Example: Position Measurement in the Two-Level system

We have seen above that a particle in the a double well potential can be described by a vector $\vert\psi\rangle$ in the Hilbert space ${\cal H} = C^2$,

$$\vert\psi\rangle = c_0 \vert L\rangle + c_1 \vert R\rangle = c_0 ... ... 1 \end{array}\right),\quad c_0,c_1 \in C, \vert c_0\vert^2+\vert c_1\vert^2=1.$$ (197)

The last condition $\vert c_0\vert^2+\vert c_1\vert^2=1$ means that the state $\vert\psi\rangle$ is normalized. What is the meaning of the coefficients $c_0$ and $c_1$?

CASE 1: $c_0=1$, $c_1=0$: The particle is in the state $\vert\psi\rangle$= $\vert L\rangle$, which means it is in the left well.

CASE 2: $c_0=0$, $c_1=1$: The particle is in the state $\vert\psi\rangle$= $\vert R\rangle$, which means it is in the right well.

We define the operator (two-by-two matrix)

$$A:=\left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right).$$ (198)

Its eigenvectors are the two basis vector $\vert L\rangle$ and $\vert R\rangle$:

$$A\vert L\rangle \equiv \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\l... ...right)=1\cdot \left(\begin{array}{c} 1 \\ 0 \end{array}\right)$$

$$A\vert R\rangle \equiv \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\l... ...ght)=-1\cdot \left(\begin{array}{c} 0 \\ 1 \end{array}\right).$$ (199)

The operator $A$ thus corresponds to the measurement `where is the particle, in the left or in the right well'. This is the simplest version of a position measurement, where we are not interested in the precise position, but only measure in which well the particle is. For example, if the particle is charged with the elementary charge $-e$, we could measure the charge in the left well. If the particle is in there, we find $-e$, if it is in the right well, we find 0.

We now apply our axiom 2d to this `position measurement': we cite it here in the form that exactly matches our two-level system:

Axiom 2d: Let $A$ have a complete system of eigenvectors $\{\vert\phi_0\rangle= \vert L\rangle, \vert\phi_1\rangle =\vert R\rangle \}$ with eigenvalues $a_0=1,a_1=-1$. The normalized state $\vert\psi\rangle$ before the measurement of $A$ can be expanded into

$$\vert\psi\rangle= \sum_{n=0}^{1}c_n\vert\phi_n\rangle = c_0 \vert L \rangle + c_1 \vert R \rangle.$$ (200)

Then, the expectation value of $A$ in $\vert\psi\rangle$ is

$$\langle \psi \vert A\vert\psi\rangle =\sum_{n=0}^{1}a_n \vert c_n\vert^2=\sum_{n=0}^{1}a_n p_n,$$ (201)

and the probability $p_n$ to find the system in the eigenstate $\vert\phi_n\rangle$ after the measurement is given by the amplitude square $p_n = \vert c_n\vert^2$.

We therefore recognize:

$$\vert c_0\vert^2=p_0 =$$ the probability to find the particle in the left well

$$\vert c_1\vert^2=p_1 = .$$ (202)

In particular, the probability to find the particle in either the left or the right well is $p_0+p_1=\vert c_0\vert^2+\vert c_1\vert^2=1$ as it must be. For example, if we perform the measurement by measuring an extra charge in the left well, we find $-e$ with probability $\vert c_0\vert^2$, and after the measurement the particle is in the left well, that is in the state $\vert L\rangle$. If, on the other hand, we find an extra charge 0 in the left well, the particle must be in the right well, that is in the state $\vert R\rangle$. The probability for this is $1-\vert c_0\vert^2=\vert c_1\vert^2$.