# Getting Even More Entangled with Wave Functions

Alright, folks!

In my previous post, I used my latest invention, quantum playing cards, to give specific examples of entangled and non-entangled quantum systems. I recommend reading that article if you have not already done so, or reread it if it has been a while, for today I am going to expand upon that example and use it to introduce a bit of mathematical formalism, which is necessary for our ultimate goal of understanding quantum teleportation.

Or, specifically, the notation is necessary for me to explain quantum teleportation—at least for now! In the meantime, bear with me, and know that if you can add, subtract, multiply, and divide, and if you are familiar with the most basic operations of algebra, then you have all the tools necessary to follow along and understand everything I am about to describe.

Here we go!

As I stated last time, our two non-entangled quantum playing cards are described by the wave functions\begin{equation} \begin{array}{c} \displaystyle \lvert \psi \rangle_A = \frac{1}{\sqrt{2}} \left( \lvert \spadesuit \rangle_A + \lvert \diamondsuit \rangle_A \right), \\[2ex] \displaystyle \lvert \psi \rangle_B = \frac{1}{\sqrt{2}} \left( \lvert \spadesuit \rangle_B + \lvert \diamondsuit \rangle_B \right), \label{300717:qpcAB} \end{array} \end{equation}and we discovered that together they describe a system of cards with the four following states:\begin{equation} \begin{array}{c|c} A & B \\ \hline \lvert \spadesuit \rangle &  \lvert \diamondsuit \rangle \\ \lvert \diamondsuit \rangle & \lvert \spadesuit \rangle \\ \lvert \spadesuit \rangle &  \lvert \spadesuit \rangle \\ \lvert \diamondsuit \rangle & \lvert \diamondsuit \rangle \end{array} \label{073017:ABne} \end{equation}

But, why is it that these two wave functions predict those four outcomes? I mean, we know that those are the four possible outcomes from our last expriment, but is it possible to use equation \eqref{300717:qpcAB} to predict table \eqref{073017:ABne} ahead of time?

Well, as a matter of fact, it is!

To demonstrate this, I must introduce the mathematical operator with which we combine quantum wave functions; a curious fellow named the tensor product:

For our purposes, the tensor product behaves like multiplication, meaning it has the following properties: for tensors $$W, X, Y,$$ and $$Z$$,

• It distributes over addition:\begin{equation} X\otimes (Y+Z)=X\otimes Y+X\otimes Z. \label{300717:dist} \end{equation}
• We can therefore FOIL the tensor product of two binomials:\begin{equation} (W+X)\otimes (Y+Z)=W\otimes Y+W\otimes Z+X\otimes Y+X\otimes Z. \label{300717:foil} \end{equation}
• It's associative, which allows us to write\begin{equation} (X \otimes Y) \otimes Z = X \otimes (Y \otimes Z) = X \otimes Y \otimes Z. \label{300717:ass} \end{equation}
• It's symmetric, meaning that\begin{equation} X \otimes Y = Y \otimes X. \label{300717:sym} \end{equation}
• "Regular numbers" (called scalars) pass right through it. If $$\alpha$$ and $$\beta$$ are such numbers, then\begin{equation} (\alpha  X)\otimes (\beta  Y)=(\alpha  \beta ) X\otimes Y. \label{300717:scale} \end{equation}

Quantum wave functions are tensors, and so with these facts in hand, it is possible to combine the two individual wave functions in \eqref{300717:qpcAB} into one single representation:\begin{equation} \lvert \psi \rangle _A\otimes \lvert \psi \rangle _B = \frac{1}{2}\left(\lvert \spadesuit \rangle _A\otimes \lvert \spadesuit \rangle _B+\lvert \spadesuit \rangle _A\otimes \lvert \diamondsuit \rangle _B + \lvert \diamondsuit \rangle _A\otimes \lvert \spadesuit \rangle _B + \lvert \diamondsuit \rangle _A\otimes \lvert \diamondsuit \rangle _B\right).\label{073017:wfAB}\end{equation}

It is my claim that equation \eqref{073017:wfAB} is a single wave function that describes both of our quantum playing cards simultaneously. To accept such a claim, we must verify that it is consistent with the outcomes of our experiments in the previous post, the main thing being that it generates table \eqref{073017:ABne}. So then, does it?

Since \eqref{073017:wfAB} contains four terms (of tensor products) added together, it describes four possible results when we turn both cards face up—those results being:\begin{equation} \begin{array}{c|c} A & B \\ \hline \lvert \spadesuit \rangle &  \lvert \spadesuit \rangle \\ \lvert \spadesuit \rangle & \lvert \diamondsuit \rangle \\ \lvert \diamondsuit \rangle &  \lvert \spadesuit \rangle \\ \lvert \diamondsuit \rangle & \lvert \diamondsuit \rangle \end{array} \label{073017:ABne2} \end{equation}But wait a minute, that's just table \eqref{073017:ABne} shuffled around! Since the orders of these tables does not matter, \eqref{073017:ABne} and \eqref{073017:ABne2} are in fact one and the same. Additionally, equation \eqref{073017:wfAB} contains an overall factor of $$1/2$$. Since $$(1/2)^2 = 1/4 = 25\%$$, we should find that there is an equal $$25\%$$ chance of finding any of table \eqref{073017:ABne2}'s values when we flip over our quantum cards. (Mad respect to those of you who actually do that experiment!)

