Practice Midterm Flashcards
A valid wavefunction must be single-valued and not go to infinity
True
(must be single-valued and not go to infinity)
A valid wavefunction must be an eigenfunction of the Hamiltonian
False
(A valid wavefunction does not have to be an eigenfunction of the Hamiltonian)
The eigenvalues of a Hermitian operator must be real numbers
True
(MUST BE REAL #’S)
The eigenfunctions of a nondegenerate Hermitian Operator must be orthogonal
True
(Hermitian = orthogonal)
Nondegenerate
quantum states that have unique energy levels, meaning that no two states share the same energy value
The most probable location of a particle in a box is always L/2
False
(depends on energy & quantum number)
For a particle in a box in a stationary state <p>=0
True
(momentum cancels itself out)
If [A,B] = 0, then A and B must have the same eigenfunctions
False.
(If they commute, they can share a common set, but not necessarily having same e-functions)
Any valid wavefunction can be constructed using the eigenfunctions of a Hermitian operator.
True.
(Hermitian is good, so yes)
A small particle will behave classically when it has low kinetic energy.
False.
(Reverse. remember the amount of oscillations and nodes contributing to a straiter line)
The energy of a rigid rotor depends on both quantum numbers l and m.
False
(depends on l more than m)
Kinetic Energy equals
(3/2)kB *T
Kinetic Energy equals
(1/2)mv^2
mass of electron
9.11 x 10^-31
If particles travel at the same energy, they must have the same de Broglie wavelength.
False
(need to have the same mass)
The expectation value <x> gives the most probable location for the particle in a box</x>
False
(average position not most probable)
A valid wavefunction must satisfy integral |Psi(x)|^2 = 1
True
(needs to satisfy the normalization condition, total probability of finding the particle somewhere in space is 1)
The Heisenburg representation of a wavefunction is a matrix.
False (they are state vectors)
The eigenfunctions of the Hamiltonian are the only possible quantum states of the system.
False
(they represent stationary states, not superpositions)
A well-behaved wavefunction cannot go to infinity on the interval where it is defined.
True
(A well-behaved wavefunction must be finite - not go to infinity)
A superposition state is also called a nonstationary state.
True.
(unlike a single energy eigenstate, a superposition of multiple eigenstates generally results in time-dependent behavior - leading to a nonstationary state)
The nondegenerate eigenstates of a Hermitian operator must be orthogonal (by default)
.
True.
(Follows properties of Hermitian operators, guarantee the eigenstates corresponding to different eigenvalues are orthogonal)
The eigenfunctions of the rigid rotor Hamiltonian are non degenerate.
False
(They exhibit degeneracy. E-values are nondegenerate for l, but the E-Functions are degenerate)
If two operators commute, there will be an uncertainty relation between their observables.
False
(There is no uncertainty. Heisenburg Uncertainty principle applies to non-commuting observables).
If two particles travel at the same velocity, they must have the same deBroglie wavelength.
False
(differing masses could travel at same v)
All Hermitian operators are real.
False.
(not necessarily real-valued operators themselves)
All eigenvalues of a Hermitian operator is a diagonal matrix
False
(doesn’t have to be, only diagonal if the basis set is eigenfunction of Hermetian operator)
An eigenfunction of the Hamiltonian is also called a stationary state.
True
(Eigen function of Hamiltonian is stationary state)
A well-behaved wavefunction must have only real values.
False.
(spherical harmonics have an imaginary part)
An eigenfunction of the Hamiltonian is also called a stationary state.
True.
The non-degenerate eigenstates of a Hermitian operator must be orthogonal.
True.
(Hermitian is usually pretty nice to work with, so you can assume orthogonality works)
A valid wavefunction must be normalized.
True.
(Physically, the total probability of the particle existing must equal to 1)
If two operators commute, there will be an uncertainty relation between their observables.
False.
(There is no uncertainty relation. Only uncertainty when they don’t commute!)
