P1 Quantum Mechanics
These problems reinforce the fundamental concepts of quantum mechanics used throughout atomic, molecular, and optical physics.
Work through the problems independently before consulting your notes. Particular attention should be paid to the physical interpretation of the mathematical results.
Quantum States and Time Evolution
Problem P1.1: The Time-Dependent Schrödinger Equation
The time-dependent Schrödinger equation for an electron, in atomic units, is
\[ i\frac{\partial}{\partial t}\psi(\vec r,t) = -\frac{\nabla^2}{2}\psi(\vec r,t) +V(\vec r,t)\psi(\vec r,t). \]
Tasks
For a time-independent potential \(V(\vec r)\), the solution can be written as a product
\[ \psi(\vec r,t)=f(\vec r)g(t). \]
Determine the form of \(g(t)\).
What is the time dependence of the probability density
\[ |\psi(\vec r,t)|^2 = \psi^*(\vec r,t)\psi(\vec r,t) \]
for such a stationary state?
Consider the superposition
\[ \psi(\vec r,t) = \frac{1}{\sqrt{2}} \left[ \psi_1(\vec r,t)-\psi_2(\vec r,t) \right], \]
where \(\psi_1\) and \(\psi_2\) are normalized energy eigenstates with distinct eigenfrequencies \(\omega_1\) and \(\omega_2\).
Derive an expression for the time-dependent probability density \(|\psi(\vec r,t)|^2\).
Explain why the probability density can now depend on time.
Write each energy eigenstate as
\[ \psi_j(\vec r,t)=\phi_j(\vec r)e^{-i\omega_j t}. \]
When calculating the probability density, pay particular attention to the interference terms.
1. Time dependence of a stationary state
For a time-independent potential, we use the separation ansatz
\[ \psi(\vec r,t)=f(\vec r)g(t). \]
Substituting into the time-dependent Schrödinger equation (in atomic units),
\[ i f(\vec r)\frac{dg}{dt} = -\frac{g(t)}{2}\nabla^2 f(\vec r) +V(\vec r)f(\vec r)g(t), \]
and dividing by \(f(\vec r)g(t)\) gives
\[ i\frac{1}{g}\frac{dg}{dt} = -\frac{1}{2f}\nabla^2 f+V(\vec r). \]
The left side depends only on time, while the right side depends only on position. Both must therefore equal a constant \(E\), the energy eigenvalue.
Thus,
\[ i\frac{dg}{dt}=Eg, \]
with solution
\[ \boxed{g(t)=e^{-iEt}} \]
up to a constant factor.
2. Probability density of a stationary state
For a stationary state,
\[ \psi(\vec r,t)=f(\vec r)e^{-iEt}. \]
The probability density is
\[ \begin{aligned} |\psi(\vec r,t)|^2 &=f^*(\vec r)e^{iEt}f(\vec r)e^{-iEt}\\ &=|f(\vec r)|^2. \end{aligned} \]
Therefore,
\[ \boxed{|\psi(\vec r,t)|^2=|f(\vec r)|^2} \]
is independent of time.
Although the wavefunction acquires a time-dependent phase, this global phase has no effect on the probability density.
3. Superposition of two energy eigenstates
Write the two energy eigenstates as
\[ \psi_j(\vec r,t)=\phi_j(\vec r)e^{-i\omega_j t}, \]
where \(\omega_j=E_j\) in atomic units.
