# B Why does parity give raise to pseudovectors?

#### Phylosopher

Summary
Why parity have vectors and pseudovectors? why not only vectors?
Summary: Why parity have vectors and pseudovectors? why not only vectors?

I am reading Griffiths "Introduction to elementary particle physics" Ed.1.

The book obviously is an undergraduate introduction.Thus, not much detail is presented, but I cannot get my head around pseudovectors (pseudoscalars as well).

Parity in the book is just an inversion, that is to say, a mirror reflection followed by a $180^{\text{o}}$ rotation. It is easy to show that the parity group is just $\{P,I\}$, Where $P$ is the inversion matrix (or $-I$). For a vector $\text{v}\in R^{3}$:

$$P\text{v}=-\text{v}$$

This is how Griffiths explains it. He follows this by saying that for a cross product $\text{w}=\text{u}\times\text{v}$, parity does not give $-\text{w}$, because $P$ is distributive over cross products:

$$P(\text{w})=P(\text{u}\times\text{v})=P(\text{u})\times P(\text{v})=(-\text{u})\times (-\text{v})=\text{u}\times\text{v}=\text{w}$$

It looks to me that he is forcing $P$ to be a linear transformation over cross products. But $\text{w}$ is a vector and the solution must be $-\text{w}$, one way or another. Unless, $\text{w}$ is not a vector (a pseudovector)**. But, in this sense, what he presented is not a cross product to begin with!

Can someone help me understand why parity have pseudo's?

**Around three years ago, I read a book about geomtric algebra (didn't complete it), it talked about bivectors and pseudovectors. Could this be the same thing? Because the explanation Griffiths presented is very incomplete in my opinion.

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#### Vanadium 50

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This is not specific to QM. The electric field is a vector, but the magnetic field is an axial vector.

• Phylosopher

#### Orodruin

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I always had a beef with pseudo-vectors, or rather how they are presented in the undergraduate curriculum. The parity transformation is usually presented as $\vec x \to -\vec x$, but this only works in odd dimensions. In a more general sense, the parity transformations are about the two disjoint classes of orthogonal transformations, the group of proper rotations $SO(n)$ and the group of improper rotations, which are rotations + a reflection.

It then all comes down to representation theory. If you have a rotation $A \in O(n)$, an object $\vec v$ in the fundamental representation transforms as $\vec v \to A\vec v$ when $\vec x \to A\vec x$, i.e., it transforms as the position vector. However, this is far from the only representation of $O(n)$. Another representation with the same dimension as the fundamental representation is obtained if you let $A \to \operatorname{det}(A) A$. An object $\vec w$ that transforms according to this representations, i.e., $\vec w \to \operatorname{det}(A)A \vec w$ when $\vec x \to A\vec x$ is called a pseudo-vector (or axial vector). In particular, in three dimensions for the transformation $\vec x \to -\vec x$, a pseudo-vector transforms according to $\vec w \to - \operatorname{det}(-I) \vec w = \vec w$ (where $I$ is the unit matrix), not $\vec w \to - \vec w$.

The cross product of two vectors is a pseudo-vector, not a vector.
But ww\text{w} is a vector and the solution must be −w−w-\text{w}, one way or another.
No, $\vec w = \vec u \times \vec v$ where $\vec u$ and $\vec v$ are vectors is a pseudo-vector. The cross product between two vectors is always a pseudo-vector because it transforms according to $\operatorname{det}(A) A$, not according to $A$.

• • vanhees71, dextercioby and Phylosopher

#### Vanadium 50

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but this only works in odd dimensions
Well, if you're going to go down that path other issues show up. In 3-d angular momentum is an axial vector, but in 2-d it is a scalar, and in 4-d it's a 6-component skew-symmetric tensor.

#### Phylosopher

This is not specific to QM. The electric field is a vector, but the magnetic field is an axial vector.
No, →w=→u×→v\vec w = \vec u \times \vec v where →u\vec u and →v\vec v are vectors is a pseudo-vector. The cross product between two vectors is always a pseudo-vector because it transforms according to det(A)A\operatorname{det}(A) A, not according to AA.
This is very interesting. It is the first time someone states to me this.

In wiki, I found these two paragraphs:
Wikipedia said:
Using the cross product requires the handedness of the coordinate system to be taken into account (as explicit in the definition above). If a left-handed coordinate system is used, the direction of the vector n is given by the left-hand rule and points in the opposite direction.

This, however, creates a problem because transforming from one arbitrary reference system to another (e.g., a mirror image transformation from a right-handed to a left-handed coordinate system), should not change the direction of n. The problem is clarified by realizing that the cross product of two vectors is not a (true) vector, but rather a pseudovector.
How can I then distinguish pseudovectors from vectors if they live in the same realm. Pardon my ignorance, but if $\vec{v}=\{1,0,0\}$ and $\vec{u}=\{0,1,0\}$ are vectors that their cross product produce $\vec{w}$ (($\vec{w}=\vec{v}\times\vec{u}=\{0,0,1\}$))

Then how can I judge, say $\vec{s}=\{0,0,1\}$? Is $\vec{s}$ a vector or a pseudovector?

#### Vanadium 50

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It can be either. The contents of a vector don't determine whether it's a polar vector or an axial vector, its transformation properties do.

If I look at the electric field in a mirror, it points the same direction: a (polar) vector. If I look at the magnetic field, it points in the opposite direction: an axial vector.

