# Berry phase and parallel transport

• A
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.

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Anybody?

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

$$\langle \psi \vert \dot{\psi} \rangle = 0$$

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.
I'm not seeing this. What specific equations in the Chang paper are you looking at?

PeterDonis
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his |Ψ> is basically an energy eigenstate
I'm not seeing this either. It looks to me like Haldane's ##\vert \Psi \rangle## is a generic state, not an energy eigenstate.

PeterDonis
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a fixed vector
It's worth noting that this term here does not really mean what one might think it means. First, it's not just one "fixed" vector but a whole set of them, one at each point in the space (or at least in the region of the space traversed by the closed curve around which the "moving" vector is parallel transported). Second, these vectors are "fixed" in the sense that their components are pre-determined functions of the coordinates (for example, equation 4 of Berry's original paper gives these functions for the components of the "fixed" vectors on the surface of the sphere). But they are not "fixed" in the sense of parallel transport--if you take the "fixed" vector at one point and parallel transport it to a different point, it will not, in general, match the corresponding "fixed" vector at the second point.

I'm not seeing this. What specific equations in the Chang paper are you looking at?
Chang analyzes the parallel transport on a sphere in pages 2-4. He then goes on to give the quantum analogue of his conclusions in p.4-5.
You can find a table with the analogy between classical geometry and quantum states up in p.5.

He states that $$\vec{n}$$ is a fixed basis (since it is a linear combination of two fixed basis vectors, u and v where these two can be the two coordinates on a spherical shell, as Berry demonstrates in the first article).

I'm not seeing this either. It looks to me like Haldane's ##\vert \Psi \rangle## is a generic state, not an energy eigenstate.
Haldane defines the Berry connection with respect to |Ψ>. But the Berry connection in the theory of Berry phase is defined with respect to |n>.
He always uses |Ψ> to define the Berry connection, curvature etc, which are always defined with respect to the fixed basis |n>.

It's worth noting that this term here does not really mean what one might think it means. First, it's not just one "fixed" vector but a whole set of them, one at each point in the space (or at least in the region of the space traversed by the closed curve around which the "moving" vector is parallel transported). Second, these vectors are "fixed" in the sense that their components are pre-determined functions of the coordinates (for example, equation 4 of Berry's original paper gives these functions for the components of the "fixed" vectors on the surface of the sphere). But they are not "fixed" in the sense of parallel transport--if you take the "fixed" vector at one point and parallel transport it to a different point, it will not, in general, match the corresponding "fixed" vector at the second point.
Yes, I am aware of this, but thanks for pointing this out.
Though, the coordinate vectors that correapond to the usual (φ,θ) coordinates on a sphere that Berry refers to is not a parallel vector field.
It is in this sense that I(and Chang) am saying that it is a fixed basis(not parallel transported) while the Ψ vector field is paralelel(see two previous replies)

PeterDonis
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from

$$\langle n \vert Dn \rangle = 0$$

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).
That equation in Haldane does not mean ##n## is parallel transported. It means that the change ##Dn## is orthogonal to the vector ##n##. That must be the case because ##n## is a unit vector, so its length cannot change, only its direction; a length change would be a change ##Dn## parallel to ##n##, while a direction change is a change ##Dn## orthogonal to ##n##.

The notation differences in these three sources invite confusion. In the Berry and Chang papers, the notations ##\dot{\psi}## and ##\dot{n}## do not refer to covariant derivatives; they refer to derivatives with respect to "moving" around the space. For example, in the case of the sphere, the dots refer to the change in a vector as you move from one point to another on the sphere. But this "change" is not parallel transport, or any other kind of transport; it doesn't tell you how the vector changes from point to point, it just refers to a change in the vector from point to point that is already defined by some other means. So for ##\psi##, the change is determined by parallel transport; while for ##n##, it's determined by the vector's components being a fixed set of predetermined functions of the coordinates. Similar remarks apply to the notations ##d\psi## and ##dn## in those papers.

PeterDonis
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the Berry connection in the theory of Berry phase is defined with respect to |n>.
I don't think this is correct. The fixed vector ##n## in the Berry and Chang papers is used to help derive equations for the Berry phase, but that doesn't mean the Berry phase is "defined" in terms of ##n##. These papers are clear that it is the vector ##\psi## that is parallel transported around a closed loop, and that the Berry phase is the change in the phase of ##\psi## during this process. No change in phase of ##n## is discussed at all.

PeterDonis
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That must be the case because ##n## is a unit vector
Note: this is assuming that your interpretation of Haldane's ##\Psi## is correct, which I'm not sure it is.

