You might think it's sufficient to just write down the state vector of each individual object -- but this doesn't really tell us the entire picture. Suppose for example we're considering the state of Schrodinger's cat, less popularly known as Schrodinger's cuckoo, where the box contains TNT and the cuckoo bird. The state of the TNT is $|\mathrm{unexploded}\rangle+|\mathrm{exploded}\rangle$ and the state of the cuckoo is $|\mathrm{alive}\rangle+|\mathrm{dead}\rangle$ -- right?
If I'm in charge of the box, the state of the cuckoo will definitely be $|\mathrm{dead}\rangle$ even if the state of the TNT is $|\mathrm{unexploded}\rangle$
Not exactly. There is a correlation between the state of the TNT and the state of the cuckoo that goes missing here -- the cuckoo is dead if and only if the TNT has exploded, and the cuckoo is alive if and only if the TNT is unexploded. In fact, defining the state vectors separately doesn't really make any sense -- we're just assigning the coefficients based on overall probabilities, and as we will see, this is really mixing up quantum and classical probabilities in a certain way whereas the state vector is supposed to only show quantum probabilities.
Well, this is really the same question as what we do when we have multiple correlated random variables in statistics -- we define a "joint" probability function on a "joint" phase space that is the Cartesian product of the original phase spaces.
You may be inclined to claim that similarly, our new Hilbert space should be the Cartesian product of the original Hilbert spaces. But Hilbert spaces are different in a fundamental sense from these phase spaces -- every point on the classical phase space is an independent state in the Hilbert space -- and general vectors are distributions on the classical phase space. So like how the cardinalities of the classical phases are multiplied, the dimensions of the Hilbert spaces are multiplied.
The reason that we sometimes draw an analogy between the classical state space and the quantum state space is that the state vectors are really the "real objects" in quantum mechanics, and the Hilbert space shows the possible configurations of the state in this sense.
The product we want of Hilbert spaces -- which is not the Cartesian product -- is called a tensor product of linear spaces -- given an orthogonal basis $(|\phi_1\rangle,|\phi_2\rangle,\ldots)$ for the first Hilbert space and $(|\psi_1\rangle,|\psi_2\rangle,\ldots)$ for the second, their tensor product is spanned by new vectors which we denote as
$$\left( {\begin{array}{*{20}{c}}{|{\phi_1}\rangle \boxtimes |{\psi_1}\rangle ,|{\phi_1}\rangle \boxtimes |{\psi_2}\rangle ,...,}\\{|{\phi_2}\rangle \boxtimes |{\psi_1}\rangle ,|{\phi_2}\rangle \boxtimes |{\psi_2}\rangle ,...,}\\ \vdots \end{array}} \right)$$
We're using $\boxtimes$ instead of $\otimes$ in the above enumeration of the basis, because we haven't yet defined the tensor product of states. The idea is that $|\phi_i\rangle\boxtimes|\psi_j\rangle$ are just placeholders, and we will shortly state that they are/can be the tensor product $|\phi_i\rangle\otimes|\psi_j\rangle$, which we will define now.
Certainly, this can represent any possible state in which the combined system of two objects can be in. What we need is a way to express the state of a combined system of two independent things in this "larger" Hilbert space -- i.e. a map from $H_1\times H_2\to H_1\otimes H_2$ that takes the (pure) states of two independent objects in $H_1$ and $H_2$ and outputs their state as a combined system in $H_1\otimes H_2$ -- we will call this product the tensor product of vectors, and denote it by the same symbol $\otimes$.
OK, so what's the map? Certainly, $|\phi_i\rangle\otimes|\psi_j\rangle$ must form an orthogonal basis for $H_1\otimes H_2$ (why? think about this for a while -- they're clearly orthogonal, as they are mutually exclusive -- you can't be in "$|\phi_i\rangle$ and $|\psi_j\rangle$" and "$|\phi_{i'}\rangle$ and $|\psi_{j'}\rangle$" unless $(i,j)=(i',j')$; spanning is proven similarly, as considering the $|\phi_i\rangle$s and $|\psi_j\rangle$s as eigenstates of some operators $X$ and $Y$ on $H_1$ and $H_2$, then if one performs the operation of "observing $X$ and $Y$" -- and we can do this because the objects are independent -- then because the objects must be found in one of $|\phi_i\rangle$ and one of $|\psi_j\rangle$, the system must be found in one of $|\phi_i\rangle\otimes|\psi_j\rangle$ -- thus its original state was a linear combination of such states).