The superposition is
\[ \psi(\vec r,t) = \frac{1}{\sqrt{2}} \left[ \phi_1(\vec r)e^{-i\omega_1 t} - \phi_2(\vec r)e^{-i\omega_2 t} \right]. \]
Taking the absolute square gives
\[ \begin{aligned} |\psi(\vec r,t)|^2 =\frac12\Big[ &|\phi_1(\vec r)|^2+|\phi_2(\vec r)|^2\\ &-\phi_1^*(\vec r)\phi_2(\vec r) e^{i(\omega_1-\omega_2)t}\\ &-\phi_2^*(\vec r)\phi_1(\vec r) e^{-i(\omega_1-\omega_2)t} \Big]. \end{aligned} \]
Combining the two interference terms,
\[ \boxed{ |\psi(\vec r,t)|^2 = \frac12\left[ |\phi_1|^2+|\phi_2|^2 -2\operatorname{Re}\left( \phi_1^*\phi_2 e^{i(\omega_1-\omega_2)t} \right) \right]. } \]
The probability density can oscillate at the beat frequency
\[ \boxed{\omega_{\mathrm{beat}}=|\omega_1-\omega_2|.} \]
Unlike a single energy eigenstate, a superposition of states with different energies has a time-dependent relative phase. This can produce observable interference effects.
The time dependence of the local density requires spatial overlap between the two eigenfunctions.
Orbital Angular Momentum
Problem P1.2: Orbital Angular Momentum and Compatible Observables
In coordinate space, the orbital angular momentum operator is
\[ \hat{\vec L} = \hat{\vec r}\times\hat{\vec p} = -i\hbar\,\vec r\times\vec\nabla, \]
with Cartesian components
\[ \hat L_x = -i\hbar\left( y\frac{\partial}{\partial z} -z\frac{\partial}{\partial y} \right), \]
\[ \hat L_y = -i\hbar\left( z\frac{\partial}{\partial x} -x\frac{\partial}{\partial z} \right), \]
\[ \hat L_z = -i\hbar\left( x\frac{\partial}{\partial y} -y\frac{\partial}{\partial x} \right). \]
The squared angular momentum operator is
\[ \hat L^2=\hat L_x^2+\hat L_y^2+\hat L_z^2. \]
Tasks
Explicitly evaluate the commutator
\[ [\hat L_x,\hat L_y] \]
and demonstrate that
\[ [\hat L_x,\hat L_y]=i\hbar\hat L_z. \]
State the corresponding commutation relations for the other two pairs of Cartesian components.
Show that
\[ [\hat L^2,\hat L_z]=0. \]
Discuss the implications of these commutation relations for simultaneous eigenstates and simultaneous measurements of
\[ (\hat L^2,\hat L_z) \qquad\text{and}\qquad (\hat L_x,\hat L_y). \]
Suppose a quantum state is an eigenstate of both \(\hat L^2\) and \(\hat L_z\).
Does this imply that it is also an eigenstate of \(\hat L_x\)? Explain briefly.
For the second task, use the operator identity
\[ [A^2,B]=A[A,B]+[A,B]A. \]
You can then apply the angular momentum commutation relations obtained in the first task.
1. Commutator \([\hat L_x,\hat L_y]\)
The angular momentum operators are
\[ \hat L_x=-i\hbar(y\partial_z-z\partial_y), \qquad \hat L_y=-i\hbar(z\partial_x-x\partial_z). \]
To evaluate their commutator, we apply it to an arbitrary sufficiently smooth function \(f(x,y,z)\):
\[ [\hat L_x,\hat L_y]f = (\hat L_x\hat L_y-\hat L_y\hat L_x)f. \]
First,
\[ \begin{aligned} \hat L_x\hat L_y f =-\hbar^2\big[ &y\partial_x f +yz\partial_z\partial_x f -xy\partial_z^2 f\\ &-z^2\partial_y\partial_x f +xz\partial_y\partial_z f \big]. \end{aligned} \]
Similarly,
\[ \begin{aligned} \hat L_y\hat L_x f =-\hbar^2\big[ &z^2\partial_x\partial_y f -yz\partial_x\partial_z f\\ &-x\partial_y f -xy\partial_z^2 f +xz\partial_z\partial_y f \big]. \end{aligned} \]
Because mixed partial derivatives commute, all second-derivative terms cancel when subtracting.