Note: this is assuming that your interpretation of Haldane's ##\Psi## is correct, which I'm not sure it is.
The definition of Haldane for the Berry connection is:
$$A_{\mu}=<\psi|\partial_{\mu}|\psi>$$
But we know that the Berry connection for the n-th energy eigenstate is given by:
$$A_{\mu}=<n|\partial_{\mu}|n>$$
Doesn't this show that his notation implies that |Ψ>=|n>?
If not, then I don't know why Haldane uses such unconventional notation.

I don't think this is correct. The fixed vector ##n## in the Berry and Chang papers is used to help derive equations for the Berry phase, but that doesn't mean the Berry phase is "defined" in terms of ##n##. These papers are clear that it is the vector ##\psi## that is parallel transported around a closed loop, and that the Berry phase is the change in the phase of ##\psi## during this process. No change in phase of ##n## is discussed at all.
I agree with what you are saying.
When |Ψ> is parallel transported along a curve(where |Ψ(0)>=|n>), then it acquires Berry phase in analogy with a parallel transported vector which rotates(anholonomy angle) in Riemannian geometry.
I totally agree with you.

When I say that the Berry phase for the n-th energy eigenstate is defined with respect to |n>, what I mean is that it is defined(or just can be computed) as:
$$\gamma_n=i\int \vec{A}_n(\vec{R})\cdot d\vec{R}$$
where R are the parameters and A is the Berry connection.
Isn't this the standard way to compute the Berry phase(this and from the Berry curvature of course)?

That equation in Haldane does not mean ##n## is parallel transported. It means that the change ##Dn## is orthogonal to the vector ##n##. That must be the case because ##n## is a unit vector, so its length cannot change, only its direction; a length change would be a change ##Dn## parallel to ##n##, while a direction change is a change ##Dn## orthogonal to ##n##.

The notation differences in these three sources invite confusion. In the Berry and Chang papers, the notations ##\dot{\psi}## and ##\dot{n}## do not refer to covariant derivatives; they refer to derivatives with respect to "moving" around the space. For example, in the case of the sphere, the dots refer to the change in a vector as you move from one point to another on the sphere. But this "change" is not parallel transport, or any other kind of transport; it doesn't tell you how the vector changes from point to point, it just refers to a change in the vector from point to point that is already defined by some other means. So for ##\psi##, the change is determined by parallel transport; while for ##n##, it's determined by the vector's components being a fixed set of predetermined functions of the coordinates. Similar remarks apply to the notations ##d\psi## and ##dn## in those papers.
As I understand it, |dn> stands for the change of |n> that is not confined to the manifold in question(i.e. covariant derivative which is the projection of the derivative on the tangent plane of the manifold). It is like the "normal" derivative with respect to the parameter with which we parametrize our curve. It is the analogue of dc/dt for a curve c which is parametrized by the parameter t. I think we are saying the same thing.

Now, on the interpretation of |Dn>, I had interpreted it as the covariant derivative of |n>(projection to the tangent plane of the manifold) because Haldane in the following presentation on slide 20, says that $$<\psi|D_{\mu}|\psi>=0$$ is a "gauge-invariant parallel transport condition":
http://wwwphy.princeton.edu/~haldane/talks/pccmsummerschool2012_haldane.pdf

I also agree with how you interpret the above condition, but what I still don't understand is why it is not a parallel transport condition for |n>(as Haldane puts it).

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Demystifier
Gold Member
Maybe it's about passive (changing the basis) vs active (changing the physical state) interpretation of the transformation?

Joker93
Maybe it's about passive (changing the basis) vs active (changing the physical state) interpretation of the transformation?
This is a gret idea, although I will have to think about it for a bit.

PeterDonis
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As I understand it, |dn> stands for the change of |n> that is not confined to the manifold in question
I'm not sure what you mean by this. ##n## is a vector in the tangent space of the manifold at a particular point. The notion of a change in ##n## that is not confined to the manifold doesn't make sense.

I'm not sure what you mean by this. ##n## is a vector in the tangent space of the manifold at a particular point. The notion of a change in ##n## that is not confined to the manifold doesn't make sense.
Take the case of the sphere again. When the sphere is embedded in R^3 then there is a distinction between the "normal" derivative of a curve on the sphere and there is also its projection to the tangent plane of the sphere, which is the covariant derivative(intrinsic derivative).
Otherwise, what is the distinction between |dn> and |Dn>?

PeterDonis
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When the sphere is embedded in R^3 then there is a distinction between the "normal" derivative of a curve on the sphere and there is also its projection to the tangent plane of the sphere, which is the covariant derivative
The covariant derivative has nothing to do with the embedding of the sphere in R^3; it's defined purely in terms of parallel transport within the sphere considered as a self-contained manifold.

