Energy Measurement in the Two-Level System

The measurement of the total energy is a bit more complicated to do in practice than the position measurement. For example, one can irradiate the system with light and measure an absorption peak when the photon energy $h\nu$ matches the energy difference $\Delta$ of the two levels. After the absorption, the particle is in the excited state $\vert 2\rangle$ with energy $\varepsilon_2$, before it was in its ground state $\vert 1\rangle$ with energy $\varepsilon_1$. The particle can go back into its ground state by, for example, spontaneous emission of another photon of energy $\Delta$.

What we would like to find out now is the following: where is the particle, if it is in its ground state $\vert 1\rangle$ or its excited state $\vert 2\rangle$? For example, after we perform the absorption experiment and know that the particle is in its excited state, we would like to find out if it is in the left or in the right well. This means, we again perform a `position' measurement, and we can apply exactly the same argument as before. The state before the position measurement is now the excited state

$$\vert 2\rangle = \frac{1}{N_2}\left[\phantom{-}2T \vert L\rangle + (\Delta -\varep... ...varepsilon_L-\varepsilon_{R},\quad \Delta:=\sqrt{\varepsilon^2+4\vert T\vert^2}$$

$$N_{2} := \sqrt{4\vert T\vert^2+(\Delta - \varepsilon)^2}.$$ (208)

Comparing the coefficients in front of the basis vectors $\vert L\rangle$ and $\vert R\rangle$, we recognize:

$$\vert c_0\vert^2=p_0= \frac{4\vert T\vert^2}{4\vert T\vert^2+(\Delta - \varepsilon)^2} = {\begin{minipage}{7cm} \rm\small the probability to find the part... ...ft well after it was in the energy eigenstate $\vert 2\rangle$. \end{minipage}}$$

$$\vert c_1\vert^2=p_1= \frac{(\Delta -\varepsilon)^2}{4\vert T\vert^2+(\Delta - \varepsilon)^2} = {\begin{minipage}{7cm} \rm\small the probability to find the part... ...ht well after it was in the energy eigenstate $\vert 2\rangle$. \end{minipage}}$$

Again, $p_0+p_1=\vert c_0\vert^2+\vert c_1\vert^2=1$ as it must be. This means: we can calculate in advance the probability to find the particle in the left well after the absorption measurement, but we can't tell where it is. If we know that the particle is in its excited state $\vert 2\rangle$, but we haven't done the position measurement yet, we still don't know if the particle is in the right or in the left well. Its state $\vert 2\rangle$ has components both in the left and the right well. For example, if $\varepsilon=0$, we find

$$\varepsilon=0\leadsto p_0=p_1= \frac{1}{2}.$$ (209)

In this case `it is completely unclear' where the particle is. Only the subsequent position measurement can give us the answer: right with probability $p_1=1/2$, and left with probability $p_0=1/2$. On the other hand, if the coupling between the two wells is zero (in the case of an infinitely high barrier), we have $T=0$. Then,

$$T=0\leadsto p_0=0,\quad p_1= 1,$$ (210)

which means that the particle is with certainty in the right well. This is obvious because for $T=0$, the energy eigenstate $\vert 2\rangle = \vert R\rangle$ is simply the basis vector $\vert R\rangle$ corresponding to the particle in the right well with probability one.