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Hello.

In the following(p.2):

https://michaelberryphysics.files.wordpress.com/2013/07/berry187.pdf

Berry uses parallel transport on a sphere to showcase the (an)holonomy angle of a vector when it is parallel transported over a closed loop on the sphere.

A clearer illustration of this can be found in(p.2):

http://phy.ntnu.edu.tw/~changmc/Paper/Berry_IFF_FF_4.pdf

where on p.5 the author(Chang) gives the geometric properties that have to do with the Berry phase through the analogy with the case of parallel transport on a sphere.

The authors say that $$\vec{\psi}$$ is parallel transported around the sphere and $$\vec{n}$$ is a fixed vector and they are related by $$\vec{\psi}=e^{-ia(t)}\vec{n}$$ where a(t) is the (an)holonomy angle between them(the change in the angle of Ψ as it is parallel transported around the curved surface).

In the quantum analogue of the above, the authors conclude that the parallel transport condition(as they call it):

$$<\psi |\dot{\psi}>=0$$

, where $$|\psi>=e^{i\int_{0}^{t}A(t')dt'}|n>$$ where |n> is an energy eigenstate and A(t) is the Berry connection) gives the anholonomy angle between the two, which is Berry's phase.

Now, if we substitute the expression for the parallel transported state into the parallel transport condition, we get:

$$<n|\dot{n}>+iA=0$$.

In the following by Haldane in slide 18:

http://wwwphy.princeton.edu/~haldane/talks/Vancouver_pitp05.pdf

Haldane defines

$$<n|Dn>=<n|\dot{n}>+iA$$ (his |Ψ> is basically an energy eigenstate, which is |n> in my notation, and he puts a minus in front of the typical definition of the Berry connection) and he calls $$|D\phi>$$ the covariant derivative of a state |φ>. The above inner product is, by construction, equal to zero.

So, finally, the problem is the following:

Berry and Chang (in the first two papers) say that it is |Ψ> that is parallel transported, while from $$<n|Dn>=0$$ of Haldane, it seems like it is |n> that is parallel transported(since, in differential geometry, the vanishing of the covariant derivative of a vector along a curve means that the vector is parallel transported along that curve).

So, what is happening here?

[I tried to continue the analogy put forth by Berry and Chang of the parallel transport along a sphere, but in no way is the following true(in their notation):

$$\vec{n}^{*} \cdot \nabla \vec{n}=0$$. Note, if it helps, that $$\vec{e_i} \cdot \nabla \vec{e_i}=0$$ if we compute the covariant derivative properly with the Chrystoffel symbols. Maybe this is the point at which the analogy between anholonomy from parallel transport along the sphere and anholonomy as in Berry phase breaks down?]

Thanks in advance and sorry for the post being so long.

In the following(p.2):

https://michaelberryphysics.files.wordpress.com/2013/07/berry187.pdf

Berry uses parallel transport on a sphere to showcase the (an)holonomy angle of a vector when it is parallel transported over a closed loop on the sphere.

A clearer illustration of this can be found in(p.2):

http://phy.ntnu.edu.tw/~changmc/Paper/Berry_IFF_FF_4.pdf

where on p.5 the author(Chang) gives the geometric properties that have to do with the Berry phase through the analogy with the case of parallel transport on a sphere.

The authors say that $$\vec{\psi}$$ is parallel transported around the sphere and $$\vec{n}$$ is a fixed vector and they are related by $$\vec{\psi}=e^{-ia(t)}\vec{n}$$ where a(t) is the (an)holonomy angle between them(the change in the angle of Ψ as it is parallel transported around the curved surface).

In the quantum analogue of the above, the authors conclude that the parallel transport condition(as they call it):

$$<\psi |\dot{\psi}>=0$$

, where $$|\psi>=e^{i\int_{0}^{t}A(t')dt'}|n>$$ where |n> is an energy eigenstate and A(t) is the Berry connection) gives the anholonomy angle between the two, which is Berry's phase.

Now, if we substitute the expression for the parallel transported state into the parallel transport condition, we get:

$$<n|\dot{n}>+iA=0$$.

In the following by Haldane in slide 18:

http://wwwphy.princeton.edu/~haldane/talks/Vancouver_pitp05.pdf

Haldane defines

$$<n|Dn>=<n|\dot{n}>+iA$$ (his |Ψ> is basically an energy eigenstate, which is |n> in my notation, and he puts a minus in front of the typical definition of the Berry connection) and he calls $$|D\phi>$$ the covariant derivative of a state |φ>. The above inner product is, by construction, equal to zero.

So, finally, the problem is the following:

Berry and Chang (in the first two papers) say that it is |Ψ> that is parallel transported, while from $$<n|Dn>=0$$ of Haldane, it seems like it is |n> that is parallel transported(since, in differential geometry, the vanishing of the covariant derivative of a vector along a curve means that the vector is parallel transported along that curve).

So, what is happening here?

[I tried to continue the analogy put forth by Berry and Chang of the parallel transport along a sphere, but in no way is the following true(in their notation):

$$\vec{n}^{*} \cdot \nabla \vec{n}=0$$. Note, if it helps, that $$\vec{e_i} \cdot \nabla \vec{e_i}=0$$ if we compute the covariant derivative properly with the Chrystoffel symbols. Maybe this is the point at which the analogy between anholonomy from parallel transport along the sphere and anholonomy as in Berry phase breaks down?]

Thanks in advance and sorry for the post being so long.

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