So, we are off to a good start, but what does equation \eqref{073017:wfAB} say about entangled states? Remember that we last discovered that the state of our non-entangled quantum cards was merely a combination of two different entangled states! Does equation \eqref{073017:wfAB} make that prediction, too? To answer that question, I must introduce examples of wave functions that describe entanglement. Recall that our cards are entangled when their values were correlated, allowing us to use the value of one card to predict that of the other. This occurs when they always have different values:\begin{equation} \begin{array}{c|c}  A_\Psi & B_\Psi \\ \hline \lvert \spadesuit \rangle & \lvert \diamondsuit \rangle \\ \lvert \diamondsuit \rangle &  \lvert \spadesuit \rangle \end{array} \label{073117:ABPsi} \end{equation}or when they always have identical values:\begin{equation} \begin{array}{c|c}  A_\Phi & B_\Phi \\ \hline \lvert \spadesuit \rangle & \lvert \spadesuit \rangle \\ \lvert \diamondsuit \rangle &  \lvert \diamondsuit \rangle \end{array} \label{073117:ABPhi} \end{equation}Using our knowledge of the relationship between tables and wave functions discovered above, it seems reasonable to say that table \eqref{073117:ABPsi} corresponds to the wave function\begin{equation} \lvert \Psi \rangle_{AB} = \frac{1}{\sqrt{2}} \left( \lvert \spadesuit \rangle_A \otimes \lvert \diamondsuit \rangle_B + \lvert \diamondsuit \rangle_A \otimes \lvert \spadesuit \rangle_B \right). \label{073117:ABwfPsi} \end{equation}Similarly,\begin{equation} \lvert \Phi \rangle_{AB} = \frac{1}{\sqrt{2}} \left( \lvert \spadesuit \rangle_A \otimes \lvert \spadesuit \rangle_B + \lvert \diamondsuit \rangle_A \otimes \lvert \diamondsuit \rangle_B \right) \label{073117:ABwfPhi} \end{equation}produces table \eqref{073117:ABPhi}. But then, if equations \eqref{073117:ABwfPsi} and \eqref{073117:ABwfPhi} describe tables \eqref{073117:ABPsi} and \eqref{073117:ABPhi}, respectively, and those tables combined produce table \eqref{073017:ABne2}, which is described by wave function \eqref{073017:wfAB}, is it possible to combine \eqref{073117:ABwfPsi} and \eqref{073117:ABwfPhi} into \eqref{073017:wfAB}?

To put that another way, if\begin{equation*} \left( \begin{array}{c|c}  A_\Psi & B_\Psi \\ \hline \lvert \spadesuit \rangle & \lvert \diamondsuit \rangle \\ \lvert \diamondsuit \rangle &  \lvert \spadesuit \rangle \end{array} \right) + \left( \begin{array}{c|c}  A_\Phi & B_\Phi \\ \hline \lvert \spadesuit \rangle & \lvert \spadesuit \rangle \\ \lvert \diamondsuit \rangle &  \lvert \diamondsuit \rangle \end{array} \right) \Longleftrightarrow \left( \begin{array}{c|c} A & B \\ \hline \lvert \spadesuit \rangle &  \lvert \spadesuit \rangle \\ \lvert \spadesuit \rangle & \lvert \diamondsuit \rangle \\ \lvert \diamondsuit \rangle &  \lvert \spadesuit \rangle \\ \lvert \diamondsuit \rangle & \lvert \diamondsuit \rangle \end{array} \right), \end{equation*}is it true that \begin{equation*} \lvert \Psi \rangle_{AB} + \lvert \Phi \rangle_{AB} \Longleftrightarrow \lvert \psi \rangle _A\otimes \lvert \psi \rangle _B ? \end{equation*}

As a matter of fact, it is!\begin{equation} \begin{aligned} \lvert \psi \rangle _A\otimes \lvert \psi \rangle _B &= \frac{1}{2}\left(\lvert \spadesuit \rangle _A\otimes \lvert \spadesuit \rangle _B+\lvert \spadesuit \rangle _A\otimes \lvert \diamondsuit \rangle _B + \lvert \diamondsuit \rangle _A\otimes \lvert \spadesuit \rangle _B + \lvert \diamondsuit \rangle _A\otimes \lvert \diamondsuit \rangle _B\right) \\[2ex] &= \frac{1}{\sqrt{2}}\left[ \frac{1}{\sqrt{2}} \left( \lvert \spadesuit \rangle _A\otimes \lvert \diamondsuit \rangle _B+\lvert \diamondsuit \rangle _A\otimes \lvert \spadesuit \rangle _B \right) + \frac{1}{\sqrt{2}} \left( \lvert \spadesuit \rangle _A\otimes \lvert \spadesuit \rangle _B + \lvert \diamondsuit \rangle _A\otimes \lvert \diamondsuit \rangle _B \right) \right] \\[2ex] &= \frac{1}{\sqrt{2}} \left( \lvert \Psi \rangle_{AB} + \lvert \Phi \rangle_{AB} \right) \end{aligned} \label{073117:ABwfres} \end{equation}

Equation \eqref{073117:ABwfres} is a very important result. It tells us that the combined wave function of our non-entangled playing cards is a superposition of the two entangled states. Amongst other things, this tells us that we can create an entangled system by performing the right action on a non-entangled state, since all we are doing is collapsing wave function \eqref{073117:ABwfres} into one of its two entangled components. In fact, this is how entanglement is produced in a laboratory, and it therefore represents the first step in the process of quantum teleportation.

Now that we have familiarized ourselves with the mathematical notation of quantum entanglement, we are ready to tackle our ultimate goal of learning how to teleport. So stay tuned—that post is on its way! In the meantime, if you have any questions, or if you see that I have made a mistake, please let me know; I really appreciate your feedback.

Thanks also to everyone who has taken the time to read this, and if you like it, share it! That really helps me out a lot.

Thanks!

Aaron