We obtain
\[ [\hat L_x,\hat L_y]f = -\hbar^2(y\partial_x-x\partial_y)f. \]
Using
\[ \hat L_z=-i\hbar(x\partial_y-y\partial_x), \]
we recognize
\[ \boxed{[\hat L_x,\hat L_y]=i\hbar\hat L_z.} \]
By cyclic permutation,
\[ \boxed{ [\hat L_y,\hat L_z]=i\hbar\hat L_x, \qquad [\hat L_z,\hat L_x]=i\hbar\hat L_y. } \]
2. Commutator \([\hat L^2,\hat L_z]\)
Using
\[ \hat L^2=\hat L_x^2+\hat L_y^2+\hat L_z^2 \]
and the identity
\[ [A^2,B]=A[A,B]+[A,B]A, \]
we find
\[ \begin{aligned} [\hat L^2,\hat L_z] ={}&[\hat L_x^2,\hat L_z] +[\hat L_y^2,\hat L_z] +[\hat L_z^2,\hat L_z]. \end{aligned} \]
The last term vanishes.
From the angular momentum commutation relations,
\[ [\hat L_x,\hat L_z]=-i\hbar\hat L_y, \qquad [\hat L_y,\hat L_z]=i\hbar\hat L_x. \]
Therefore,
\[ [\hat L_x^2,\hat L_z] = -i\hbar(\hat L_x\hat L_y+\hat L_y\hat L_x) \]
and
\[ [\hat L_y^2,\hat L_z] = i\hbar(\hat L_y\hat L_x+\hat L_x\hat L_y). \]
The two terms cancel:
\[ \boxed{[\hat L^2,\hat L_z]=0.} \]
3. Simultaneous eigenstates and measurements
Since
\[ [\hat L^2,\hat L_z]=0, \]
the two operators possess a common eigenbasis. Their eigenstates can be labeled by the quantum numbers \(\ell\) and \(m\):
\[ \hat L^2|\ell,m\rangle = \hbar^2\ell(\ell+1)|\ell,m\rangle, \]
\[ \hat L_z|\ell,m\rangle = \hbar m|\ell,m\rangle. \]
Thus, the magnitude of the angular momentum and its \(z\)-component can have simultaneously well-defined values.
In contrast,
\[ [\hat L_x,\hat L_y]=i\hbar\hat L_z\neq 0 \]
as an operator identity.
Consequently, \(\hat L_x\) and \(\hat L_y\) cannot be simultaneously diagonalized in a complete common eigenbasis.
Their uncertainties satisfy
\[ \boxed{ \Delta L_x\,\Delta L_y \geq \frac{\hbar}{2}|\langle\hat L_z\rangle|. } \]
4. Does an eigenstate of \(\hat L^2\) and \(\hat L_z\) also have a definite \(L_x\)?
In general, no.
The state \(|\ell,m\rangle\) has definite values of \(L^2\) and \(L_z\), but is generally a superposition of different \(L_x\) eigenstates.
For example, if \(m\neq 0\), the uncertainty relation implies that \(L_x\) and \(L_y\) cannot both be sharply defined.
An important exception is \(\ell=0\), where
\[ \hat L_x|0,0\rangle = \hat L_y|0,0\rangle = \hat L_z|0,0\rangle=0. \]
The \(\ell=0\) state is therefore a simultaneous eigenstate of all three components.
The fact that two operators do not commute does not exclude the existence of special common eigenstates.
Problem P1.3: Angular Momentum in Coordinate Space
In spherical coordinates, the \(z\)-component of the orbital angular momentum operator is
\[ \hat L_z=-i\hbar\frac{\partial}{\partial\phi}. \]
Tasks
Show that
\[ \Phi_m(\phi)=Ae^{im\phi} \]
is an eigenfunction of \(\hat L_z\), and determine its eigenvalue.
A physical orbital wavefunction must be single-valued. Apply the condition
\[ \Phi_m(\phi+2\pi)=\Phi_m(\phi) \]
and determine the allowed values of \(m\).
Consider the state
\[ \psi(\phi) = \frac{1}{\sqrt{2}} \left(e^{i\phi}+e^{-i\phi}\right). \]
Here the exponentials represent normalized angular-momentum basis states, with their common normalization factor suppressed.