OK, so
$$
(p_1|\phi_1\rangle+p_2|\phi_2\rangle+\ldots)\otimes(q_1|\psi_1\rangle+q_2|\psi_2\rangle+\ldots)\\
\begin{align}
=\ & r_{11}|\phi_1\rangle\otimes|\psi_1\rangle + r_{12}|\phi_1\rangle\otimes|\psi_2\rangle+\ldots+\\
&r_{21}|\phi_2\rangle\otimes|\psi_1\rangle + r_{22}|\phi_2\rangle\otimes|\psi_2\rangle+\ldots+\\
& \vdots
\end{align}$$
What are the coefficients $r_{ij}$?
Well, it's fairly obvious that $|r_{ij}|^2=|p_{i}|^2|q_{j}|^2$ -- that the probabilities are multiplicative, this is tautological given what we want our product to represent -- the probability that the system is found in the state $|\phi_i\rangle\otimes|\psi_j\rangle$ is the probability that the objects are found in states $|\phi_i\rangle$ and $|\psi_j\rangle$, which is the product of the respective probabilities, as they are independent objects.
Is it also true that the probability amplitudes are multiplicative, i.e. $r_{ij}=p_iq_j$?
This may seem hard to prove, but the idea is quite simple: suppose we observe the state with the observables $X$ and $Y$, and find it in the state $|\phi_i\rangle\otimes|\psi_j\rangle$. Well, then if $r_{ij}=u_{ij}|r_{ij}|$ for some unit complex number $u_{ij}$, then from the right-hand-side, we must have collapsed to $u_{ij}|\phi_i\rangle\otimes|\psi_j\rangle$. So we must have $u_{ij}=1$.
So indeed the product we're looking for is exactly the tensor product from tensor algebra.
Here's a thing worth noting -- we've been referring to "systems" and "objects" as if they are somehow completely distinct things. But are they? The cat's state is itself a tensor product of a massive number of different states belonging to each elementary particle in its body, and lives already in a massive Hilbert space, because the "object" is itself a system. We will use the term subsystem instead of object from now.
Alright: so we now know that elements of the tensored Hilbert space are all states, and only the ones that are factorable into an element of $H_1$ and $H_2$ represent subsystems that are independent. This is precisely how only factorable probability mass/density functions represent independent variables in statistics. Otherwise the variables are correlated -- not necessarily linearly correlated, but correlated.
Such correlations can, of course, exist in our quantum mechanical theory, too -- like the cuckoo-TNT system we mentioned earlier. These are called quantum correlations or quantum entanglement.
Why the fancy name? Because its consequences may seem superficially kinda "surprising". It's also a demonstration of quantum mechanics being different from classical mechanics, because without entangled states, the dimension of $H_1\otimes H_2$ would indeed be the sum of those of $H_1$ and $H_2$ rather than their product, like with phase spaces in classical mechanics.
OK, what kind of surprising consequences?
They're basically all of the following nature: suppose we have a state given by:
$$\frac1{\sqrt2} (|\phi\rangle\otimes|\psi\rangle+|\psi\rangle\otimes|\phi\rangle)$$
I.e. two entangled particles where we know that they are in two distinct states, but we don't know which is which. Such a state can certainly be produced -- how? Just put two identical independent particles in a box then do a "partial" measurement -- a "peek" -- (which can be achieved, e.g. by some logic gates) that checks if they're in the same state or not, and uncovers no other information.
Now separate the particles spatially -- there's nothing wrong with this, they're still a system, which still has a state -- and give one to Alice and the other to Bob. Now if Bob looks at his particle and sees it in $|\psi\rangle$, he immediately knows that Alice could only observe her particle to be in state $|\phi\rangle$ -- there's nothing Alice can do to change this outcome.
(you may worry that spatially separating the particles alters the state in some important way -- but it doesn't: the states $|\phi\rangle$ and $|\psi\rangle$ are both transformed individually that doesn't change the entangled structure of the combined state -- make sure this makes sense to you. But if it makes you happy, you could imagine the particles were already spatially separated when they were first entangled.)
