Thanks! I think I’m confused on a more basic step, which is, like, what exactly is the purported invariance? Consider a Shrödinger-style wave function on a configuration-space with two particles (that have positions, so 6 dimensions total). I know what it means that this wave-function is phase-invariant (if I rotate all amplitudes at once then the dynamics don’t change). What exactly would it mean for it to be “locally phase-invariant”, though?
As a sub-question, what exactly is the data of the local phase invariance? A continuous function from space to U(1)? Two continuous functions, because we have two particles? Am I not even supposed to try to name the data before switching over to thinking of a whole particle field, instead of “two particles”?
As a second sub-question, how exactly do I apply this data to the wave-function? I know how to multiply a wave-function by a single (global) phase, but I don’t know how to multiply a wave-function by (say) a function-from-space-to-U(1).
(I am aware that these questions aren’t quite in the usual ontology of QFT; this is part of why I haven’t been able to quickly extract the answers to my questions from the texts.)
You don’t need QFT here, gauge invariance is a thing even for non-relativistic quantum charged particles moving in a background electromagnetic field. The gauge transformation group consists of (sufficiently regular) functions g:R4→U(1). The transformation law of the n-particle wavefunction is:
ψ′(x1…xn,t)=n∏i=1g(xi,t)qiψ(x1…xn,t)
Here, qi is the electric charge of the i-th particle, in units of positron charge.
In math-jargony terms, the wavefunction is a section of the line bundle
W:=n⨂i=1(pri×idR)∗L⊗qi
Here, pri:(R3)n→R3 is the projection to the position of the i-th particle and L is the “standard” line bundle on R4 on which the electromagnetic field (the 4-potential A, which is useful here even though the setting is non-relativistic) is a connection.W has an induced connection, and the electromagnetic time-dependent Shroedinger equation is obtained from the ordinary time-dependent Shroedinger equation by replacing ordinary derivatives with covariant derivatives.
...actually, having n factors of ψ feels surprising to me; like, this map doesn’t seem to be the identity when g is trivial; did you forget some parentheses? (Or did I misunderstand the parenthesis-insertion conventions?)
re: bundles, I don’t get it yet, and I’d appreciate some clarification. Perhaps I’m simply being daft, but on my current understanding, ψ:(R3)n×R→C, and I’m not quite seeing which bundle it’s supposed to be a section of.
Like, my understanding of bundles is that a bundle constitutes some total space T and some base space B and a projection π:T→B, with a section being a map back σ:B→T such that π∘σ=idB.
Presumably W here is supposed to be the projection π, and ψ is supposed to be the section σ, with the total space and base space left implicit. But presumably the total space isn’t C, and so I’m struggling to see how ψ could have type B→T, so I’m clearly missing something. What’s the base space? What’s the total space? How do we re-view ψ as being a map from the base space back up to the total space? In what sense is ψ inverted by W?
Your guess is exactly what I meant. The ψ is outside the product, otherwise this expression is not even a valid group action.
Now, about bundles.
As you said, a bundle over a manifold B is another manifold T with a projection π:T→B s.t. locally it looks like a product. Formally, every p∈B should have an open neighborhood U s.t. there is a diffeomorphism between π restricted to π−1(U) and a projection pr:U×F→U for some manifold F (the “fiber”).
A vector bundle is a bundle equipped with additional structure that makes every fiber a vector space. Formally, we need to have a smooth addition mapping +:T×BT→T and a multiplication-by-scalar mapping ⋅:C×T→T which are (i) morphisims of bundles and (ii) make every fiber (i.e. the inverse π-image of every point in B) into a vector space. Here, ×B stands for the fibre product (the submanifold of T×T given by π(x)=π(y)). I’m using C here because we will need complex vector bundles.
A line bundle is just a vector bundle s.t. every fiber is 1-dimensional.
A Hermitian vector bundle is a vector bundle equipped with a smooth mapping of bundles ⟨⟩:T×BT which makes every fibre into an inner product space.
