Quantum Physics I Practice

The probability is the integral of $|\psi(x)|^2$ over the interval:

So the probability is $\frac{1}{2}$.

Use the normalization condition:

Here that becomes

So a normalized choice is

up to an overall phase.

The note gives

Substitute $n=2$:

So

The momentum operator in one dimension is

This is the operator used for momentum measurements in the position representation.

Use the de Broglie relation

Solve for $p$:

Use the photoelectric equation:

Substitute the values:

So the maximum kinetic energy is $2\ \text{eV}$.

An electron has spin $1/2$, so the only allowed measurement outcomes are

along the chosen axis.

The Heisenberg uncertainty principle says

Solve for $\Delta p$:

The Hamiltonian is

This is the energy operator used in the one-dimensional Schrodinger equation.

Check the sum of the squared magnitudes of the coefficients:

So the state is normalized.

First normalize:

so

Now compute the probability on $[0,\frac{1}{2}]$:

For a uniform distribution on $[0,L]$, the probability density is

So

Since $E_n \propto \frac{1}{L^2}$, doubling $L$ multiplies the energy by

So each energy level becomes one-fourth as large.

No. The note gives the canonical commutation relation

Because the operators do not commute, the observables cannot both be sharply defined at the same time. This is summarized by

Let $T_a$ be the transmission for width $a$ and $T_{3a}$ for width $3a$.

Then

and

So the ratio is

For a time-independent potential, separation of variables leads to the time-independent Schrodinger equation:

So the spatial part is an eigenfunction of the Hamiltonian, and $E$ is the corresponding allowed energy.

Using the inverse-square scaling,

Because the hydrogen energies are negative, $E_4$ is still negative but closer to zero than $E_1$.

Substitute $n=0$:

Substitute $n=1$:

The spacing is

So the ground state has nonzero energy, and the levels are evenly spaced.

Since

electrons are emitted.

Use

So

The pattern shows that electrons have wave behavior, captured by the de Broglie relation

The effect becomes noticeable when the wavelength is comparable to the size of the slit spacing, crystal spacing, or other relevant length scale. That is why pure classical trajectories are not enough.

The appropriate idea is the WKB approximation.

Its intuition is semiclassical: connect classical motion with quantum phase accumulation when the wavelength changes slowly, rather than solving the full problem exactly.

The result along $x$ is not predetermined by the first measurement along $z$.

Measurements along different axes generally do not commute, so the first measurement does not fix the second one with certainty. The second outcome is probabilistic.

Use perturbation theory.

The idea is to start from the known eigenstates and energies of $\hat{H}_0$ and compute small corrections from the additional term $\lambda \hat{V}$.

The probabilities of measuring $|0\rangle$ or $|1\rangle$ are the same because the coefficient magnitudes are the same.

However, the relative phase is different. That phase matters when states interfere, so the two states can behave differently under other measurements or later evolution even though the basis probabilities match.

Normalize the wave function:

So

Use symmetry:

Thus

and we may take

Now compute the expectation value:

Since $|\psi(x)|^2$ is even, the integrand $x|\psi(x)|^2$ is odd, so the integral is zero:

You vary $\alpha$ to minimize the expected energy.

That is the variational method. The minimized value is an estimate of the ground-state energy, and the note says it gives an upper bound to that energy.

The uncertainty principle prevents both position and momentum from being exactly known at the same time:

So the ground state cannot have both $x=0$ and $p=0$ with zero spread.

The oscillator energies are

so the ground-state energy is

which is nonzero.