# Partial Measurement and the Math Behind it

$\newcommand{\ket}{\left| #1 \right>}$ $\newcommand{bra}{\left< #1 \right|}$

Talking about the partial measurement the professor defines the state $\ket \psi$ to be

$$\ket{\psi} = \sum_{i,j} a_{ij} \ket{e_i} \otimes \ket{f_j}$$

where $\ket{e_i} \in V$ and $\ket{f_j} \in W$ are orthonormal bases. Then he rewrites the state $\ket \psi$ as

$$\ket \psi = \sum_i \ket {e_i} \otimes \ket{w_i}$$

where $\ket{w_i}= \sum_j a_{ij}\ket{f_j}$. I'm find until now. However he does the following:

$$\bra{e_i}\sum_j \ket{e_j} \otimes \ket{w_j}$$

and says, I quote, "You should understand this equation as $\bra{e_i}$ only talks to $\ket{e_j}$. You could have written the $\bra{e_i}$ in different ways maybe you could have written $\bra{e_i} \otimes \mathbf 1$" and he says that writing $\bra{e_i} \otimes \mathbf 1$ is a little strange because $\bra{e_i}$ is a bra and $\mathbf 1$ is an operator.

Assuming that what we mean by the bra $\bra{e_i}$ is $\bra{e_i} \otimes \mathbf 1$, I have difficulty understanding what the following mathematical object is:

$$\bra{e_i} \otimes \mathbf 1 \sum_j \ket{e_j} \otimes \ket{w_j} = \sum_j \delta_{ij} \otimes \ket{w_j}$$

I can think of $\delta_{ij}$ as a number but then I don't know what $1 \otimes \ket{w_i}$ (note that $1$ is a number in this expression) should mean.

If on the other hand, I think of $\delta_{ij}$ as a tensor then I cannot simplify this any further and I should write the last expression as:

$$\delta_i^{\;j} \otimes \ket{e_j}$$

where there is a funny summation over $j$. Either way I cannot reduce it any further in order to get to the equation that he has at the very last which is the following:

$$\bra{e_i}\sum_j \ket{e_j} \otimes \ket{w_j} = \ket{w_i}$$

I don't understand how we have gone from a space with dimension $v\cdot w$, where $\mathrm{dim}(V)=v$ and $\mathrm{dim}(W)=w$, to a space with dimension $w$. What is the meaning of $s\otimes \ket{e_i}$, where $s$ is a scalar, if such an object really exists and lastly is the mathematics behind the above calculation is correct?

I can understand your confusion. Let me start by saying that this is the "correct" way of writing down what you want to have in physics notation:

$$\langle e_i|\otimes \mathbf{1} \sum_j |e_j\rangle\otimes |w_j\rangle = \sum_j \delta_{ij} |w_j\rangle$$

Note the absence of the tensor product sign. Otherwise, you will never have this reduction of dimension, as you noted. In physics notation, the idea is that $\langle e_i|e_j\rangle$ is now a number, it is no longer a vector in some Hilbert space, so the tensor product makes no sense anymore.

Mathematically, just define $$\langle e_i|\otimes \mathbf{1}: \mathcal{H}\otimes \mathcal{H}^{\prime}; |e_j\rangle\otimes |f_k\rangle \mapsto \delta_{ij} |f_k\rangle.$$ If you take this as a definiton, the rest will work out.

To be honest though, I don't really like this answer, because I do not think this is the ideal formalism. In particular, it's maybe not totally clear why the definition of $\langle e_i|\otimes \mathbf{1}$ is as above.

Let us consider a state $|\psi\rangle \in\mathcal{H}$ (some - for sake of simplicity finite dimensional - Hilbert space). We know that we can write $|\psi\rangle= \sum_i a_i |e_i\rangle$ for some basis of the Hilbert space. Let us now suppose that we measure an observable $A$ with eigenstates $|e_i\rangle$ according to eigenvalues $a_i$. Let's suppose we measured $a_1$ which happens with probability $|\langle e_1|\psi\rangle|^2$ (suppose that this is not zeor), then we know that our state after measurement is in state $|e_1\rangle$. This means that we projected onto the eigenspace of $a_1$. In other words, we applied the operatore $P=|e_1\rangle\langle e_1|$ to $|\psi\rangle$ and renormalized the state. We could call this $P$ the "measurement of state $|e_1\rangle$".