OK, perhaps you don't find this particularly surprising or unintuitive -- I don't either. But perhaps you do -- perhaps you think there's a violation of locality -- and the reason you do is because you haven't yet fully accepted logical positivism. Let's consider what locality entails for each observer in the set-up, and see if it's violated:
- Alice: From Alice's perspective, Bob opening his box is just another way to observe her particle -- or rather, she can observe Bob's brain that contains the information, which collapses the state from her perspective. But this is perfectly local -- it takes time for information to propagate from Bob to her. Alternatively, if she doesn't observe Bob's brain and just observes her own box later, that's when her state collapses to $|\phi\rangle$, and she then learns that Bob had collapsed his state into $|\psi\rangle$ -- but as Bob cannot choose what his state vector collapses to, so he can't send her any information through entanglement. Even if there were a large number of entangled systems this way, the distribution of the states Alice can observe is the same whether or not Bob has collapsed his states (you can confirm this -- this is an idea called the no-communication theorem which we will discuss later in more mathematical detail).
- Bob: Certainly, Bob acquires knowledge of something far away, but no information actually propagated from Alice to him -- he just observed his own box.
- another observer: Charlie, who stands somewhere between Alice and Bob, too takes time to observe Bob's brain.
So there really isn't a violation of locality. This isn't surprising at all -- certainly one could have classical correlations too. You could just juggle two distinct particles in a box and give them to each person, and Bob discovering his particle allows him to determine Alice's particle.
The difference between the classical case and the quantum case is that in the classical case you could pretend that there's some hidden truth that is just not known to the observers. Quantum mechanics forbids any such hidden truth (as confirmed by commutator relations), and forces you to accept logical positivism, and there cannot be a "universal observer" as such a notion is inherently non-local. But the fact that correlation isn't non-local doesn't depend on whether you have metaphysical notions of hidden truths in classical physics -- it is a physical question, and is the same in the classical and quantum cases.
Are we done writing down our algebra of tensor products? We still haven't discussed how inner products and projections of tensor products behave. The basic question is "how do we upgrade/combine operators from $H_1$ and $H_2$ to $H_1\otimes H_2$? Let's start with the simple case of a factorable state in the form $|\phi\rangle\otimes|\varphi\rangle$. Suppose we apply a projection operator $X$ on the first particle. Have we made any observation on the second state? No -- just an identity projection. Or we could make an observation, a projection $Y$. So we can say that for the combined observation $X\otimes Y$,
$$(X\otimes Y)(|\phi\rangle\otimes|\varphi\rangle)=(X|\phi\rangle)\otimes(Y|\varphi\rangle)$$
And an upgrade from $H_1$ to $H_1\otimes H_2$ is just tensoring with the identity $X\otimes 1$.
But the full range of operators on $H_1\otimes H_2$ is a lot more complicated. We could consider entangled states. We could consider operators that are entangled ("partial measurement" operators like we described -- think about what these are). How would measurements on linear combinations of states look like (we know they should apply linearly, but let's show that)?
Suppose we have a state in the form $\frac1{\sqrt2}|\phi\rangle\otimes|\varphi\rangle+\frac1{\sqrt2}|\varphi\rangle\otimes|\phi\rangle$. What exactly is this? We had two independent subsystems each in state $\frac1{\sqrt2}|\phi\rangle+\frac1{\sqrt2}|\varphi\rangle$, then we made an observation that showed they were in two distinct states -- we don't know which is in which. Now we make an observation and collapse the first subsystem to $|\chi\rangle$.
How does this alter the state of sub-system 2?
OK, so "was" (quotation marks! quotation marks!) the system in $|\phi\rangle\otimes|\varphi\rangle$ or $|\varphi\rangle\otimes|\phi\rangle$? This is a question for Bayes' theorem.
So the "probability that the system was in $|\phi\rangle\otimes|\varphi\rangle$" (quotation marks! quotation marks!) is:
$$\frac{\frac12|\langle\chi|\phi\rangle|^2}{\frac12|\langle\chi|\phi\rangle|^2 + \frac12|\langle\chi|\varphi\rangle|^2}$$
(which if $\langle\phi|\varphi\rangle=0$ is just $|\langle\chi|\phi\rangle|^2$) And analogously for the other possibility. So the collapse of sub-system 1 to $|\chi\rangle$ collapses the entire state to
$$|\chi\rangle\otimes\left(\frac1{\sqrt2}\langle\chi|\phi\rangle\cdot|\varphi\rangle+\frac1{\sqrt2}\langle\chi|\varphi\rangle\cdot|\phi\rangle\right)$$
Or some normalisation thereof if $\langle\phi|\varphi\rangle\ne0$. You can confirm that if $|\chi\rangle=|\phi\rangle$ or $|\chi\rangle=|\varphi\rangle$, this reduces to $|\phi\rangle\otimes|\varphi\rangle$ or $|\varphi\rangle \otimes |\phi\rangle$ respectively as we expect.
