Wavefunctions & Probability Density - Physical Chemistry 2

A particle in a box has wavefunction $\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$ for $n = 1, 2, 3, ...$ If measurements show that the probability of finding the particle in the left half of the box $[0, L/2]$ is exactly 0.5, which quantum numbers $n$ are consistent with this observation?
Answer: All values of $n$ since the particle-in-a-box wavefunctions are symmetric about $x = L/2$ For all particle-in-a-box states, $\sin^2(n\pi x/L)$ is symmetric about $x = L/2$ . The substitution $u = x/L$ shows that $\int_0^{1/2} \sin^2(n\pi u) du = 1/2$ for any integer $n$ . Choice A incorrectly associates antisymmetry with probability distribution. Choice B misunderstands the relationship between wavefunction parity and probability symmetry. Choice D incorrectly invokes uncertainty principles that don't apply to position probability calculations.
Consider a wavefunction $\psi(x) = \begin{cases} Ax(L-x) & \text{if } 0 \leq x \leq L \\ 0 & \text{elsewhere} \end{cases}$ . After normalization, which region has the highest probability density?
Answer: At $x = L/2$ where the wavefunction amplitude is maximum The probability density is $|\psi(x)|^2 \propto x^2(L-x)^2$ . To find the maximum, we differentiate: $\frac{d}{dx}[x^2(L-x)^2] = 2x(L-x)(L-2x)$ . This equals zero when $x = 0$ , $x = L$ , or $x = L/2$ . Since the function is zero at the boundaries, the maximum occurs at $x = L/2$ . Choice A confuses boundary conditions with probability maxima. Choice B incorrectly identifies inflection points. Choice D is false since this is clearly a parabolic-like distribution.
A normalized wavefunction $\psi(x) = A\sin(\pi x/L)$ is defined over the interval $0 \leq x \leq L$ . If the probability of finding the particle in the region $0 \leq x \leq L/4$ is 0.091, what is the probability of finding the particle in the region $3L/4 \leq x \leq L$ ?
Answer: 0.091 Due to the symmetry of $\sin^2(\pi x/L)$ about $x = L/2$ , the probability density at position $x$ equals the probability density at position $(L-x)$ . Therefore, the probability of finding the particle in $[0, L/4]$ equals the probability in $[3L/4, L]$ . Choice B represents twice the given probability. Choice C would be the probability for $[L/4, 3L/4]$ . Choice D represents a much larger fraction inconsistent with the small probability in the tail regions.
Consider the wavefunction $\psi(x,t) = A\left[\psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}\right]$ where $\psi_1$ and $\psi_2$ are real energy eigenfunctions with $E_2 > E_1$ . At time $t = 0$ , the probability density $|\psi(x,0)|^2$ has a node (zero) at $x = x_0$ . This node will:
Answer: Disappear and reappear periodically with period $2\pi\hbar/(E_2 - E_1)$ When you encounter a superposition of energy eigenstates, you're dealing with quantum interference that creates time-dependent behavior. The key insight is that while individual energy eigenstates have time-independent probability densities, their superposition creates oscillating interference patterns. Let's analyze what happens to the probability density. At any time t, we have: $|\psi(x,t)|^2 = |A|^2[\psi_1^2(x) + \psi_2^2(x) + 2\psi_1(x)\psi_2(x)\cos((E_2-E_1)t/\hbar)]$ The crucial term is the interference term $2\psi_1(x)\psi_2(x)\cos((E_2-E_1)t/\hbar)$ , which oscillates with frequency $(E_2-E_1)/\hbar$ . At $t=0$ , there's a node at $x_0$ , meaning the total probability density equals zero there. As time evolves, the cosine term changes, causing the interference pattern to shift. The node will disappear when the interference becomes constructive, then reappear when it becomes destructive again, completing a full cycle every $2\pi\hbar/(E_2-E_1)$ . Option A is wrong because nodes in superposition states aren't conserved—only the individual eigenfunctions maintain their form. Option B incorrectly suggests oscillation about a fixed point rather than complete disappearance/reappearance. Option C is wrong because nodes don't translate with constant velocity; they vanish and reform. Remember: superposition states create time-dependent interference patterns with period $2\pi\hbar/\Delta E$ . The probability density oscillates between different spatial distributions, causing nodes to appear and disappear periodically rather than maintaining fixed positions.
A wavefunction $\psi(r,\theta,\phi) = R(r)Y_\ell^m(\theta,\phi)$ describes a hydrogen-like atom where $Y_\ell^m$ are spherical harmonics. If the radial probability density $P(r) = r^2|R(r)|^2$ has a maximum at $r = r_{max}$ , and the angular probability density $|Y_\ell^m(\theta,\phi)|^2$ has $\ell$ nodal lines, what is the total probability of finding the electron in a spherical shell of thickness $\Delta r$ centered at $r_{max}$ ?
