# A question about quantum measurement and associating a linear self adjoint operator to it

I have a question about the concept of measurement and observable in quantum mechanics. I'd like to fist explain my understanding of it and then ask the question.

First we have a system and its state function $\psi(x)$ and naturally it means there are two mutually incompatible observables $x$ and $\omega$ (momentum). Now we'd like to set up an experimental setup and verify whether this state function actually desribes the given system.

Consider an example of tossing an unbiased coin. We want to verify whether this coin is unbiased or not. There are two mutually exclusive outputs heads and tails. We repeat this experiment infinite number of times (actually only very high number of times) and record how many heads and tails resulted there by we know the probability of heads and tails.

Similarly we set up the quantum system, and perform the experiment repeatedly and every time we note the value of $x$ (position of particle) and there by we get the probability distribution of $x$ and hence can verify the validity of the state function $\psi(x)$.

But this is a very naive way of doing the experiment as we have to set up particle sensors every where in space (all $x$) and it they should be able to detect particles at any velocity. This kind of experiment is not easy to set up. What if we already have some equipment and experimental set up to verify a certain type of state functions. We may use them to our advantage.

Suppose we have measuring equipment which can directly verify the occurence of certain state functions $f_1$,$f_2$,...

Assume the experiment gives out to a system into different possible states $f_1$,$f_2$,$f_3$,... which are mutually exclusive that is the systems cannot jump to two state $f_1$ and $f_2$ simultaneously. which means if system jumps to $f_1$ then it will not jump to any state $f_k$, $k \ne 1$. Let us assume that the probability of jumping from a state $psi_1$ to a state $\psi_2$ is given by the inner product $<\psi_1|\psi_2>$. Now that the states $f_1$,$f_2$... are mutually exclusive outcomes, and probability of transitioning from one state to other is given by their inner product, the set of states $f_1$,$f_2$,... form a orthogonal set. Since the probabilty of transitioning from one state to another is given by the inner product, these states form an orthogonal set and $\psi$ can be expressed as a linear combination of these states.( Lets us assume this set is a complete orthonormal set so that we can generalize it for any $\psi(x)$).

Now what we are actually doing is we have an experimental setup to detect certain set of states which form a complete orthogonal basis of the Hilbert space. We can repeat the experiment to find the probability of occurrences of these states.

We have chosen the experimental set up for detecting $f_1$,$f_2$... as we already have such an equipment, means we are doing a certain type of measurement defined by the set $f_1$,$f_2$,... Which means the measurement is defined by choice of complete orthogonal set $f_1$,$f_2$,... and evry such set defines a linear, self adjoint operator $L$ on hilbert space, we can associate/define every measurement by a linear self adjoint operator.

Hence measurement = linear self adjoint operator. It can be mathematically easily verified that the set of states $f_1$,$f_2$,... are eigen vectors of the linear operator, and since we choose inner product as the probability of jumping from state to another state, it can be easily seen that the eigen values associated with these vectors give the probability of their occurence.

Question

Why do we always need to choose a a orthogonal set which is complete? I mean Why we need to choose a measurement defined by a linear operator.

What I say is can't we choose a set of orthogonal set which is not complete but it be used to express the our state vector as a linear combination. I mean why cant we choose a set of orthonormal vectors which is not complete, but $\psi$ lies in its span. After all thats wht we want. We can choose this set based on the $\psi$ we have and also our equipment.

We just don't want to generalize a measurement to every $\psi$, but we choose measurement based on $\psi$ we have.

(Added after answer by SMeznaric) Thanks for the illustrous answer. My question is not against linearity or orthogonality, but regarding why formalize and give importance to a universal orthogonal basis. I mean why should we even talk about them. On the contrary Lets say we decide the orthogonal basis purely on the ψ(x) at hand. Unless you give ψ(x) we will not talk about anything. We use linear orthogonal basis states, but our measurement/basis will totally be adapted to the given ψ(x) and we will not talk anything unless you give the ψ(x) to be analyzed.

Consider this example of a state function $\psi(x)$ of an electron. Now it tells me that the best way to verify this is to put a electron detector and a filter (allows only electrons of 16 units/s) covering the region of (0,1) and place a similar one but with a filter of 32 units/s covering (1,2), now you are done. But this is based on common sense, now we need a mathematical model for this, which chooses the basis based on the state function, and yet make very useful in practical terms. (suitability to our equipment).

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So you have two questions: Why linear and why complete.

In short, linearity is required because probabilistic mixtures of state must give a probabilistic mixtures of results. Completeness is required so that probabilities sum to one for all states. Here are more formal details.

Linearity: Let $M_k$ be a measurement operator corresponding to outcome $k$. Given a mixed state $\rho$, $M_k(\rho)$ gives us the post-measurement state. Now suppose you are given $\rho_1$ with probability $p_1$ and $\rho_2$ with probability $p_2$. Then $\rho = p_1 \rho_1 + p_2 \rho_2$. Then you want to have that the post measurement state is $M_k(\rho_1)$ with probability $p_1$ and $M_k(\rho_2)$ with probability $p_2$. Hence, $M_k(p_1 \rho_1 + p_2 \rho_2) = p_1 M_k(\rho_1) + p_2 M_k(\rho_2)$. This is linearity. So from now on I will write it as $M_k \rho M_k^\dagger$, where $M_k$ is a matrix. (As a curiosity, you might want to know that there are more general linear operators on the space of density matrices, but they correspond to measurements where some information is lost. For simplicity I will not consider these here, but see http://en.wikipedia.org/wiki/Quantum_operation).

If $M_k$ correspond to simple states you can write them as $|f_k\rangle \langle f_k|$, since this is the unique linear operator that maps all states with support on $|f_k\rangle$ to something $\propto |f_k\rangle$.

Completeness: This comes from requiring that the norm of the post-measurement state is the probability of its detection, ie. $Tr[M_k \rho M_k^\dagger] = p_k$. Then you have that $\sum_k Tr[M_k \rho M_k^\dagger] = 1$ for all $\rho$, which implies $\sum_k M_k^\dagger M_k = 1$. For simple states as you require, where $M_k = |f_k\rangle \langle f_k|$, you have for orthonormal $|f_k\rangle$ that $\sum_k |f_k\rangle \langle f_k| = 1$ if and only if $|f_k\rangle$ are complete.

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Thanks for the illustrous answer. My question is not against linearity or orthogonality, but regarding why formalize and give importance to a universal orthogonal basis. I mean why should we even talk about them. On the contrary Lets say we decide the orthogonal basis purely on the $\psi(x)$ at hand. Unless you give $\psi(x)$ we will not talk about anything. We use linear orthogonal basis states, but our measurement/basis will totally be adapted to the given $\psi(x)$ and we will not talk anything unless you give the $\psi(x)$ to be analyzed. – Rajesh Dachiraju Jun 7 '13 at 14:56
I don't quite follow what you are saying. In particular, I don't see how you can deduce a basis by being given a single vector. – SMeznaric Jun 7 '13 at 18:21
The basis are adapted to the signal. I agree I can't express it mathematically right now, but I'll try to give an example with some figure.(adding to the question) in a while. – Rajesh Dachiraju Jun 7 '13 at 19:52
sorry i mistakenly added my edit to answer instead of question. Now I have removed it. Hope someone takes care of it. My bad. – Rajesh Dachiraju Jun 7 '13 at 20:17
the basis we deduce works only for that state vector. Need not work for any $\psi$. So its not actually a basis for the entire space, but its suitable to our problem and equipment we usually have in our lab. – Rajesh Dachiraju Jun 7 '13 at 20:23