If we are given a wave function written in terms of harmonic oscillator energy eigenfunctions how can we determine the maximum possible momentum expectation value? It's a combination of the first two energy states, weighted with equal probability – why is there even a range of possible momentum expectation values?
[Physics] Harmonic Oscillator Energy to Momentum Expectation Value
eigenvaluehomework-and-exercisesmomentumoperatorsquantum mechanics
Related Solutions
Let's write the eigenstates for the infinite well as $$ \psi_n(x,t) = \phi_n(x)e^{-i\omega_n t} = |n\rangle e^{-i\omega_n t} , $$
where $\phi_n(x) = A \sin\left( n \pi x /L \right)$.
If your linear combination is $$ |\psi_0\rangle = \alpha |1\rangle e^{-i\omega_1 t}+ \beta |2\rangle e^{-i\omega_2 t}. $$ Then $$ \begin{eqnarray} \langle \psi_0 | \hat{p} | \psi_0 \rangle &=& \left( \alpha^* \langle 1| e^{i\omega_1 t}+ \beta^* \langle 2| e^{i\omega_2 t} \right)\hat{p} \left( \alpha |1\rangle e^{-i\omega_1 t}+ \beta |2\rangle e^{-i\omega_2 t} \right) \\ &=& |\alpha|^2 \langle 1 | \hat{p} | 1 \rangle + |\beta|^2 \langle 2 | \hat{p} | 2 \rangle + \alpha^*\beta \langle 1 | \hat{p} | 2 \rangle e^{-i\Delta\omega t} + \alpha\beta^* \langle 2 | \hat{p} | 1 \rangle e^{i\Delta\omega t} \end{eqnarray} $$ The $\hat{p}$ operator does not affect the time-depentent part. However, it will convert $\sin$ to $\cos$ making $\langle 1 | \hat{p} | 1 \rangle = \langle 2 | \hat{p} | 2 \rangle = 0$.
Now, the terms that remain are not zero and are complex conjugates of each other so, as with the expectation value of the position, it is proportional to $\cos\Delta\omega t$.
This is a fantastic question! Let's get started.
I will assume that we already have defined the ladder operators $a$ and $a^{\dagger}$ and have defined a "ground state" $|0\rangle$ (we still have not proved it is the ground state) such that $a|0\rangle=0$. We will also assume that we already know that the Hamiltonian of the Harmonic oscillator can be written in the form
$$H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right).$$
(Note that the ground state $|0\rangle$ is trivially an eigenstate with $E_0=\hbar\omega/2$.) Finally, I will assume that we have already shown the commutation relations of the ladder operators. Namely,
$$[a,a^{\dagger}]=1.$$
With this, we have enough for a proof.
We can define a state $|n\rangle$ (let's forget about normalization for now) as
$$|n\rangle=(a^{\dagger})^n|0\rangle,$$
where $n$ is a nonnegative integer. The state $|n\rangle$ is an eigenstate of the Hamiltonian with energy $E_n=\hbar\omega(n+1/2)$. We wish to show that the set $\{|n\rangle\}_{n\in\mathbb{Z}^+}$ are all of the possible normalizable eigenstates of the Hamiltonian.
Recall that in the position representation, if we have a potential $V(x)$, then we cannot have a normalizable eigenstate $|\psi\rangle$ whose energy satisfies $E_{\psi}\leq\min V(x)$. That is, we can not have an energy less than the minimum potential energy of the system (ie the kinetic energy must be positive).
Now, we finish off with a proof by contradiction. Consider an eigenstate $|\psi\rangle$ whose energy is given by $E_{\psi}=\hbar\omega(n+1/2+\epsilon)$, with $\epsilon\in(0,1)$. Such a state would essentially describe any of the "other" states that $H$ could permit. Now, consider the state $|\psi^{(1)}\rangle=a|\psi\rangle$. By the commutator algebra, it is not hard to show that $|\psi^{(1)}\rangle$ has energy
$$E_{\psi^{(1)}}=\hbar\omega\left((n-1)+\frac{1}{2}+\epsilon\right).$$
Now, we can induct and define a state $|\psi^{(m)}\rangle\equiv(a^m)|\psi\rangle$. Clearly, its energy is given by
$$E_{\psi^{(m)}}=\hbar\omega\left((n-m)+\frac{1}{2}+\epsilon\right).$$
Thus, unless this process terminates at some point (that is, $a|\psi^{(m)}\rangle=0$ for some $m$), we can achieve an arbitrarily low energy. However, this process could never terminate, since the ground state $|0\rangle$ is unique (it's defined in terms of a position operator and a single derivative operator, so $a|0\rangle=0$ simply defines a first order differential equation in position space) and has energy $\hbar\omega/2$, this cannot be achieved for any $\epsilon$ in the given range. Thus, no such state $|\psi\rangle$ can occur. Similarly, we cannot have a state with energy $E_{\psi}\in(0,\hbar\omega/2)$ by the same logic.
Thus, we have (very rigorously) shown that the only normalizable of $H$ are those with energy $\hbar\omega(n+1/2)$, which are uniquely made from the action of ladder operators on the ground state.
I hope this helped!
(TL;DR -- If another state did exist, it would have an energy not of the form of those given by ladder operators. However, acting on this state many times with $a$ would produce an arbitrarily low energy, and is thus such a state could not exist.)
Best Answer
Writing $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, where $|0\rangle$a and $|1\rangle$ are the first energy eigenvectors, and $\alpha$ and $\beta$ are complex, and with $\hat P = i \sqrt{\frac{m\omega\hbar}{2}} (a^\dagger-a)$, we one gets :
$\langle \psi|\hat P|\psi\rangle = i \sqrt{\frac{m\omega\hbar}{2}}(\alpha^*\langle 0| + \beta^*\langle 1|)(\alpha|1\rangle + \beta(\sqrt{2}|2\rangle -|0\rangle) \\=i \sqrt{\frac{m\omega\hbar}{2}}(\beta^*\alpha -\alpha^*\beta)$.
So, finally, $\langle\hat P\rangle = \frac{\langle \psi|\hat P|\psi\rangle}{\langle \psi|\psi\rangle} = i \sqrt{\frac{m\omega\hbar}{2}} \frac{\beta^*\alpha -\alpha^*\beta}{|\alpha|^2 + |\beta|^2}$
The maximum of $|\frac{\beta^*\alpha -\alpha^*\beta}{|\alpha|^2+ |\beta|^2}|$ is $1$, and is reached, for instance for $\alpha$ real and $\beta = \pm i\alpha$ pure imaginary (or vice-versa). For these values, we have $\langle\hat P\rangle = \pm \sqrt{\frac{m\omega\hbar}{2}}$
The minimum is reached for $\alpha$ and $\beta$ both real, or both pure imaginary, in these cases, $\langle\hat P\rangle = 0$