In my previous post, I promised to do something on symmetries. Something simple but then… Well… You know how it goes: one question always triggers another one. 🙂

Look at the situation in the illustration *on the left* below. We suppose we have something *real *going on there: something is moving **from left to right** (so that’s **in** **the 3 o’clock direction**), and then something else is going around **clockwise** (so that’s *not* the direction in which we measure angles (which also include the argument θ of our wavefunction), because that’s always *counter*-clockwise, as I note at the bottom of the illustration). To be precise, we should note that the angular momentum here is all about the *y*-axis, so the angular momentum vector **L** points in the (positive) *y*-direction. We get that direction from the familiar right-hand rule, which is illustrated in the top right corner.

Now, suppose someone else is looking at this from the other side – or just think of yourself going around *a full 180°* to look at the same thing from the back side. You’ll agree you’ll see the same thing going from *right *to *left *(so that’s **in the 9 o’clock direction **now – or, if our clock is transparent, the 3 o’clock direction of our *reversed *clock). Likewise, the thing that’s turning around will now go *counter*-clockwise.

Note that both observers – so that’s me and that other person (or myself after my walk *around* this whole thing) – use a regular coordinate system, which implies the following:

- We’ve got regular 90° degree angles between our coordinates axes.
- Our
*x*-axis goes from negative to positive from left to right, and our*y*-axis does the same going*away*from us. - We also both define our
*z*-axis using, once again, the ubiquitous right-hand rule, so our*z*-axis points upwards.

So we have two observers looking at the same *reality* – some *linear *as well as some *angular *momentum – but from opposite sides. And so **we’ve got a reversal of both the linear as well as the angular momentum**. *Not *in reality, of course, because we’re looking at the same thing. But we *measure *it differently. Indeed, if we use the subscripts 1 and 2 to denote the *measurements* in the two coordinate systems, we find that **p**_{2} = –**p**_{1}. Likewise, we also find that **L**_{2} = –**L**_{1}*.*

Now, when you see these two equations, you will probably not worry about that **p**_{2} = –**p**_{1} equation – although you should, because it’s actually only valid for this rather particular orientation of the linear momentum (I’ll come back to that in a moment). It’s the **L**_{2} = –**L**_{1} equation which should surprise you most. Why? Because you’ve always been told there is a big difference between (1) *real *vectors (aka polar vectors), like the momentum **p**, or the velocity** v**, or the force

**F**, and (2)

*pseudo*-vectors (aka axial vectors), like the

*angular*momentum

**L**. You may also remember how to distinguish between the two:

**if you change the direction of the axes of your reference frame, polar vectors will change sign too, as opposed to axial vectors: axial vectors do**

*not*swap sign if we swap the coordinate signs.So… Well… How does that work here? In fact, what we should ask ourselves is: why does that *not *work here? Well… It’s simple, really. We’re *not* changing the direction of the axes here. Or… Well… Let me be more precise: we’re only swapping the sign of the *x*– and *y*-axis. **We did not flip the z-axis**. So we turned things

*around*, but we didn’t turn them

*upside down*. It makes a huge difference. Note, for example, that if

*all*of the linear momentum would have been in the

*z*-direction only (so our

**p**vector would have been pointing in the

*z*-direction, and in the

*z*-direction

*only*), it would

*not*swap sign. The illustration below shows what really happens with the coordinates of some vector when we’re doing a

*rotation*. It’s, effectively, only the

*x*– and

*y*-coordinates that flip sign.

It’s easy to see that this *rotation *about the *z*-axis here preserves our deep sense of ‘up’ versus ‘down’, but that it swaps ‘left’ for ‘right’, and vice versa. Note that this is *not *a reflection. We are *not *looking at some mirror world here. The difference between a reflection (a mirror world) and a rotation (the real world seen from another angle) is illustrated below. It’s quite confusing but, unlike what you might think, a reflection does *not* swap left for right. It does turn things inside out, but that’s what a rotation does as well: near becomes far, and far becomes near.