Is this state an eigenstate of \(\hat L_z\)?
If \(\hat L_z\) is measured, what values can be obtained, and with what probabilities?
Consider instead
\[ \chi(\phi)=Ae^{i(3/2)\phi}. \]
Show that \(\chi(\phi)\) is formally an eigenfunction of the differential operator \(\hat L_z\).
Why is it nevertheless not an allowed orbital-angular-momentum eigenfunction?
For the last task, evaluate
\[ \frac{\chi(\phi+2\pi)}{\chi(\phi)}. \]
Remember that \(\phi\) and \(\phi+2\pi\) describe the same spatial direction.
1. Eigenfunctions of \(\hat L_z\)
The operator is
\[ \hat L_z=-i\hbar\frac{\partial}{\partial\phi}. \]
Applying it to
\[ \Phi_m(\phi)=Ae^{im\phi} \]
gives
\[ \begin{aligned} \hat L_z\Phi_m &=-i\hbar\frac{\partial}{\partial\phi} \left(Ae^{im\phi}\right)\\ &=-i\hbar(im)Ae^{im\phi}\\ &=\hbar m\Phi_m. \end{aligned} \]
Therefore,
\[ \boxed{\hat L_z\Phi_m=\hbar m\Phi_m,} \]
with eigenvalue
\[ \boxed{L_z=\hbar m.} \]
2. Allowed values of \(m\)
A physical orbital wavefunction must be single-valued:
\[ \Phi_m(\phi+2\pi)=\Phi_m(\phi). \]
Substituting the exponential form,
\[ Ae^{im(\phi+2\pi)}=Ae^{im\phi}. \]
For a nonzero wavefunction, this requires
\[ e^{i2\pi m}=1. \]
Hence,
\[ \boxed{m=0,\pm1,\pm2,\ldots} \]
The orbital magnetic quantum number must be an integer.
3. Superposition of \(m=+1\) and \(m=-1\)
The state is
\[ \psi(\phi) = \frac{1}{\sqrt2} \left(e^{i\phi}+e^{-i\phi}\right), \]
where the normalized angular basis functions are understood.
Applying \(\hat L_z\),
\[ \hat L_z\psi = \frac{\hbar}{\sqrt2} \left(e^{i\phi}-e^{-i\phi}\right). \]
This is not proportional to the original state, so
\[ \boxed{\psi\text{ is not an eigenstate of }\hat L_z.} \]
The state is an equal-weight superposition of two angular momentum eigenstates:
\[ \psi=\frac{1}{\sqrt2} \left(|m=1\rangle+|m=-1\rangle\right). \]
The possible measurement outcomes are
\[ \boxed{ L_z=+\hbar \quad (P=1/2), \qquad L_z=-\hbar \quad (P=1/2). } \]
The expectation value is
\[ \langle L_z\rangle = \frac12(+\hbar)+\frac12(-\hbar)=0. \]
A vanishing expectation value does not mean that every measurement produces zero.
4. The half-integer orbital state
Consider
\[ \chi(\phi)=Ae^{i(3/2)\phi}. \]
Applying \(\hat L_z\) gives
\[ \hat L_z\chi = -i\hbar\frac{\partial}{\partial\phi} \left(Ae^{i(3/2)\phi}\right) = \frac{3\hbar}{2}\chi. \]
Thus, formally,
\[ \boxed{\hat L_z\chi=\frac{3\hbar}{2}\chi.} \]
However,
\[ \begin{aligned} \chi(\phi+2\pi) &=Ae^{i(3/2)(\phi+2\pi)}\\ &=Ae^{i(3/2)\phi}e^{i3\pi}\\ &=-\chi(\phi). \end{aligned} \]
The wavefunction changes sign after a full rotation in the azimuthal coordinate.
Since \(\phi\) and \(\phi+2\pi\) represent the same spatial point, this violates the required single-valuedness of a scalar orbital wavefunction.
