So what is circularly polarized light?

那麼，什麼是圓偏振光呢？

Well, in the last video we talked about linearly polarized light.

在上一個視頻中，我們談到了線性偏振光。

In other words, the direction of the electric field doesn't change as you go throughout space.

換句話說，電場的方向在整個太空中不會改變。

So let's say that this is the propagation direction of our plane wave and then we've got our coordinate axis.

假設這是平面波的傳播方向，然後我們就有了座標軸。

A linearly polarized light, if it's polarized initially in X, so it's got some X component, that X component will change throughout space.

線性偏振光，如果最初是以 X 偏振的，那麼它就有一些 X 分量，X 分量會在整個空間發生變化。

So it'll get bigger and smaller and bigger and smaller because this is a sinusoidal wave, this is a plane wave.

所以它會越來越大，越來越小，因為這是正弦波，這是平面波。

So we expect the magnitude to go up and down as we go throughout space.

是以，我們預計，在整個太空中，幅度會時大時小。

But it doesn't change direction.

但它不會改變方向。

The direction stays the same.

方向不變。

It's always pointed along the X axis.

它總是沿著 X 軸指向。

Now, circularly polarized light doesn't behave this way.

現在，圓偏振光的表現並非如此。

It actually changes the direction that this arrow is pointing as you go throughout space.

實際上，當你在太空中飛行時，這個箭頭所指向的方向會發生變化。

So this is our Z axis.

這就是我們的 Z 軸。

Now, you might ask, how on earth is that possible?

你可能會問，這到底是怎麼做到的？

That seems just bizarre.

這似乎太奇怪了。

And the answer has to do with the wave nature of light and the fact that we can... add two polarized waves together.

答案與光的波性有關，我們可以......將兩個偏振光波相加。

So let's say we have this polarized wave, which is polarized in the X direction.

假設我們有一個偏振波，它在 X 方向上偏振。

Let's say we also want to add a wave that's polarized in the Y direction.

假設我們還想添加一個 Y 方向的偏振波。

So this is our Y axis.

這就是我們的 Y 軸。

But let's say that instead of having the maximum of the Y polarization match up with the maximum of the X polarization, so that would look something like this, instead of doing that, why don't we offset it?

但是，與其讓 Y 偏振的最大值與 X 偏振的最大值相匹配，讓它看起來像這樣，我們為什麼不把它抵消掉呢？

So let's say that the maximum of our Y polarization happens when our X polarization is equal to zero.

是以，假設當我們的 X 偏振等於零時，我們的 Y 偏振就會達到最大值。

So this is the maximum of our Y polarization.

這就是我們 Y 極化的最大值。

And similarly, it decays to zero, or the sine wave goes to zero when the X polarization is at a maximum.

同樣，當 X 偏振達到最大值時，正弦波會衰減為零。

So if we were to trace that out in the Y axis, it would look something like this.

是以，如果我們在 Y 軸上進行追蹤，就會看到這樣的結果。

So we've got arrows pointing along the Y axis, and then our arrows start pointing backwards.

是以，我們有箭頭沿著 Y 軸指向，然後我們的箭頭開始向後指向。

And most importantly, at this point, so let's call this Z equals zero, we have an entirely X polarized wave, or what looks like purely X polarized light.

最重要的是，在這一點上，讓我們稱之為 Z 等於零，我們有一個完全 X 偏振的波，或者看起來像純粹的 X 偏振光。

So if we were to add up these two polarizations, this is X and this is Y, then initially, our wave is just polarized in X.

是以，如果我們把這兩種偏振加起來，這是 X，這是 Y，那麼最初，我們的波只是偏振於 X。

And I'm going to draw this total in a different color, let's do blue.

我將用不同的顏色繪製這個總數，就用藍色吧。

So initially, this is our total electric field, it's polarized in X.

是以，最初，這是我們的總電場，它的極化方向是 X。

But as we go along some distance, our total field ends up being polarized in Y.

但是，當我們走過一段距離後，我們的總場最終會被偏振成 Y。

And then if we go some distance more, it's polarized in negative X, and then negative Y, and then X, and so on and so on.