Onward to quantum mechanics. Let X be physical space and Y:=X×R physical spacetime. In the non-relativistic setting, Y is isomorphic to R4, so all Hermitian line bundles over Y are isomorphic. So, in principle any one of them can be identified with the trivial bundle: total space Y×C with π being the canonical projection. However, it is often better to imagine some Heremitian line bundle L without such an identification. In fact, choosing an identification precisely corresponds to choosing a gauge. This is like how all finite dimensional real vector spaces are isomorphic to Rn but it is often better not to fix a particular isomorphism (basis), because that obscures the underlying symmetry group of the problem. For finite dimensional vector spaces, the symmetry group is the automorphisms of the vector space (a group isomorphic to GL(n)), for bundles it is the automorphism group of the bundle (= the group of gauge of transformations).
So, let’s fix a Hermitian line bundle L on Y. This allows constructing a Hermitian line bundle W on Xn×R (where n is the number of particles) using the equation I gave before. That equation involves the operations of tensor product and pullback-by-mapping for bundles. I can explain, but maybe you can guess how they are defined (just imagine what it should do to every fibre, and then there is only one reasonable way to “glue” it together). If we fix an isomorphism between L and the trivial bundle over Y (=gauge) then it induces an isomorphism between W and the trivial bundle over Xn×R. In this picture, saying that ψ is a section of W amounts to saying it is a mapping Xn×R→Xn×R×C which is compatible with the projection. The latter condition just means it is the identity on the Xn×R component of the output, so all the information is in the C component on the output, reducing it to a mapping Xn×R→C.
This way, in every particular gauge the wavefunction is just a complex function, but there is a sense in which it is better to avoid fixing a gauge and think of the wavefunction as a section of the somewhat abstract bundle W. Just like a vector in a finite dimensional vector space can be thought of as a column of numbers, but often it’s better to think of it as just an abstract vector.
I’m not entirely sure that I follow the construction of W yet.
Let’s figure out the total space. If you just handed me a line bundle L on Y=X×R, and were like “make a bundle on Xn×R”, then the construction that I’d consider most obvious would be to make the total space be the pullback of total-space(L)n such that all of the time-coordinates agree...
...ah, but that wouldn’t be a line bundle; the tangent space would be n-dimensional. I see.
You suggested starting by considering what happens to an individual fiber, which… is an easier operation to do when I already know what the total space is, but whatever, a particular fiber is isomorphic to C, and it’s over a (x1,…,xn,t):Xn×R, and… yeah I’m not seeing how this helps me figure out the total space.
A third stab at figuring out the total space of W: maybe we take Xn×L, and now we have (in some sense) an “extra” Y that we want to somehow cancel out. If X were equipped with some binary operation ⋅, then the projection from Xn×L to Xn×R could, like, first project out (x1,…,xn) from the left, and then project out (x,t) from L, and then produce (x1⋅x,…,xn⋅x,t). But I don’t see any reasonable “multiplication” operation to fill that role.
I could maybe figure it out if I stared at it longer (as does sound fun), but my first three idiot ideas off the top of my head have not succeeded at extracting an answer to my question “what is the total space of W?” from your text.
(Though, for the record, it makes sense to me how a wave function ψ is a section of the trivial line bundle Xn×R×C, and how we can more generally ask for sections of line bundles over Xn×R and thereby get a more abstract theory. The part that’s not immediately-clicking for me is the part where we can extend a line bundle over Y to a line bundle over Xn×R in an obvious way.)
There are two operations involved in the definition of W: pullback and tensor product.
Pullback is defined for arbitrary bundles. Given a mapping f:X→Y (these X and Y are arbitrary manifolds, not the specific ones from before) and a bundle B over Y with total space TotB and projection mapping prB:Tot(B)→Y, the pullback of B w.r.t. f (denoted f∗B) is the bundle over X with total space Tot(B)×YX and the obvious projection mapping. I remind that Tot(B)×YX is the fibre product, i.e. the submanifold of Tot(B)×X defined by prB(t)=f(x). Notice that the fibre of f∗B over any x∈X is canonically isomorphic to the fibre of B over f(x). The word “canonical” means that there is a particular isomorphism that we obtain from the construction.