Before we move on, let me say a few things about the *mirror world *and, more in particular, about the obvious question: could it possibly *exist*? Well… **What do you think?** Your first reaction might well be: “Of course! What nonsense question! We just walk around whatever it is that we’re seeing – or, what amounts to the same, we just turn it around – and there it is: that’s the mirror world, right? So

*of course*it exists!” Well… No. That’s

*not*the mirror world. That’s just the

*real*world seen from the opposite direction, and that world… Well… That’s just the real world. 🙂 The mirror world is, literally, the world

*in the mirror*– like the photographer in the illustration below. We don’t swap left for right here: some object going from left to right in the real world is still going from left to right in the mirror world!Of course, you may now involve the photographer in the picture above and observe – note that you’re now an observer of the observer of the mirror 🙂 – that, if he would move his

*left*arm in the real world, the photographer in the mirror world would be moving his

*right*arm. But… Well… No. You’re saying that because you’re now

*imaging*that you’re the photographer in the mirror world yourself now, who’s looking at the real world from inside, so to speak. So you’ve rotated the perspective

*in your mind*and you’re saying it’s his right arm because you

*imagine*yourself to be the photographer in the mirror. We usually do that because… Well… Because we look in a mirror every day, right? So we’re used to seeing ourselves that way and we always think it’s us we’re seeing. 🙂 However, the illustration above is correct: the mirror world only swaps near for far, and far for near, so it only swaps the sign of the

*y-*axis.

So the question *is *relevant: could the mirror world actually exist? What we’re *really *asking here is the following: can we swap the sign of *one* coordinate axis *only *in all of our physical laws and equations and… Well… Do we then still get the same laws and equations? Do we get the same Universe – because that’s what those laws and equations describe? If so, our mirror world can exist. If not, then not.

Now, I’ve done a post on that, in which I explain that mirror world can only exist if it would consist of *anti*-matter. So if our real world and the mirror world would actually meet, they would annihilate each other. 🙂 But that post is quite technical. Here I want to keep it *very *simple: I basically only want to show what the *rotation *operation implies for the wavefunction. There is no doubt whatsoever that the *rotated *world exists. In fact, the rotated world is just *our *world. We walk around some object, or we turn it around, but so we’re still watching the *same* object. So we’re not thinking about the mirror world here. We just want to know how things look like when adopting some other perspective.

So, back to the starting point: we just have two observers here, who look at the same thing but from opposite directions. Mathematically, this corresponds to a rotation of our reference frame *about *the *z*-axis of 180°. Let me spell out – somewhat more precisely – what happens to the linear and angular momentum here:

- The direction of the linear momentum
**in the**swaps direction.*xy*-plane - The angular momentum
**about the**, as well as*y*-axis**about the**, swaps direction too.*x*-axis

Note that the illustration only shows angular momentum about the *y*-axis, but you can easily verify the statement about the angular momentum about the *x*-axis. In fact, the angular momentum about *any *line in the *xy*-plane will swap direction.

Of course, the *x*-, *y*-, *z*-axes in the other reference frame are different than mine, and so I should give them a subscript, right? Or, at the very least, write something like *x’*, *y’*, *z’*, so we have a *primed *reference frame here, right? Well… Maybe. Maybe not. Think about it. 🙂 A coordinate system is just a mathematical thing… Only the momentum is real… Linear or angular… Equally real… And then Nature doesn’t care about our position, does it? So… Well… No subscript needed, right? Or… Well… **What do you think? **🙂

It’s just funny, isn’t it? It looks like we can’t really separate reality and perception here. Indeed, note how our **p**_{2} = –**p**_{1 }and **L**_{2} = –**L**_{1} equations already mix reality with how we *perceive *it. It’s the same thing *in reality *but the coordinates of **p**_{1} and **L**_{1 }are positive, while the coordinates of **p**_{2} and **L**_{2 }are negative. To be precise, these coordinates will look like this:

