potential is defined by,

$$ \mathbf{A}_{n} (\mathbf{R}) \ = \ i \langle n | \mathbf{\nabla}_{\mathbf{R}}|n\rangle$$

Here, $\mathbf{R}$ is the vector of parameters that captures the time dependence in the Hamiltonian, i.e., $H(t) = H(\mathbf{R}(t))$ and $\mathbf{\nabla}_{\mathbf{R}}$ is the gradient defined "with respect to $\mathbf{R}$".

Now, I am not sure what explicitly this gradient means (i.e.$\mathbf{\nabla}_{\mathbf{R}} = ?$) nor how to calculate it. However, my text says that this is just the gradient defined in the $\mathbf{R}$-space; therefore, I expect it to be like this:

$$ \mathbf{\nabla}_{\mathbf{R}} = \hat{e}_{i}^{\mathbf{R}} \frac{\partial}{\partial R_{i}}$$

Here, $R_{i}$ is the $i$-th component of the vector and $\hat{e}_{i}^{\mathbf{R}}$ denotes the unit vector in that direction.

Now, I have 2 questions:

  1. Am I correct in explicitly writing out $\mathbf{\nabla}_{\mathbf{R}}$ the way I have done?

  2. Consider the case of the spin - 1/2 system in a magnetic field (which is $\mathbf{R}$ in this case). The Hamiltonian is $H = -\alpha\ \mathbf{R}\cdot \mathbf{S}$.

Here, while explicitly calculating the Berry potential through its definition, the standard argument is that the eigenstates are independent of the magnitude of the field, and depend only on the angular coordinates. Since, the field itself is parametrized by our real world spatial coordinates, this just means that the states depend on the $\theta, \ \phi$ configurations of the field in our 3D space. The gradient is therefore just the normal gradient in 3D. Now, the normal gradient in spherical coordinates is just,

$$ \nabla = \hat{r} \frac{\partial}{\partial r} + \frac{\hat{\theta}}{r}\frac{\partial}{\partial \theta} + \frac{\hat{\phi}}{r sin\theta} \frac{\partial}{\partial \phi}$$

However, the text sets $r = 1$ without any reason (i.e., $\nabla_{B_{\theta}} = \hat{\theta}\frac{\partial}{\partial \theta} $ and $\nabla_{B_{\phi}} = \frac{\hat{\phi}}{sin\theta}\frac{\partial}{\partial \phi} $). I cannot understand why this should be so.

Further, if I assume that the answer to (1) is "yes", then shouldn't we have something like the following? $$ \nabla_{B_{x_{1}}} = \hat{x_{1}} \frac{\partial}{\partial B_{x_{1}}} = \hat{x_{1}} \frac{\partial}{\partial x_{i}} \frac{\partial x_{i}}{\partial B_{x_{1}}}$$

Thanks in advance!

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1 Answer 1

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  1. Yes
  2. To quote the relevant passage of Sakurai:

Let us continue our discussion by taking as an example, the problem of the spin in a magnetic field mentioned earlier (spin magnitude $S$, and magnetic moment $g\mu\mathbf S/\hbar$. When the magnetic field $\mathbf B(t)$ slowly changes its direction while keeping the magnitude constant [...]

Writing $\mathbf B = B \mathbf R$, the parameter space of all $\mathbf R$'s is not $\mathbb R^3$ (the set of all possible magnetic field vectors) but rather $\mathrm S^2$ (the set of all unit vectors). In the standard coordinate system $(\theta,\phi)$ on the 2-sphere, the gradient is $$\nabla_\mathbf R = \hat \theta \frac{\partial}{\partial \theta} + \frac{\hat \phi}{\sin(\theta)} \frac{\partial}{\partial \phi}$$

which, as you say, can be obtained from the gradient in spherical coordinates by setting $r=1$.

Further, if I assume that the answer to (1) is "yes", then shouldn't we have something like the following? $$ \nabla_{B_{x_{1}}} = \hat{x_{1}} \frac{\partial}{\partial B_{x_{1}}} = \hat{x_{1}} \frac{\partial}{\partial x_{i}} \frac{\partial x_{i}}{\partial B_{x_{1}}}$$

No - what is $x_1$? The components of $\mathbf R$ (or $\mathbf B$, if you prefer) are the coordinates in the parameter space. There are no spatial coordinates $x_i$ involved.

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  • $\begingroup$ So, when we write $\mathbf{B} = B\mathbf{R}$, we are implying that $\mathbf{R} = (\hat{R}, \hat{\theta}, \hat{\phi})$? Also, I wrote down that last bit because the field components depend on the spatial coordinates, therefore if we wanted to express the gradient in terms of spatial coordinates (which is what I felt we were doing when taking the $\theta$, $\phi$ derivatives), we could use the chain rule. $\endgroup$
    – ShKol
    May 13, 2023 at 18:49
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    $\begingroup$ @ShKol No. The Hamiltonian is (proportional to) $\mathbf B \cdot \mathbf S$, so the parameter space is the set of all $\mathbf B$'s with fixed magnitude, which is a 2-sphere of radius $|\mathbf B|$. An element of that space is a $3$-vector with components $(B_x,B_y,B_z)$, but because we are keeping $|\mathbf B|$ fixed, these components are not independent of one another. The standard way to write the gradient on the 2-sphere of radius $B$ is $\nabla_{\mathbf B} = \frac{1}{B}(\nabla_\mathbf R)$, where $\nabla_\mathbf R$ is the expression above with the angular derivatives. $\endgroup$
    – J. Murray
    May 13, 2023 at 19:58
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    $\begingroup$ "Also, I wrote down that last bit because the field components depend on the spatial coordinates [...]" Not if you're talking about the cartesian components $(B_1,B_2,B_3)$. This is crucial - the $\mathbf B$ field you are applying to your system is spatially uniform. The angles $\theta,\phi$ do not refer to the location of a point in physical space around your particle - they are parameters which set the direction of the $\mathbf B$ field. They don't depend on any spatial location - they are independent parameters which you, the experimenter, set and manipulate by turning some knobs. $\endgroup$
    – J. Murray
    May 13, 2023 at 20:02
  • $\begingroup$ Thanks for the help! I had overlooked the fact that the field was uniform - so, when we are taking derivatives here, the field doesn't change anywhere, but we are looking at the change in the Hamiltonian with respect to the field (which is specified by those angles because the magnitude is kept fixed). $\endgroup$
    – ShKol
    May 14, 2023 at 3:53
  • $\begingroup$ @ShKol Yes, that's right. $\endgroup$
    – J. Murray
    May 14, 2023 at 4:13

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