#### Orodruin

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Well, if you're going to go down that path other issues show up. In 3-d angular momentum is an axial vector, but in 2-d it is a scalar, and in 4-d it's a 6-component skew-symmetric tensor.
I was referring to pseudo-vectors in general, not necessarily angular momentum. Angular momentum generally is a 2-form, which in three dimensions happens to transform according to the pseudo-vector representation, but this does not hold in other dimensions (wrong dimension for the representation).

• weirdoguy

#### Orodruin

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It can be either. The contents of a vector don't determine whether it's a polar vector or an axial vector, its transformation properties do.
This is a very important lesson. A vector is not just a collection of three numbers, those are just the components of the vector in some given basis. A vector is an object that also implies particular transformation properties of the components, i.e., how the components in different bases relate to each other.

If I look at the electric field in a mirror, it points the same direction: a (polar) vector. If I look at the magnetic field, it points in the opposite direction: an axial vector.
This is slightly misleading though and depends on which components you look at. If by "looking in a mirror" you mean doing a reflection such that the direction perpendicular to the mirror is flipped, then you would have (taking the third direction as the direction perpendicular to the mirror) $\vec e_3 \to -\vec e_3$, $\vec e_i \to \vec e_i$ for $i = 1$ and $2$. Under this transformation, a vector $\vec v = v^i \vec e_i$ would transform to $v^1 \vec e_1 + v^2 \vec e_2 - v^3 \vec e_3$ and to minus this if it were a pseudo-vector. Only if you do $\vec x \to -\vec x$ do you get $\vec v \to -\vec v$ for a proper vector and $\vec v \to \vec v$ for a pseudo vector. As a student, this had me stumped for a long time before I realised what was actually going on. This is partially why I find the "standard approach" of introducing parity as $\vec x \to -\vec x$ and at the same time talking about it as a "mirror" to be misleading.

Edit: Also note that a vector changes sign under $\vec x \to -\vec x$ whereas a pseudo-vector (axial vector) does not.

#### Vanadium 50

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This is a very important lesson.
Thank you,
This is slightly misleading though
I don't think so. (Obviously, because I wrote it) I think it's "mostly right" to associate a mirror with the parity operator, and the fact that E is even and B is odd is not just an example, it's a helpful example: B flips sign because the current loop that generates it flips sign. It is true that this is not completely precise, but this is a B-level thread after all.

#### Orodruin

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I don't think so. (Obviously, because I wrote it) I think it's "mostly right" to associate a mirror with the parity operator, and the fact that E is even and B is odd is not just an example, it's a helpful example: B flips sign because the current loop that generates it flips sign. It is true that this is not completely precise, but this is a B-level thread after all.
No, this is wrong. In fact, note that the parity transformation that flips all spatial directions flips a vector (i.e., the electric field) but not an axial vector (i.e., the magnetic field is unchanged under this transformation). For a pure reflection (i.e., not flipping all spatial directions, just a mirror in a plane), it is true that, for a vector, the directions within the plane of reflection do not flip. However, the direction perpendicular to the plane of reflection does flip. This can be intuitively understood as the reflection makes the point under consideration lie on the other side of the charge and therefore the component must flip under the reflection.

For the pseudo-vector, e.g., magnetic field, it depends on how you orient the loop. The magnetic field in the centre of the loop of course points normal to the loop. If the loop lies planar parallel to the plane of reflection, then the pure reflection does nothing with the direction of the current and the magnetic field in the loop centre remains the same. If the loop lies orthogonal to the plane of reflection, then the pure reflection does change the direction of the current and the magnetic field in the loop centre flips sign. All of this is in perfect accordance with what I described above for the transformation properties of vectors and pseudo-vectors under pure reflections.

It is also an important lesson that a pure reflection does not just amount to possibly flipping a sign. It is not only the determinant, but also the transformation $A$ itself, that is important in the pseudo-vector representation $\operatorname{det}(A) A$.

• weirdoguy

#### Vanadium 50

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Well, I guess we disagree on what the appropriate degree of "wrongness" is for a B-level thread.

#### Orodruin

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Well, I guess we disagree on what the appropriate degree of "wrongness" is for a B-level thread.
Well, first of all, I do not consider this a B-level thread. If the OP is advanced enough to read about vectors vs pseudo-vectors it is enough for me to consider him I-level. Second, the misconception that parity just "flips the sign" of vectors but not pseudo-vectors is very confusing to anyone who just starts thinking about looking at a wheel in a mirror. If the wheel is parallel to the plane of reflection, it will have the same angular momentum in the mirror world as in the "ordinary" world because both the wheel and the mirror wheel spin in the same direction. If it is orthogonal to the plane of reflection the angular momentum changes sign because the wheel and the mirror wheel now spin in the opposite directions. (This example is what first had me stumped and wondering what was going on.)

Also, note that it is vectors, not pseudo-vectors that flip sign under $\vec x \to -\vec x$. There is no improper rotation that has $\vec v \to \vec v$ for all proper vectors $\vec v$. Therefore I do think it is misleading, even at B-level, to state that the E-field remains the same after reflection.

• weirdoguy

#### Vanadium 50

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I'll let you have the last word on this.

#### vanhees71

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Well, I guess we disagree on what the appropriate degree of "wrongness" is for a B-level thread.
Only, because something is labeled as B-level, it's no excuse for inexact explanations. To the contrary, for a beginner it's very helpful to be correct, and @Orodruin did a great job!

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