Why can't the ground state of a quantum harmonic oscillator have zero total energy?
AThe potential energy at the equilibrium position is nonzero, so total energy cannot be zero
BZero energy would require simultaneously zero position uncertainty and zero momentum uncertainty, violating the Heisenberg uncertainty principle
CQuantum systems always have more energy than their classical counterparts due to quantization
DThe energy eigenvalues of the Hamiltonian are all positive by construction
Zero total energy would require zero kinetic energy (so zero momentum) and zero potential energy (so exact location at x = 0). But Δx = 0 and Δp = 0 simultaneously violates ΔxΔp ≥ ℏ/2. The uncertainty principle forces a trade-off: confining the particle near equilibrium requires nonzero Δp, which means nonzero kinetic energy. The zero-point energy ½ℏω is precisely the minimum energy consistent with the uncertainty constraint. Option A is wrong — V = 0 at the equilibrium position x = 0. Option D restates the result algebraically without explaining the physical reason.
Question 2 Multiple Choice
Liquid helium remains liquid under atmospheric pressure all the way to absolute zero (unlike every other element). The primary physical reason is:
AHelium atoms repel each other too strongly at short range to form a crystal lattice
BHelium's large zero-point kinetic energy keeps atoms in constant motion, preventing them from localizing into a fixed lattice
CHelium has the lowest boiling point of any element, placing it in a special quantum liquid regime
DQuantum mechanics prohibits noble gases from forming the covalent bonds needed for solidification
Helium atoms are very light (mass number 4), which means high ω = √(k/m) and thus large zero-point energy ½ℏω. This zero-point kinetic energy is large enough that atoms cannot localize into a fixed crystal lattice — the quantum fluctuations overwhelm the weak van der Waals attraction between helium atoms. You must apply ~25 atm of pressure to force it to solidify. Option A (repulsion) is incorrect — the issue is the kinetic energy of localization, not interatomic repulsion. Option C describes a consequence, not the cause.
Question 3 True / False
In the ground state of the quantum harmonic oscillator, the average kinetic energy and average potential energy are equal, each contributing ℏω/4 to the total zero-point energy of ½ℏω.
TTrue
FFalse
Answer: True
This follows from the virial theorem for the harmonic oscillator: ⟨T⟩ = ⟨V⟩ for any energy eigenstate. In the ground state, ⟨T⟩ = ⟨p²⟩/2m = ℏω/4 and ⟨V⟩ = ½mω²⟨x²⟩ = ℏω/4, summing to ½ℏω. The equal partition mirrors classical equipartition — but here the contribution is quantum mechanical, present even at T = 0 when all thermal energy has been removed. Neither the kinetic nor potential contribution can be zero.
Question 4 True / False
Zero-point energy is a theoretical prediction with no directly observable physical consequences.
TTrue
FFalse
Answer: False
Zero-point energy has multiple measurable consequences. The Casimir effect — an attractive force between two uncharged parallel conducting plates in vacuum — arises because the plates restrict which vacuum modes can exist between them, reducing the zero-point energy density relative to outside and creating a net inward force. The kinetic isotope effect (deuterium substitution slows chemical reactions) occurs because heavier deuterium has lower zero-point energy, raising the effective activation barrier. Liquid helium's refusal to solidify under atmospheric pressure is a macroscopic consequence of zero-point motion. These are experimentally verified phenomena.
Question 5 Short Answer
Explain why the uncertainty principle makes zero-point energy unavoidable for a particle confined in a potential well.
Think about your answer, then reveal below.
Model answer: A particle in a potential well is spatially confined, so its position uncertainty Δx is finite. The uncertainty principle requires ΔxΔp ≥ ℏ/2, so a finite Δx forces a nonzero Δp — the particle cannot have precisely zero momentum. Nonzero Δp means ⟨p²⟩ > 0, which means nonzero kinetic energy ⟨T⟩ = ⟨p²⟩/2m > 0. The minimum total energy compatible with the uncertainty constraint is the zero-point energy. A classical particle can sit still at the bottom of a well with zero kinetic and zero potential energy; a quantum particle cannot be simultaneously localized and at rest.
The Gaussian ground-state wavefunction is not arbitrary — it is the unique shape that saturates the uncertainty inequality (ΔxΔp = ℏ/2), minimizing total energy subject to the constraint. Any narrower wavefunction (smaller Δx) would have larger Δp and more kinetic energy. Any broader wavefunction (larger Δx) would have more potential energy. The Gaussian balances the two to find the global minimum — which is ½ℏω, not zero.