如果我們再往前走一段距離，就會出現負 X 極化，然後是負 Y 極化，然後是 X 極化，以此類推。

And if we were to trace out what this looks like, this would actually be, if they have the same magnitude, a circle.

如果我們描繪出它的樣子，如果它們的大小相同，這實際上就是一個圓。

So I've done my best to draw the, this is what it would look like, where we've got, initially we're in front of the axis, then we go behind the axis, and we sort of curve around, we continually curve around this axis.

所以，我已經盡我所能畫出了它的樣子，最初我們在軸線的前面，然後我們走到軸線的後面，然後我們繞著軸線彎曲，我們不斷地繞著這個軸線彎曲。

And this is what's known as right-hand circularly polarized light.

這就是所謂的右旋圓偏振光。

So if you stuck your thumb out in this direction, and you were to curve your fingers of your right hand, you'd get that they travel in the same direction as this wave curves around the axis.

是以，如果你把拇指伸向這個方向，並彎曲你右手的手指，你會發現它們的移動方向與這個波繞軸的彎曲方向相同。

And so here's a visualization more of what this looks like, this curving light, or this curving polarization.

是以，這裡有一個更直觀的視覺效果，這個彎曲的光線，或者說這個彎曲的偏振。

Notice it doesn't change its magnitude, so it stays constant in magnitude, it doesn't go from zero to one, it just changes its direction.

請注意，它不會改變量級，是以量級保持不變，不會從 0 變為 1，只是改變了方向。

So you can kind of see how it curves to the right, or sort of a right-handed helix.

是以，你可以看到它是如何向右彎曲的，或者說是一種右手螺旋。

And if we advance the time, we can watch this wave propagate as you increase the time.

如果我們把時間提前，就能看到這個波隨著時間的增加而傳播。

So we can see it's propagating down the Z-axis, and the electric field is sort of curving out this circle as we propagate it.

是以，我們可以看到它正沿著 Z 軸向下傳播，而電場在傳播的過程中就像畫了一個圓。

And if we freeze the helix in time, we can see that it's a right-handed helix, so it follows the curve of our right-handed fingers as we point our thumb in the direction of propagation.

如果我們把螺旋線定格在時間上，就會發現它是一個右手螺旋線，所以當我們把拇指指向傳播方向時，它就會沿著我們右手手指的曲線移動。

So let's get rid of that.

所以，讓我們把它去掉吧。

So that's fine and good, but how do we mathematically model this?

這很好，但我們如何用數學模型來模擬呢？

And the answer is going to be with our Jones vectors.

答案就是我們的瓊斯矢量。

So how do we represent a wave that's shifted, that's delayed with respect to the other wave by a quarter of the wavelength?

那麼，我們該如何表示一個偏移的波，即相對於另一個波延遲了四分之一波長的波呢？

So this distance would be lambda over four, this distance here, whoop, what the fuck?

是以，這個距離是λ大於4，這個距離在這裡，嗚呼，搞什麼鬼？

So this distance is lambda over four, our wavelength over four.

所以這個距離是 lambda 乘以 4，也就是我們的波長乘以 4。

This total distance, a full half period, would be lambda over two, and this would be our full wavelength, so we've completed a full cycle.

這個總距離，一個完整的半週期，將是 lambda 乘以 2，這將是我們的完整波長，所以我們已經完成了一個完整的週期。

So when we write down our X-polarized wave, we can write out a formula for that, we know what it is, and then we write down our Y-polarized wave, we need to delay this wave by lambda over four.

是以，當我們寫下 X 偏振波時，我們可以寫出一個公式，我們知道它是什麼，然後我們寫下 Y 偏振波，我們需要將這個波延遲 lambda 超過 4。

So we need to set Z to Z minus lambda over four.

是以，我們需要將 Z 設為 Z 減 lambda 大於 4。

And let's see what happens when we do that.

讓我們看看這樣做會發生什麼。

So our X-polarized wave, we can just write as, let's say, some magnitude E-naught in the X direction, e to the j omega t minus kz, and now our Y-polarized wave, wow, that looks like an X, that's actually a Y, our Y-polarized wave has the same magnitude, and we're assuming that, so this is what you need to assume for circularly polarized light, e to the j omega t minus k, and now we said we need to replace Z with Z minus lambda over four, so Z minus lambda over four, and we can rewrite that, so E-naught Y e to the j omega t minus kz, plus, so what's k times lambda over four?