Therefore,
\[ \boxed{ m=\frac32\text{ is not allowed for orbital angular momentum.} } \]
Half-integer angular momentum does occur for spin, but spin states are not described by scalar orbital wavefunctions of this form.
Quantum-to-Classical Correspondence
Problem P1.4: From Quantum to Classical Behavior
According to Schrödinger’s wave mechanics, a massive particle passing through a diffracting object, such as a grating or a double slit, can exhibit constructive and destructive interference.
The correspondence principle states that quantum theory reproduces classical physics in appropriate limits, for example for large quantum numbers, large energies, or large masses.
Tasks
Give two distinct reasons why interference becomes increasingly difficult to observe for very massive or macroscopic objects.
Quantum mechanics allows systems to exist in superposition states.
Does quantum mechanics itself forbid a macroscopic object from being in a superposition?
Discuss why coherent superpositions of macroscopically distinguishable states are nevertheless extremely difficult to prepare and observe.
You may use Schrödinger’s cat as a conceptual example.
Consider both the de Broglie wavelength and the interaction of a quantum system with its environment.
Distinguish between what quantum mechanics permits in principle and what is experimentally observable.
1. Why is interference difficult to observe for macroscopic objects?
Two important reasons are:
(a) Extremely small de Broglie wavelength
The de Broglie wavelength is
\[ \lambda=\frac{h}{p}=\frac{h}{mv}. \]
At a given velocity, increasing the mass decreases the wavelength.
For a macroscopic object, the de Broglie wavelength becomes extraordinarily small. Consequently, the characteristic interference angles and fringe spacings become too small to resolve with ordinary experimental apparatus.
For example, a particle passing through a diffraction grating with spacing \(d\) has a characteristic first-order diffraction angle of approximately
\[ \theta\approx\frac{\lambda}{d}, \]
when \(\lambda\ll d\).
As \(\lambda\) decreases, the diffraction angle becomes correspondingly smaller.
(b) Decoherence through environmental interactions
Macroscopic objects interact strongly with their environment through processes such as
- scattering of air molecules,
- emission and absorption of thermal radiation,
- interactions with surrounding electromagnetic fields.
These interactions can entangle the object’s quantum state with environmental degrees of freedom.
Information about the different possible paths becomes encoded in the environment, reducing the coherence necessary for observable interference.
The interference visibility can therefore decrease extremely rapidly.
2. Are macroscopic superpositions forbidden?
No. Quantum mechanics does not impose a fundamental mass or size limit beyond which superpositions become impossible.
The superposition principle applies to macroscopic systems as well as microscopic ones.
For example, one may formally construct a state
\[ |\Psi\rangle = \frac{1}{\sqrt2} \left( |\text{macroscopic state A}\rangle + |\text{macroscopic state B}\rangle \right). \]
The main difficulty is preserving the relative phase coherence between the two components.
When the system interacts with its environment, the combined state may evolve into
\[ |\Psi_{\mathrm{total}}\rangle = \frac{1}{\sqrt2} \left( |A\rangle|E_A\rangle + |B\rangle|E_B\rangle \right), \]
where \(|E_A\rangle\) and \(|E_B\rangle\) describe environmental states correlated with the two alternatives.
If the environmental states become nearly orthogonal,
\[ \langle E_A|E_B\rangle\approx0, \]
the interference terms in measurements of the macroscopic system alone become strongly suppressed.
This is the process of decoherence.
The resulting system behaves approximately like a classical statistical mixture, even though the combined system and environment can remain in a pure quantum state.
Physical interpretation
The classical behavior of macroscopic objects is not simply a consequence of quantum mechanics ceasing to apply.
Rather, classical behavior emerges because quantum interference is often extremely difficult to resolve and because coherence between macroscopically distinct alternatives is rapidly lost to the environment.
Schrödinger’s cat illustrates precisely this tension between the linear superposition principle and the classical appearance of macroscopic objects.