You can check that this is precisely what you get from applying the projection operator $|\chi\rangle\langle\chi|\otimes 1$ as a linear operator to the original state. The above argument can be repeated for a general vector in the tensored space, yielding the linearity of the tensored operator.
There was the other case we mentioned -- we may have operators that are themselves entangled. What does this mean? Suppose we start with the factorable state:
$$\frac12|\phi\rangle\otimes|\phi\rangle+\frac12|\phi\rangle\otimes|\varphi\rangle+\frac12|\varphi\rangle\otimes|\phi\rangle+\frac12|\varphi\rangle\otimes|\varphi\rangle$$
Then perform the observation corresponding to "are the states different from each other?" This is a projection onto the plane spanned by $(|\phi\rangle\otimes|\varphi\rangle,|\varphi\rangle\otimes|\phi\rangle)$, perpendicular to the plane spanned by $(|\phi\rangle\otimes|\phi\rangle,|\varphi\rangle\otimes|\varphi\rangle)$ (confirm that this is true based on our discussion above of applying inner products to tensored states) -- we can write this as:
$$\left(|\phi\rangle\otimes|\varphi\rangle\right)\left(\langle\phi|\otimes\langle\varphi|\right) +
\left(|\varphi\rangle\otimes|\phi\rangle\right)\left(\langle\varphi|\otimes\langle\phi|\right)
$$
(check that this, and the system of "distributing bras" makes sense). Or alternatively:
$$|\phi\rangle\langle\phi|\otimes|\varphi\rangle\langle\varphi|+|\varphi\rangle\langle\varphi|\otimes|\phi\rangle\langle\phi|$$
One can check that applying this operator to the factorable state indeed results in the entangled state (up to normalisation). For better insight, perform the operation on the factored form of the state.
Are we done writing down our algebra of tensor products? We still haven't discussed how inner products and projections of tensor products behave. The basic question is "how do we upgrade/combine operators from $H_1$ and $H_2$ to $H_1\otimes H_2$? Let's start with the simple case of a factorable state in the form $|\phi\rangle\otimes|\varphi\rangle$. Suppose we apply a projection operator $X$ on the first particle. Have we made any observation on the second state? No -- just an identity projection. Or we could make an observation, a projection $Y$. So we can say that for the combined observation $X\otimes Y$,
$$(X\otimes Y)(|\phi\rangle\otimes|\varphi\rangle)=(X|\phi\rangle)\otimes(Y|\varphi\rangle)$$
And an upgrade from $H_1$ to $H_1\otimes H_2$ is just tensoring with the identity $X\otimes 1$.
But the full range of operators on $H_1\otimes H_2$ is a lot more complicated. We could consider entangled states. We could consider operators that are entangled ("partial measurement" operators like we described -- think about what these are). How would measurements on linear combinations of states look like (we know they should apply linearly, but let's show that)?
Suppose we have a state in the form $\frac1{\sqrt2}|\phi\rangle\otimes|\varphi\rangle+\frac1{\sqrt2}|\varphi\rangle\otimes|\phi\rangle$. What exactly is this? We had two independent subsystems each in state $\frac1{\sqrt2}|\phi\rangle+\frac1{\sqrt2}|\varphi\rangle$, then we made an observation that showed they were in two distinct states -- we don't know which is in which. Now we make an observation and collapse the first subsystem to $|\chi\rangle$.
How does this alter the state of sub-system 2?
OK, so "was" (quotation marks! quotation marks!) the system in $|\phi\rangle\otimes|\varphi\rangle$ or $|\varphi\rangle\otimes|\phi\rangle$? This is a question for Bayes' theorem.
The above diagram is for illustration only. There is no real hidden truth (as we will see a few articles from now) of whether the state was initially $|\phi\rangle\otimes|\varphi\rangle\otimes|\phi\rangle$ or otherwise. But the probabilities still obey all the standard laws, such as Bayes's theorem, so tree diagrams make sense to illustrate this.