It is easy enough to see that the pullback of a vector bundle is a vector bundle, the pullback of a line bundle is a line bundle, and the pullback of a Hermitian vector bundle is a Hermitian vector bundle.
Tensor product is an operation over vector bundles. There are different ways to define it, corresponding to the different ways to define a tensor product of vector spaces. Specifically for line bundles there is the following shortcut definition. Let L1 and L2 be line bundles over X. Then, the total space of L1⊗L2 is the quotient of L1×XL2 by the equivalence relation given by: (v1,v2)∼(w1,w2) iff v1⊗v2=w1⊗w2. Here, I regard v1,w1 as vectors in the vector space which is the corresponding fibre fo L1 and similarly for v2,w2 and L2. The quotient of a manifold by an equivalence relation is not always a manifold, but in this case it is.
I notice that you wrote “a particular fiber is isomorphic to C”. Your error here is, it doesn’t matter what it’s isomorphic to, you should still think of it as an abstract vector space. So, if e.g.V1 and V2 are 1-dimensional vector spaces, then V1⊗V2 is yet another “new” vector space. Yes, they are all isomorphic, but they are not canonically isomorphic.
Thanks! Cool, it makes sense to me how we can make the pullback of L with Xn×R, in n different ways to get n different line bundles, and then tensor them all together. (I actually developed that hypothesis during a car ride earlier today :-p.)
(I’m still not quite sure what the syntax L⊗qi means, but presumably the idea is that there’s an automorphism on 1D vector fields that flips the sign, and we flip the sign of the negative-charge line bundles before tensoring everything together?)
(Also, fwiw, when I said “they’re all isomorphic to C”, I meant that I didn’t expect to figure much out by looking at a single fiber in isolation, and did not mean to imply that there was a canonical isomorphism; it’s clear to me that lacking access to a particular isomorphism is kinda the whole point. That said, I appreciate the pedagogy anyway! I prefer overexplanations to underexplanations whenever my convo-partner is up for generating them.)
The syntax L⊗q means ”L to the tensor power of q”. For q>0, it just means tensoring L with itself q times. For q=0, L⊗q is just the trivial line bundle with total space Y×C (and, yes, all line bundles are isomorphic to the trivial line bundle, but this one just is the trivial bundle… or at least, canonically isomorphic to it). For q<0, we need the notion of a dual vector bundle. Any vector bundle V has a dual V∗, and for a line bundle the dual is also the inverse, in the sense that L⊗L∗ is canonically isomorphic to the trivial bundle. We can then define all negative powers by L⊗q:=(L∗)⊗−q. Notice that non-negative tensor powers are defined for all vector bundles, but negative tensor powers only make sense for line bundles.
It remains to explain what is V∗. But, for our purposes we can take a shortcut. The idea is, for any finite-dimensional complex vector space U with an inner product, there is a canonical isomorphism between U∗ and ¯U, where ¯U is the complex-conjugate space. What is the complex-conjugate space? It is a vector space that (i) has the same set of vectors (ii) has the same addition operation and (iii) has its multiplication-by-scalar operation modified, so that multiplying u by z in ¯U is the same thing as multiplying u by ¯z in U, where ¯z is just the complex number conjugate to z.
Equipped with this observation, we can define the dual of a Hermitian line bundle L to be ¯L, where ¯L is the bundle obtained for L by changing its multiplication-by-scalar mapping in the obvious way.
Vanessa’s reply is excellent, and Steven clearly descibed the gauge transforms of EM that I was gesturing at. That said, if you want to see some examples of fibre bundles in physics other than those in non-relativistic QM, nLab has a great article on the topic.
Also, if you know some GR, the inability to have more than local charts in general spacetimes makes clear that you do need to user bundles other than trivial ones i.e. ones of the form M×T. In my view, GR makes the importance of between global structure = fibre bundle clearer than QM, but maybe you have different intuitions.