所以我們的 X 偏振光波，我們可以寫成，比方說，X 方向上的某個量級的 E 偏振光波，e 到 j 歐米茄 t 減 kz，現在我們的 Y 偏振光波，哇，看起來像 X，實際上是 Y，我們的 Y 偏振光波有相同的量級，我們假設是這樣、所以這就是圓偏振光需要假設的，e 到 j 歐米茄 t 減 k，現在我們說需要用 Z 減 lambda 大於 4 來代替 Z，所以 Z 減 lambda 大於 4，我們可以重寫，所以 E 偏 Y e 到 j 歐米茄 t 減 kz，加上，那麼 k 乘以 lambda 大於 4 是多少？

Well, k is two pi over lambda times lambda over four, lambdas cancel, and we get pi over two, so plus pi over two.

好吧，k 是兩個 pi 乘以 lambda 再乘以 lambda 乘以 4，lambdas 相抵消，我們得到 pi 乘以 2，所以加上 pi 乘以 2。

And now the really clever part, you might notice we can just take this out of our wave, so we can factor this out front, and we'll have E-naught e to the j pi over two times our traveling wave that we had before, omega t minus kz, and this remembers the part that we're not interested in, so at least not after this point, at least not after this point, because we know that it's a traveling wave, we know that it's traveling in the Z direction, we only care about the components, so if we just write down the X and the Y components into our lazy man's vector, which is almost a Jones vector, it's just E-naught X plus E-naught e to the j pi over two Y, now we could write this in column vector form, and factor out the E-naught, and in that case we'll just have one, and e to the j pi over two, which is the imaginary number I, and so for the Jones vector, we just drop the amplitude, and we normalize the vector, so you need to divide by the square root of two, and this is our Jones vector for right-hand polarized light, and this contains all the information that we had before, but it's super, super simple, so rather than writing down this plus this, we just write down this super simple column vector, and this will actually allow us to do some really cool things, now I should make one note on conventions, so I've been using the convention that a traveling wave is represented as e to the j omega t minus kz, there's some people that do e to the j kz minus omega t, and in this case right-hand polarized light would be one minus I, there are also different conventions on whether this is right-hand polarized, or left-hand polarized, and the convention that I follow, and what seems to be the most common one, is the handedness convention, so stick your right hand out, curve your fingers, if your fingers follow the path of the light, then the light is right-hand polarized, if it follows the path of your left hand, it's left-hand polarized, so if instead we had shifted the wave, instead of shifting the y polarized wave in this direction, we could have also shifted it in this direction, and that would have given us left-hand polarized light, which the Jones vector, we could just represent as one minus I, and so these are two circularly polarized forms of light, and they correspond to helices, that are either right or left-handed, and are propagating along the z direction, but you might wonder, what if we hadn't shifted by a perfect lambda over four, what if we had shifted by a little bit less, or a little bit more, and that's a perfectly reasonable question, and we'll answer that in the next video, on elliptically polarized light, and you'll get the same exact thing, if you instead had different amplitudes, in front of this x and y, but we'll go over that in the future video, so I hope you enjoyed this one, if you did please give it a like down below, and subscribe to my channel, also if you have any questions or comments, please feel free to post those down below, and I'll try to get back to you as soon as I can, and thanks for watching, I'll see you next time.

現在是真正聰明的部分，你可能會注意到 我們可以把這個從我們的波中拿出來 所以我們可以把這個因式分解 我們就會得到 E-naught e 到 j pi 超過我們之前的行波的兩倍 Omega t 減去 kz 記住我們不感興趣的部分我們只關心分量，所以如果我們把 X 分量和 Y 分量寫入我們的懶人矢量，這幾乎是一個瓊斯矢量，它就是 E-naught X 加上 E-naught e to the j pi over two Y，現在我們可以把它寫成列矢量形式，然後把 E-naught 因子除去，這樣我們就只有一個