I'm not sure what you mean by the "normal" derivative of a curve on the sphere. Derivative with respect to what?

The covariant derivative has nothing to do with the embedding of the sphere in R^3; it's defined purely in terms of parallel transport within the sphere considered as a self-contained manifold.

I'm not sure what you mean by the "normal" derivative of a curve on the sphere. Derivative with respect to what?
For a curve on a sphere, a(t) where t is the parameter with which we parametrize the curve and a'(t) the tangent vector to the curve, we have da'/dt which can have non-zero component along the direction that is normal to the local tangent plane and we also have the covariant derivative of c, denoted as Da'/dt which is the projection of da'/dt on the local tangent plane, thus giving Da'/dt its intrinsic nature. But, we can only talk about da'/dt when considering the manifold(here a sphere) as being embedded in another space(here R^3).

You can see it here:
where a''(t)=d(a')/dt=d/dt(da/dt) and D(a')/dt is the covariant derivative.

PeterDonis
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For a curve on a sphere, a(t) where t is the parameter with which we parametrize the curve and a'(t) the tangent vector to the curve, we have da'/dt which can have non-zero component along the direction that is normal to the local tangent plane
There seems to be a confusion here between a parameterization of the curve in the higher dimensional space (in this case R^3), and a parameterization of the curve within the manifold itself (in this case the sphere). They are not the same. I see what is meant by a projection into the tangent plane of the derivative with respect to a curve parameter in the higher dimensional space; but the observation that the covariant derivative can be viewed as such a projection is just an observation (and only holds in the case where an isometric embedding into a higher dimensional space exists). It is not the definition of the covariant derivative; that's entirely intrinsic to the manifold.

Jimster41
There seems to be a confusion here between a parameterization of the curve in the higher dimensional space (in this case R^3), and a parameterization of the curve within the manifold itself (in this case the sphere). They are not the same. I see what is meant by a projection into the tangent plane of the derivative with respect to a curve parameter in the higher dimensional space; but the observation that the covariant derivative can be viewed as such a projection is just an observation (and only holds in the case where an isometric embedding into a higher dimensional space exists). It is not the definition of the covariant derivative; that's entirely intrinsic to the manifold.
I see.
But, if you think about it, in the example that Berry and Chang use, they are talking about normal vectors to the local tangent plane. So, they are building their whole analogy based on a sphere in ##R^3##, right?

Also, if we are not to view the covariant derivative in this way, then what is the conceptual distinction between ##|dn>## and ##|Dn>##? Haldane seems to distinguish them from each other by stating that ##|D\psi>=|d\psi>-iA|\psi>## is the covariant derivative.

Lastly, note that what motivates me to speak about the two "kinds" of derivatives in terms of an ambient space is due to the structure of ##|DΨ>## as given above:
In Riemannian geometry, the covariant derivative has the analogous definition, which is just the "normal" derivative minus its component that is normal to the tangent plane. This is illustrated as follows:

Let ##\vec{u}=u^{\nu}\vec{e}_{\nu}## be a vector that belongs to a local tangent plane on a manifold and ##\vec{e}_i## be the local orthonormal basis of that tangent plane.
The derivative of the above vector is given by:
$$\partial_{\mu}\vec{u}=\partial_{\mu}u^{\nu}\vec{e}_{\nu} + u^{\nu}\partial_{\mu}(\vec{e}_{\nu})$$
Now, we can decompose ## \partial_{\mu}\vec{e}_{\nu}## in tangential and normal components as follows:
$$\partial_{\mu}\vec{e}_{\nu}=\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}+K_{\mu \nu}\vec{n}$$ where ##\vec{n}## is the normal vector to the tangent plane.
So, $$\partial_{\mu}\vec{u}=\partial_{\mu}u^{\nu} + u^{\nu}\partial_{\mu}(\vec{e}_{\nu})=\partial_{\mu}u^{\nu}\vec{e}_{\nu}+u^{\nu}\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}+u^{\nu}K_{\mu \nu}\vec{n}$$
So, now we can define the covariant derivative as the "derivative of the ambient space(here ##R^3##) minus its component that is normal to the local tangent plane:
$$D_{\mu}\vec{u}=\partial_{\mu}\vec{u}-u^{\nu}K_{\mu \nu}\vec{n}=\partial_{\mu}u^{\nu}\vec{e}_{\nu}+u^{\nu}\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}=(\partial_{\mu}u^{\sigma}+u^{\nu}\Gamma_{\mu \nu}^{\sigma})\vec{e}_{\sigma}$$
With the above definition, we can see that ##D_{\mu}\vec{e}_{\nu}=\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}##, which agrees with the what we get from Riemannian geometry.