Thanks! I think I’m confused on a more basic step, which is, like, what exactly is the purported invariance? Consider a Shrödinger-style wave function on a configuration-space with two particles (that have positions, so 6 dimensions total). I know what it means that this wave-function is phase-invariant (if I rotate all amplitudes at once then the dynamics don’t change). What exactly would it mean for it to be “locally phase-invariant”, though?
As a sub-question, what exactly is the data of the local phase invariance? A continuous function from space to U(1)? Two continuous functions, because we have two particles? Am I not even supposed to try to name the data before switching over to thinking of a whole particle field, instead of “two particles”?
As a second sub-question, how exactly do I apply this data to the wave-function? I know how to multiply a wave-function by a single (global) phase, but I don’t know how to multiply a wave-function by (say) a function-from-space-to-U(1).
(I am aware that these questions aren’t quite in the usual ontology of QFT; this is part of why I haven’t been able to quickly extract the answers to my questions from the texts.)
You don’t need QFT here, gauge invariance is a thing even for non-relativistic quantum charged particles moving in a background electromagnetic field. The gauge transformation group consists of (sufficiently regular) functions g:R4→U(1). The transformation law of the n-particle wavefunction is:
ψ′(x1…xn,t)=n∏i=1g(xi,t)qiψ(x1…xn,t)
Here, qi is the electric charge of the i-th particle, in units of positron charge.
In math-jargony terms, the wavefunction is a section of the line bundle
W:=n⨂i=1(pri×idR)∗L⊗qi
Here, pri:(R3)n→R3 is the projection to the position of the i-th particle and L is the “standard” line bundle on R4 on which the electromagnetic field (the 4-potential A, which is useful here even though the setting is non-relativistic) is a connection.W has an induced connection, and the electromagnetic time-dependent Shroedinger equation is obtained from the ordinary time-dependent Shroedinger equation by replacing ordinary derivatives with covariant derivatives.
Thanks! My top guess was
(n∏i=1g(xi,t)qi)ψ(x1…xn,t),
so I much appreciate the correction.
...actually, having n factors of ψ feels surprising to me; like, this map doesn’t seem to be the identity when g is trivial; did you forget some parentheses? (Or did I misunderstand the parenthesis-insertion conventions?)
re: bundles, I don’t get it yet, and I’d appreciate some clarification. Perhaps I’m simply being daft, but on my current understanding, ψ:(R3)n×R→C, and I’m not quite seeing which bundle it’s supposed to be a section of.
Like, my understanding of bundles is that a bundle constitutes some total space T and some base space B and a projection π:T→B, with a section being a map back σ:B→T such that π∘σ=idB.
Presumably W here is supposed to be the projection π, and ψ is supposed to be the section σ, with the total space and base space left implicit. But presumably the total space isn’t C, and so I’m struggling to see how ψ could have type B→T, so I’m clearly missing something. What’s the base space? What’s the total space? How do we re-view ψ as being a map from the base space back up to the total space? In what sense is ψ inverted by W?
Your guess is exactly what I meant. The ψ is outside the product, otherwise this expression is not even a valid group action.
Now, about bundles.
As you said, a bundle over a manifold B is another manifold T with a projection π:T→B s.t. locally it looks like a product. Formally, every p∈B should have an open neighborhood U s.t. there is a diffeomorphism between π restricted to π−1(U) and a projection pr:U×F→U for some manifold F (the “fiber”).
A vector bundle is a bundle equipped with additional structure that makes every fiber a vector space. Formally, we need to have a smooth addition mapping +:T×BT→T and a multiplication-by-scalar mapping ⋅:C×T→T which are (i) morphisims of bundles and (ii) make every fiber (i.e. the inverse π-image of every point in B) into a vector space. Here, ×B stands for the fibre product (the submanifold of T×T given by π(x)=π(y)). I’m using C here because we will need complex vector bundles.
A line bundle is just a vector bundle s.t. every fiber is 1-dimensional.
A Hermitian vector bundle is a vector bundle equipped with a smooth mapping of bundles ⟨⟩:T×BT which makes every fibre into an inner product space.
