Expectation value for the superposition of the two states (meaning of the imaginary part)

Question

The wave function $\Psi$ of an electron that can exist in both states $n$ and $m$ is $$ \Psi = a\Psi_n + b\Psi_m \tag{6.28} $$ where $a^*a$ is the probability that the electron is in state $n$ and $b^*b$ is the probability that it is in state $m$. Of course, it must always be true that $a^*a+b^*b=1$. Initially $a=1$ and $b=0$; when the electron is in the excited state, $a=0$ and $b=1$; and ultimately $a=1$ and $b=0$ once more. While the electron is in either state, there is no radiation, but when it is in the midst of the transition from $m$ to $n$ (that is, when both $a$ and $b$ have nonvanishing values), electromagnetic waves are produced.

The expectation value $\langle x \rangle$ that corresponds to the composite wave function of Eq. (6.28) is \begin{align} \langle x \rangle & = \int_{-\infty}^\infty x(a^* \Psi_n^*+b^*\Psi_m^*)(a\Psi_n + b\Psi_m)dx \\ & = \int_{-\infty}^\infty x(a^2 \Psi_n^*\Psi_n +b^*a \Psi_m^*\Psi_n + a^*b \Psi_n^*\Psi_m +b^2 \Psi_m^*\Psi_m)dx \tag{6.29} \end{align} Here, as before, we let $a^*a=a^2$ and $b^*b=b^2$. The first and last integrals do not vary with time, so the second and third integrals are the only ones able to contribute to a time variation in $\langle x\rangle$. With the help of Eqs. (6.26) we expand Eq. (6.29) to give \begin{align} \langle x \rangle & = a^2 \int_{-\infty}^\infty x \psi_n^*\psi_n dx +b^*a \int_{-\infty}^\infty x \psi_m^*e^{+(iE_m/\hbar)t}\psi_ne^{-(iE_n/\hbar)t} dx \\ & \qquad +a^*b \int_{-\infty}^\infty x \psi_n^*e^{+(iE_n/\hbar)t}\psi_me^{-(iE_m/\hbar)t} dx +b^2 \int_{-\infty}^\infty x \psi_m^*\psi_m dx \tag{6.30} \end{align} Because $$ e^{i\theta} = \cos\theta + i \sin \theta \qquad \text{and} \qquad e^{-i\theta} = \cos\theta - i \sin \theta $$ the two middle terms of Eq. (6.30), which are functions of time, become \begin{align} \cos\left(\frac{E_m-E_n}{\hbar}\right) t \int_{-\infty}^\infty & x [b^*a \psi_m^*\psi_n + a^*b \psi_n^*\psi_m]dx \\ & \quad + i \sin\left(\frac{E_m-E_n}{\hbar}\right) t \int_{-\infty}^\infty x [b^*a \psi_m^*\psi_n - a^*b \psi_n^*\psi_m]dx \tag{6.31} \end{align}

Beiser, A. , Concepts of Modern Physics, 6th Edition, p.219

In the last equation (6.31) there is an imaginary component of the expectation value of the superposition of the two states. Our teacher said "we" don't know what it really means. Is this really true? Does it have a physical meaning or its just a result of the mathematical calculations?

Answer 1

In the last equation (6.31) there is an imaginary component of the expectation value of the superposition of the two states. Our teacher said "we" don't know what it really means.

To be level with you, the only honest way I can find to characterize your teacher's statement is as complete rubbish. At best, that "we" includes your teacher only. We understand those terms perfectly.

Your expression,

\begin{align} \cos\left(\frac{E_m-E_n}{\hbar}\right) t \int_{-\infty}^\infty & x [b^*a \psi_m^*\psi_n + a^*b \psi_n^*\psi_m]dx \\ & \quad + i \sin\left(\frac{E_m-E_n}{\hbar}\right) t \int_{-\infty}^\infty x [b^*a \psi_m^*\psi_n - a^*b \psi_n^*\psi_m]dx \tag{6.31} \end{align}

is best understood as an expression of the form

\begin{align} \langle x \rangle = A \cos\left(\frac{E_m-E_n}{\hbar}\right) t + B\sin\left(\frac{E_m-E_n}{\hbar}\right) t, \end{align}

with the coefficients

\begin{align} A & = \int_{-\infty}^\infty x [b^*a \psi_m^*\psi_n + a^*b \psi_n^*\psi_m]dx \qquad \text{and} \\ B & = i \int_{-\infty}^\infty x [b^*a \psi_m^*\psi_n - a^*b \psi_n^*\psi_m]dx. \end{align}

The integrand in $A$ contains the sum $b^*a \psi_m^*\psi_n + a^*b \psi_n^*\psi_m$ of a complex number and its conjugate, so it is always real-valued. Similarly, the integrand in $B$ includes the difference $b^*a \psi_m^*\psi_n - a^*b \psi_n^*\psi_m$ of those same two complex-conjugate terms, which means that the integrand is always pure imaginary; once you add the global factor of $i$, you get a real-valued $B$.

As with all sums of the form $A \cos\left(\frac{E_m-E_n}{\hbar}\right) t + B\sin\left(\frac{E_m-E_n}{\hbar}\right) t$, the relative weight and sign of the two coefficients gives the phase of the oscillation. And, frankly, your book's formalism goes a good long way to make that phase extremely hard to understand. For an easier form, consider instead the following identity for the oscillating terms: \begin{align} \mathrm{osc.} & = b^*a \int_{-\infty}^\infty x \psi_m^*e^{+(iE_m/\hbar)t}\psi_ne^{-(iE_n/\hbar)t} dx +a^*b \int_{-\infty}^\infty x \psi_n^*e^{+(iE_n/\hbar)t}\psi_me^{-(iE_m/\hbar)t} dx \\ & = 2\mathrm{Re}\mathopen{}\left[ b^*a \ e^{i\frac{E_m-E_n}{\hbar}t} \int_{-\infty}^\infty x \psi_m^*\psi_n dx \right] , \end{align} which is valid since any term plus its complex conjugate gives twice the real part of both. Now, for most problems, the dipole-moment integral $d_{mn} = \int_{-\infty}^\infty x \psi_m^*\psi_n dx$ can be chosen as real-valued, so it can go straight out of the real part, and more importantly we can represent the product of the amplitudes as $$ b^*a = |b||a| e^{i\phi}, $$ where $\phi = \arg(a)-\arg(b)$ is the relative complex phase between the two amplitudes. This means, then, that the oscillating terms can be expressed as \begin{align} \mathrm{osc.} & = 2|b||a| d_{mn}\mathrm{Re}\mathopen{}\left[ e^{i\frac{E_m-E_n}{\hbar}t}e^{i\phi} \right] \\ & = 2|b||a| d_{mn}\cos\mathopen{}\left( \frac{E_m-E_n}{\hbar}t + \phi \right), \end{align} which (as opposed to the actively-obscure form in your book) makes it very clear what the relationship is between the relative phase of the coefficients and the phase of the oscillations in the dipole moment.

For more on the subject, see my answer to Is there oscillating charge in a hydrogen atom?.

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Oh, and while I'm bashing your book:

Here, as before, we let $a^*a=a^2$ and $b^*b=b^2$.

Never, ever do this. This is lazy writing and the only thing that it will achieve is to confuse your reader.

Beiser is acceptable as a stepping stone to bigger and better books (say, Cohen-Tannoudji, Sakurai, Shankar, take your pick), and as soon as you're able to digest the better ones you should move on from Beiser.