**p**_{1}= (p, 0, 0) and**L**_{1 }= (0, L, 0)**p**_{2}= (−p, 0, 0) and**L**_{1 }= (0, −L, 0)

So are they two different things or are they not? 🙂 Think about it. I’ll move on in the meanwhile. 🙂

Now, you probably know a thing or two about *parity *symmetry, or P-symmetry: if if we flip the sign of *all* coordinates, then we’ll still find the same physical laws, like **F** = m·** a** and what have you. [It works for all physical laws, including quantum-mechanical laws – except those involving the

*weak*force (read: radioactive decay processes).] But so here we are talking rotational symmetry. That’s

*not*the same as P-symmetry. If we flip the signs of

*all*coordinates, we’re also swapping ‘up’ for ‘down’, so we’re not only turning

*around*, but we’re also getting

*upside down*. The difference between

*rotational*symmetry and P-symmetry is shown below.

As mentioned, we’ve talked about P-symmetry at length in other posts, and you can easily *google *a lot more on that. The question we want to examine here – just as a fun exercise – is the following:

**How does that rotational symmetry work for a wavefunction? **

The very first illustration in this post gave you the functional form of the *elementary *wavefunction *e*^{i}^{θ} = *e*^{i}^{·}^{(E·t – p·x)/ħ}. We should actually use a *bold type x* = (

*x*,

*y*,

*z*) in this formula but we’ll assume we’re talking something similar to that

**p**vector: something moving in the

*x*-direction only – or in the

*xy-plane*only. The

*z*-component doesn’t change. Now, you know that we can reduce all

*actual*wavefunctions to some linear combination of such elementary wavefunctions by doing a

*Fourier*decomposition, so it’s fine to look at the

*elementary*wavefunction only – so we don’t make it too complicated here. Now think of the following.

The energy E in the *e ^{i}*

^{θ}=

*e*

^{i}^{·}

^{(E·t – p·x)/ħ }function is a

*scalar*, so it doesn’t have any direction and we’ll measure it the same from both sides – as kinetic or potential energy or, more likely, by adding both. But… Well… Writing

*e*

^{i}^{·}

^{(E·t – p·x)/ħ }or

*e*

^{i}^{·}

^{(E·t + p·x)/ħ }is not the same, right? No, it’s not. However, think of it as follows: we won’t be changing the

*direction of time*, right? So it’s OK to

*not*change the sign of E. In fact, we can re-write the two expressions as follows:

*e*^{i}^{·}^{(E·t – p·x)/ħ }=*e*^{i}^{·}^{(E/ħ)·t}·*e*^{–}^{i}^{·(}^{p/ħ)·x}*e*^{i}^{·}^{(E·t + p·x)/ħ }=*e*^{i}^{·}^{(E/ħ)·t}·*e*^{i}^{·(}^{p/ħ)·x}

The first wavefunction describes some particle going in the *positive* *x*-direction, while the second wavefunction describes some particle going in the *negative* *x*-direction, so… Well… That’s *exactly *what we see in those two reference frames, so there is no issue whatsoever. 🙂 It’s just… Well… I just wanted to show the wavefunction *does *look different too when looking at something from another angle.

So why am I writing about this? Why am I being fussy? Well.. It’s just to show you that those *transformations *are actually quite natural – just as natural as it is to see some particle go in one direction in one reference frame and see it go in the other in the other. 🙂 It also illustrates another point that I’ve been trying to make: the wavefunction is something *real*. It’s not just a figment of our imagination. The real and imaginary part of our wavefunction have a precise geometrical meaning – and I explained what that might be in my more speculative posts, which I’ve brought together in the *Deep Blue *page of this blog. But… Well… I can’t dwell on that here because… Well… You should read that page. 🙂