Onward to quantum mechanics. Let X be physical space and Y:=X×R physical spacetime. In the non-relativistic setting, Y is isomorphic to R4, so all Hermitian line bundles over Y are isomorphic. So, in principle any one of them can be identified with the trivial bundle: total space Y×C with π being the canonical projection. However, it is often better to imagine some Heremitian line bundle L without such an identification. In fact, choosing an identification precisely corresponds to choosing a gauge. This is like how all finite dimensional real vector spaces are isomorphic to Rn but it is often better not to fix a particular isomorphism (basis), because that obscures the underlying symmetry group of the problem. For finite dimensional vector spaces, the symmetry group is the automorphisms of the vector space (a group isomorphic to GL(n)), for bundles it is the automorphism group of the bundle (= the group of gauge of transformations).
So, let’s fix a Hermitian line bundle L on Y. This allows constructing a Hermitian line bundle W on Xn×R (where n is the number of particles) using the equation I gave before. That equation involves the operations of tensor product and pullback-by-mapping for bundles. I can explain, but maybe you can guess how they are defined (just imagine what it should do to every fibre, and then there is only one reasonable way to “glue” it together). If we fix an isomorphism between L and the trivial bundle over Y (=gauge) then it induces an isomorphism between W and the trivial bundle over Xn×R. In this picture, saying that ψ is a section of W amounts to saying it is a mapping Xn×R→Xn×R×C which is compatible with the projection. The latter condition just means it is the identity on the Xn×R component of the output, so all the information is in the C component on the output, reducing it to a mapping Xn×R→C.
This way, in every particular gauge the wavefunction is just a complex function, but there is a sense in which it is better to avoid fixing a gauge and think of the wavefunction as a section of the somewhat abstract bundle W. Just like a vector in a finite dimensional vector space can be thought of as a column of numbers, but often it’s better to think of it as just an abstract vector.
Thanks!
I’m not entirely sure that I follow the construction of W yet.
Let’s figure out the total space. If you just handed me a line bundle L on Y=X×R, and were like “make a bundle on Xn×R”, then the construction that I’d consider most obvious would be to make the total space be the pullback of total-space(L)n such that all of the time-coordinates agree...
...ah, but that wouldn’t be a line bundle; the tangent space would be n-dimensional. I see.
You suggested starting by considering what happens to an individual fiber, which… is an easier operation to do when I already know what the total space is, but whatever, a particular fiber is isomorphic to C, and it’s over a (x1,…,xn,t):Xn×R, and… yeah I’m not seeing how this helps me figure out the total space.
A third stab at figuring out the total space of W: maybe we take Xn×L, and now we have (in some sense) an “extra” Y that we want to somehow cancel out. If X were equipped with some binary operation ⋅, then the projection from Xn×L to Xn×R could, like, first project out (x1,…,xn) from the left, and then project out (x,t) from L, and then produce (x1⋅x,…,xn⋅x,t). But I don’t see any reasonable “multiplication” operation to fill that role.
I could maybe figure it out if I stared at it longer (as does sound fun), but my first three idiot ideas off the top of my head have not succeeded at extracting an answer to my question “what is the total space of W?” from your text.
(Though, for the record, it makes sense to me how a wave function ψ is a section of the trivial line bundle Xn×R×C, and how we can more generally ask for sections of line bundles over Xn×R and thereby get a more abstract theory. The part that’s not immediately-clicking for me is the part where we can extend a line bundle over Y to a line bundle over Xn×R in an obvious way.)
There are two operations involved in the definition of W: pullback and tensor product.
Pullback is defined for arbitrary bundles. Given a mapping f:X→Y (these X and Y are arbitrary manifolds, not the specific ones from before) and a bundle B over Y with total space TotB and projection mapping prB:Tot(B)→Y, the pullback of B w.r.t. f (denoted f∗B) is the bundle over X with total space Tot(B)×YX and the obvious projection mapping. I remind that Tot(B)×YX is the fibre product, i.e. the submanifold of Tot(B)×X defined by prB(t)=f(x). Notice that the fibre of f∗B over any x∈X is canonically isomorphic to the fibre of B over f(x). The word “canonical” means that there is a particular isomorphism that we obtain from the construction.