So the "probability that the system was in $|\phi\rangle\otimes|\varphi\rangle$" (quotation marks! quotation marks!) is:
$$\frac{\frac12|\langle\chi|\phi\rangle|^2}{\frac12|\langle\chi|\phi\rangle|^2 + \frac12|\langle\chi|\varphi\rangle|^2}$$
(which if $\langle\phi|\varphi\rangle=0$ is just $|\langle\chi|\phi\rangle|^2$) And analogously for the other possibility. So the collapse of sub-system 1 to $|\chi\rangle$ collapses the entire state to
$$|\chi\rangle\otimes\left(\frac1{\sqrt2}\langle\chi|\phi\rangle\cdot|\varphi\rangle+\frac1{\sqrt2}\langle\chi|\varphi\rangle\cdot|\phi\rangle\right)$$
Or some normalisation thereof if $\langle\phi|\varphi\rangle\ne0$. You can confirm that if $|\chi\rangle=|\phi\rangle$ or $|\chi\rangle=|\varphi\rangle$, this reduces to $|\phi\rangle\otimes|\varphi\rangle$ or $|\varphi\rangle \otimes |\phi\rangle$ respectively as we expect.
You can check that this is precisely what you get from applying the projection operator $|\chi\rangle\langle\chi|\otimes 1$ as a linear operator to the original state. The above argument can be repeated for a general vector in the tensored space, yielding the linearity of the tensored operator.
What about inner products of tensored states? Convince yourself that the inner product of $|\phi_1\rangle\otimes|\phi_2\rangle$ and $|\chi_1\rangle \otimes |\chi_2\rangle$ is $\langle\phi_1|\chi_1\rangle\langle\phi_2|\chi_2\rangle$.
There was the other case we mentioned -- we may have operators that are themselves entangled. What does this mean? Suppose we start with the factorable state:
$$\frac12|\phi\rangle\otimes|\phi\rangle+\frac12|\phi\rangle\otimes|\varphi\rangle+\frac12|\varphi\rangle\otimes|\phi\rangle+\frac12|\varphi\rangle\otimes|\varphi\rangle$$
Then perform the observation corresponding to "are the states different from each other?" This is a projection onto the plane spanned by $(|\phi\rangle\otimes|\varphi\rangle,|\varphi\rangle\otimes|\phi\rangle)$, perpendicular to the plane spanned by $(|\phi\rangle\otimes|\phi\rangle,|\varphi\rangle\otimes|\varphi\rangle)$ (confirm that this is true based on our discussion above of applying inner products to tensored states) -- we can write this as:
$$\left(|\phi\rangle\otimes|\varphi\rangle\right)\left(\langle\phi|\otimes\langle\varphi|\right) +
\left(|\varphi\rangle\otimes|\phi\rangle\right)\left(\langle\varphi|\otimes\langle\phi|\right)
$$
(check that this, and the system of "distributing bras" makes sense). Or alternatively:
$$|\phi\rangle\langle\phi|\otimes|\varphi\rangle\langle\varphi|+|\varphi\rangle\langle\varphi|\otimes|\phi\rangle\langle\phi|$$
One can check that applying this operator to the factorable state indeed results in the entangled state (up to normalisation). For better insight, perform the operation on the factored form of the state.
In the representation shown in the above diagram, a factorable state is one given by a single square region, while a factorable projection operator is one that selects a square region out of the maximally distributed state. With this notion, it becomes clear that an entangled operator (one that cannot be factored into operators on each Hilbert space) is the only kind of operator that projects a factorable state into an entangled state.
We're not saying that entangled operators always project factorable states onto entangled ones -- you can just measure an irrelevant property of the system (e.g. you know each particle is either in the UK or France, and you check "are they both in India?"). But they are the only operators that can, and for any such operator, there exist factorable states that it entangles (almost by definition).
Note that we're only talking about projection operators above. We could certainly have factorable observables that enforce a partial measurement -- e.g. $X_1\otimes X_2$, which measures the product of the positions of the two particles -- but the projection operators onto each of the eigenstates of this operator are not factorable (check this).
The following rules then determine the action of our operations on the tensored Hilbert space.
- $(A\otimes B)(|\phi\rangle\otimes|\varphi\rangle)=(A|\phi\rangle)\otimes(B|\varphi\rangle)$ -- the tensored operators associate with the corresponding states.
- The tensor product of two linear operators is linear.
- The image of a linear combination of operators is the linear combination of the images, i.e. $(A+B)|\psi\rangle=A|\psi\rangle+B|\psi\rangle$.