Now suppose our system was in a tensor product $|\psi\rangle \otimes |\phi\rangle\in \mathcal{H}\otimes \mathcal{H}^{\prime}$. Since the two systems are uncorrelated, we would say that making a measurement on the first system should not disturb the second system. This means that if we measure $P$ as above, we should in fact measure $P\otimes \mathbf{1}$ and the state after this partial measurement would be $(P\otimes \mathbf{1}) |\psi\rangle \otimes |\phi\rangle\propto |e_1\rangle\otimes |\phi\rangle$ (the state is not normalized, hence we don't have equality in the last step - this can be corrected by rewriting the state with the $|w_i\rangle$). This is already nearly what was written above. The only difference is that you don't care about the first system anymore. Since you know what state it is in, you can just erase it and obtain $|\phi\rangle$. For entangled states, you'll now do the same thing - only you'll have to introduce a few more sums. In the end, you'll arrive at the outcome you wrote in your question.

Putting this together, it is easier to define the process of "measuring $P$ on the first system and then forgetting it" by defining the operator $\langle e_i|\otimes \mathbf{1}$ as above - the outcome is the same.

To give yet another idea how to make sense of this, you'll have to consider density matrices. More precisely, given a density matrix $\rho\in \mathcal{B}(\mathcal{H}\otimes \mathcal{H}^{\prime})$, and given a pure state $|\Psi\rangle$ with projector $P=|\Psi\rangle\langle\Psi|$, you can write the (unnormalized) state on the second subsystem after projection onto $P$ as $$\rho_{post~measure}=\operatorname{tr}_A((P\otimes \mathbf{1}) \rho)$$ where $\operatorname{tr}_A$ is the partial trace of the first subsystem. The partial trace is a well-defined object, but when you write it down in coordinates, you'll have to write down something like $\langle e_i|\otimes \mathbf{1}$ with the definition given above.

Maybe it helps if we can place this in a context.

You can have spaces like $A$ and $B$ and then you can make product spaces like $A \otimes B$. You can make linear operators like $S:A\rightarrow C$ and $T:B\rightarrow D$ and then since an arbitrary thing in $A \otimes B$ is spanned by things like $a \otimes b$ (with $a\in A$ and $b\in B$) you can clearly make the object $s=S(a)$ and $t=T(b)$ and hence get the object $s\otimes t\in C\otimes D.$ Those are all things we can do. By linearity we can now send any object in $A \otimes B$ to an object in $C\otimes D.$

So now you can consider the map $S\otimes T:A\otimes B\rightarrow C\otimes D$ that maps via $a\otimes b\mapsto S(a)\otimes T(b),$ so we can take the tensor product of operators and have them act on the tensor product of the corresponding spaces, like $S\otimes T:A\otimes B\rightarrow C\otimes D:a\otimes b\mapsto S(a)\otimes T(b).$

Nothing weird. Now, what is a bra? It is a map from $A$ to $\mathbb C.$ And $\mathbb C$ is nothing except a special (but simple) Hilbert Space so we can let $A=B=D=H,$ let $C=\mathbb C,$ $T=\mathbf 1:H\rightarrow H:h\mapsto h$ and let $S=\bra e :A\rightarrow C:\ket a \mapsto \bra e \ket a.$

So it isn't weird and since $\bra e:H\rightarrow \mathbb C$ and $\mathbf 1:H\rightarrow H$ we can clearly have $\bra e \otimes \mathbf 1 :H\otimes H\rightarrow \mathbb C \otimes H.$ But if you look at the tensor product $\mathbb C \otimes H$ you'll notice that there is an obvious isomorphism (i.e. $h\mapsto 1\otimes h$) between $H$ and $\mathbb C \otimes H$ so that's the map they are talking about.

It's just the map $\bra e \otimes \mathbf 1$ with the result in $\mathbb C \otimes H$ identified with $H.$

Now, you specifically asked about what $\mathbb C \otimes H$ is mathematically. And the tensor product of any hilbert spaces is the hilbert space spanned elementary products of basis vectors from the original hilbert spaces (if you tensor product an infinite number of hilbert spaces together at once you might have to spend more time worrying about making the result complete as a metric space and even putting an inner product on it). And $\mathbb C$ is a hilbert space, with only one orthonormal basis element (it is one dimensional as a Hilbert Space), just pick any number like $e^{i\theta},$ and any element can be written as a linear combination of that one orthonormal element. So the tensor product $\mathbb C \otimes H$ (when $H$ has a basis $\mathcal B=\{\ket n\}$) is the set $\{e^{i\theta}\otimes \ket n: \ket n \in \mathcal B\}.$ Just like any other tensor product. And as a Hilbert Space it happens to not look or act any differently than $H.$