It is easy enough to see that the pullback of a vector bundle is a vector bundle, the pullback of a line bundle is a line bundle, and the pullback of a Hermitian vector bundle is a Hermitian vector bundle.
Tensor product is an operation over vector bundles. There are different ways to define it, corresponding to the different ways to define a tensor product of vector spaces. Specifically for line bundles there is the following shortcut definition. Let L1 and L2 be line bundles over X. Then, the total space of L1⊗L2 is the quotient of L1×XL2 by the equivalence relation given by: (v1,v2)∼(w1,w2) iff v1⊗v2=w1⊗w2. Here, I regard v1,w1 as vectors in the vector space which is the corresponding fibre fo L1 and similarly for v2,w2 and L2. The quotient of a manifold by an equivalence relation is not always a manifold, but in this case it is.
I notice that you wrote “a particular fiber is isomorphic to C”. Your error here is, it doesn’t matter what it’s isomorphic to, you should still think of it as an abstract vector space. So, if e.g.V1 and V2 are 1-dimensional vector spaces, then V1⊗V2 is yet another “new” vector space. Yes, they are all isomorphic, but they are not canonically isomorphic.
Thanks! Cool, it makes sense to me how we can make the pullback of L with Xn×R, in n different ways to get n different line bundles, and then tensor them all together. (I actually developed that hypothesis during a car ride earlier today :-p.)
(I’m still not quite sure what the syntax L⊗qi means, but presumably the idea is that there’s an automorphism on 1D vector fields that flips the sign, and we flip the sign of the negative-charge line bundles before tensoring everything together?)
(Also, fwiw, when I said “they’re all isomorphic to C”, I meant that I didn’t expect to figure much out by looking at a single fiber in isolation, and did not mean to imply that there was a canonical isomorphism; it’s clear to me that lacking access to a particular isomorphism is kinda the whole point. That said, I appreciate the pedagogy anyway! I prefer overexplanations to underexplanations whenever my convo-partner is up for generating them.)
Thanks again!
The syntax L⊗q means ”L to the tensor power of q”. For q>0, it just means tensoring L with itself q times. For q=0, L⊗q is just the trivial line bundle with total space Y×C (and, yes, all line bundles are isomorphic to the trivial line bundle, but this one just is the trivial bundle… or at least, canonically isomorphic to it). For q<0, we need the notion of a dual vector bundle. Any vector bundle V has a dual V∗, and for a line bundle the dual is also the inverse, in the sense that L⊗L∗ is canonically isomorphic to the trivial bundle. We can then define all negative powers by L⊗q:=(L∗)⊗−q. Notice that non-negative tensor powers are defined for all vector bundles, but negative tensor powers only make sense for line bundles.
It remains to explain what is V∗. But, for our purposes we can take a shortcut. The idea is, for any finite-dimensional complex vector space U with an inner product, there is a canonical isomorphism between U∗ and ¯U, where ¯U is the complex-conjugate space. What is the complex-conjugate space? It is a vector space that (i) has the same set of vectors (ii) has the same addition operation and (iii) has its multiplication-by-scalar operation modified, so that multiplying u by z in ¯U is the same thing as multiplying u by ¯z in U, where ¯z is just the complex number conjugate to z.
Equipped with this observation, we can define the dual of a Hermitian line bundle L to be ¯L, where ¯L is the bundle obtained for L by changing its multiplication-by-scalar mapping in the obvious way.
Vanessa’s reply is excellent, and Steven clearly descibed the gauge transforms of EM that I was gesturing at. That said, if you want to see some examples of fibre bundles in physics other than those in non-relativistic QM, nLab has a great article on the topic.
Also, if you know some GR, the inability to have more than local charts in general spacetimes makes clear that you do need to user bundles other than trivial ones i.e. ones of the form M×T. In my view, GR makes the importance of between global structure = fibre bundle clearer than QM, but maybe you have different intuitions.