The point to note is the following: we *do* have *different* wavefunctions in different reference frames, but these wavefunctions describe the *same physical reality*, and they also do respect the symmetries we’d expect them to respect, except… Well… The laws describing the *weak *force don’t, but I wrote about that a *very *long time ago, and it was *not *in the context of trying to explain the relatively simple basic laws of quantum mechanics. 🙂 If you’re interested, you should check out my post(s) on that or, else, just *google *a bit. It’s really exciting stuff, but not something that will help you much to understand the basics, which is what we’re trying to do here. 🙂

The *second* point to note is that those *transformations *of the wavefunction – or of quantum-mechanical *states *– which we go through when rotating our reference frame, for example – are really quite natural. There’s nothing special about them. We had such transformations in classical mechanics too! But… Well… Yes, I admit they do *look *complicated. But then that’s why you’re so fascinated and why you’re reading this blog, isn’t it? 🙂

**Post scriptum**: It’s probably useful to be somewhat more precise on all of this. You’ll remember we visualized the wavefunction in some of our posts using the animation below. It uses a left-handed coordinate system, which is rather unusual but then it may have been made with a software which uses a left-handed coordinate system (like RenderMan, for example). Now the rotating arrow at the center moves with time and gives us the polarization of our wave. Applying our customary *right-hand *rule,you can see this beam is *left*-circularly polarized. [I know… It’s quite confusing, but just go through the motions here and be consistent.]Now, you know that *e*^{–}^{i}^{·(}^{p/ħ)·x }and *e*^{–}^{i}^{·(}^{p/ħ)·x} are each other’s complex conjugate:

*e*^{–}^{i}^{·k}^{·x }=*cos*(k·x) +*i*·*sin*(k·x)*e*^{–}^{i}^{·k}^{·x }=*cos*(-k·x) +*i*·*sin*(-k·x) =*cos*(k·x) −*i*·*sin*(k·x)

Their real part – the cosine function – is the same, but the imaginary part – the sine function – has the opposite sign.* *So, assuming the direction of propagation is, effectively, the *x*-direction, then what’s the polarization of the mirror image? Well… The wave will now go from right to left, and its polarization… Hmm…* Well… What? *

Well… If you can’t figure it out, then just forget about those signs and just imagine you’re effectively looking *at the same thing *from the backside. In fact, if you have a laptop, you can push the screen down and go around your computer. 🙂 There’s no shame in that. In fact, I did that just to make sure I am *not *talking nonsense here. 🙂 If you look at this beam from the backside, you’ll effectively see it go from right to left – instead of from what you see on this side, which is a left-to-right direction. And as for its polarization… Well… The angular momentum vector swaps direction too but **the beam is still left-circularly polarized**. So… Well… That’s consistent with what we wrote above. 🙂 The real world is real, and axial vectors are as real as polar vectors. This

*real*beam will only appear to be

*right*-circularly polarized

*in a mirror*. Now, as mentioned above, that mirror world is not

*our*world. If it would exist – in some other Universe – then it would be made up of anti-matter. 🙂

So… Well… Might it actually exist? Is there some other world made of anti-matter out there? I don’t know. We need to think about that reversal of ‘near’ and ‘far’ too: as mentioned, a mirror turns things inside out, so to speak. So what’s the implication of *that*? When we walk *around *something – or do a *rotation *– then the reversal between ‘near’ and ‘far’ is something *physical*: we go near to what was far, and we go away from what was near. But so how would we get into our mirror world, so to speak? We may say that this anti-matter world in the mirror is entirely *possible*, but then how would we get there? We’d need to turn ourselves, literally, inside out – like short of shrink to the zero point and then come back out of it to do that *parity *inversion along our line of sight. So… Well… I don’t see that happen, which is why I am a fan of the One World hypothesis. 🙂 So *I *think the mirror world is just what it is: the mirror world. Nothing real. But… Then… Well… **What do you think? **🙂