As you can see, ##D_{\mu}\vec{u}=\partial_{\mu}\vec{u}-u^{\nu}K_{\mu \nu}\vec{n}## has very similar "structure" to ## |Dn>=|dn>-iA|n> ##. So, this is what motivated me to think in these lines.Also, from the definition of the Berry connection, ##A_{\mu}=-i<n|\partial_{\mu}|n>## or just ##A=-i<n|dn>, |Dn>## can be geometrically interpreted as just ##|dn>## minus the part of ##|dn>## that is along ##|n>##.

P.S. Sorry for the long post!

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Jimster41
I see.
But, if you think about it, in the example that Berry and Chang use, they are talking about normal vectors to the local tangent plane. So, they are building their whole analogy based on a sphere in ##R^3##, right?

Also, if we are not to view the covariant derivative in this way, then what is the conceptual distinction between ##|dn>## and ##|Dn>##? Haldane seems to distinguish them from each other by stating that ##|D\psi>=|d\psi>-iA|\psi>## is the covariant derivative.

Lastly, note that what motivates me to speak about the two "kinds" of derivatives in terms of an ambient space is due to the structure of ##|DΨ>## as given above:
In Riemannian geometry, the covariant derivative has the analogous definition, which is just the "normal" derivative minus its component that is normal to the tangent plane. This is illustrated as follows:

Let ##\vec{u}=u^{\nu}\vec{e}_{\nu}## be a vector that belongs to a local tangent plane on a manifold and ##\vec{e}_i## be the local orthonormal basis of that tangent plane.
The derivative of the above vector is given by:
$$\partial_{\mu}\vec{u}=\partial_{\mu}u^{\nu}\vec{e}_{\nu} + u^{\nu}\partial_{\mu}(\vec{e}_{\nu})$$
Now, we can decompose ## \partial_{\mu}\vec{e}_{\nu}## in tangential and normal components as follows:
$$\partial_{\mu}\vec{e}_{\nu}=\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}+K_{\mu \nu}\vec{n}$$ where ##\vec{n}## is the normal vector to the tangent plane.
So, $$\partial_{\mu}\vec{u}=\partial_{\mu}u^{\nu} + u^{\nu}\partial_{\mu}(\vec{e}_{\nu})=\partial_{\mu}u^{\nu}\vec{e}_{\nu}+u^{\nu}\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}+u^{\nu}K_{\mu \nu}\vec{n}$$
So, now we can define the covariant derivative as the "derivative of the ambient space(here ##R^3##) minus its component that is normal to the local tangent plane:
$$D_{\mu}\vec{u}=\partial_{\mu}\vec{u}-u^{\nu}K_{\mu \nu}\vec{n}=\partial_{\mu}u^{\nu}\vec{e}_{\nu}+u^{\nu}\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}=(\partial_{\mu}u^{\sigma}+u^{\nu}\Gamma_{\mu \nu}^{\sigma})\vec{e}_{\sigma}$$
With the above definition, we can see that ##D_{\mu}\vec{e}_{\nu}=\Gamma_{\mu \nu}^{\sigma}\vec{e}_{\sigma}##, which agrees with the what we get from Riemannian geometry.

As you can see, ##D_{\mu}\vec{u}=\partial_{\mu}\vec{u}-u^{\nu}K_{\mu \nu}\vec{n}## has very similar "structure" to ## |Dn>=|dn>-iA|n> ##. So, this is what motivated me to think in these lines.Also, from the definition of the Berry connection, ##A_{\mu}=-i<n|\partial_{\mu}|n>## or just ##A=-i<n|dn>, |Dn>## can be geometrically interpreted as just ##|dn>## minus the part of ##|dn>## that is along ##|n>##.

P.S. Sorry for the long post!
Sorry, forget about the last paragraph.
What I wanted to show is that ##D_{\mu}\vec{u}=(\partial_{\mu}u^{\sigma}+u^{\nu}\Gamma_{\mu \nu}^{\sigma})\vec{e}_{\sigma}## is the analogous of ## |D\phi>=|d\phi>+iA|\phi> ## for a state ##|\phi>## if we define ##A_{\mu}=i<n|\partial_{\mu}|n>##.
I was wrong to say that it is ##D_{\mu}\vec{u}=\partial_{\mu}\vec{u}-u^{\nu}K_{\mu \nu}\vec{n}## that is analogous to ##|D\phi>=|d\phi>+iA|\phi> ##.
I was also trying to show what I said about the derivative and the covariant derivative through the relation ##D_{\mu}\vec{u}=\partial_{\mu}\vec{u}-u^{\nu}K_{\mu \nu}\vec{n}##

Jimster41