Answer: $4\pi r_{max}^2 |R(r_{max})|^2 \Delta r$ from integrating over all angles When you encounter probability calculations for hydrogen-like atoms, remember that the wavefunction separates into radial and angular parts, and finding an electron requires integrating over the appropriate volume element. To find the total probability in a spherical shell, you need to integrate the probability density $|\psi|^2 = |R(r)|^2|Y_\ell^m(\theta,\phi)|^2$ over the entire shell volume. The volume element in spherical coordinates is $r^2 \sin\theta \, dr \, d\theta \, d\phi$ , so for a thin shell of thickness $\Delta r$ at radius $r_{max}$ , you integrate: $\int_0^{2\pi} \int_0^{\pi} |R(r_{max})|^2 |Y_\ell^m(\theta,\phi)|^2 r_{max}^2 \sin\theta \, d\theta \, d\phi \, \Delta r$ Since spherical harmonics are normalized ( $\int |Y_\ell^m|^2 d\Omega = 1$ where $d\Omega = \sin\theta \, d\theta \, d\phi$ ), the angular integration gives $4\pi$ . This yields $4\pi r_{max}^2 |R(r_{max})|^2 \Delta r$ . Answer A incorrectly uses the radial probability density $P(r) = r^2|R(r)|^2$ , which already has the $r^2$ factor built in but lacks the $4\pi$ from angular integration. Answer C omits the $4\pi$ factor entirely, representing only a differential probability element rather than the total shell probability. Answer D incorrectly divides by $(2\ell + 1)$ , confusing orbital degeneracy with probability normalization—the degeneracy doesn't affect the probability calculation for a specific orbital. The key insight: always include the full volume element when calculating spatial probabilities, and remember that normalized spherical harmonics integrate to give the surface area of a unit sphere, $4\pi$ .
A quantum particle has a wavefunction $\psi(x) = A\sin(kx)e^{-\lambda|x|}$ where $A$ , $k$ , and $\lambda$ are positive constants. For this wavefunction to be normalizable, which condition must be satisfied?
Answer: $\lambda > 0$ only, with no restrictions on $k$ When analyzing whether a wavefunction is normalizable, you need to determine if the normalization integral $\int_{-\infty}^{\infty} |\psi(x)|^2 dx$ converges to a finite value. For the given wavefunction $\psi(x) = A\sin(kx)e^{-\lambda|x|}$ , the probability density is $|\psi(x)|^2 = A^2\sin^2(kx)e^{-2\lambda|x|}$ . The key insight is that the exponential decay term $e^{-2\lambda|x|}$ dominates the behavior at large $|x|$ . Since $\lambda > 0$ , this exponential factor decays rapidly as $x \to \pm\infty$ , ensuring the integral converges regardless of the oscillatory $\sin^2(kx)$ term. The exponential decay is strong enough that even though $\sin^2(kx)$ oscillates between 0 and 1, the overall integrand approaches zero fast enough for normalization. No additional constraints on $k$ are needed for convergence. Answer A is incorrect because quantization conditions like $k = n\pi/L$ arise from boundary conditions in confined systems, not normalizability requirements. Answer C is wrong because there's no mathematical requirement that $\lambda^2 > k^2$ for convergence—the exponential decay with any positive $\lambda$ is sufficient to make the integral finite. Answer D adds an unnecessary restriction that $k$ be non-zero; even $k = 0$ would give a normalizable wavefunction $Ae^{-\lambda|x|}$ . Study tip: For normalizability questions, focus on the behavior at infinity. Exponential decay with positive exponents always ensures convergence, while polynomial or oscillatory terms alone typically don't affect the overall convergence when combined with proper decay.
Consider a wavefunction $\psi(x,y) = A\psi_1(x)\psi_2(y)$ where $\psi_1(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{\pi x}{a}\right)$ for $0 \leq x \leq a$ and $\psi_2(y) = \sqrt{\frac{2}{b}}\sin\left(\frac{\pi y}{b}\right)$ for $0 \leq y \leq b$ . If the probability of finding the particle in the region $0 \leq x \leq a/2$ and $0 \leq y \leq b/2$ is exactly $1/4$ , what is the value of the normalization constant $A$ ?
Answer: $A = 1$ and the given probability condition is automatically satisfied When dealing with separable wavefunctions in quantum mechanics, you need to check both normalization and probability calculations systematically. The key insight is recognizing when individual wavefunctions are already properly normalized. First, let's verify that $\psi_1(x)$ and $\psi_2(y)$ are normalized. For $\psi_1(x)$ : $\int_0^a \psi_1^2(x)dx = \int_0^a \frac{2}{a}\sin^2\left(\frac{\pi x}{a}\right)dx = 1$ Similarly, $\psi_2(y)$ is normalized. Since both individual wavefunctions are normalized, the total wavefunction $\psi(x,y) = A\psi_1(x)\psi_2(y)$ requires $A = 1$ for proper normalization. Now for the probability condition. The probability of finding the particle in the specified region is: $P = \int_0^{a/2}\int_0^{b/2}|\psi(x,y)|^2 dx dy = \int_0^{a/2}\psi_1^2(x)dx \times \int_0^{b/2}\psi_2^2(y)dy$ For particle-in-a-box wavefunctions, by symmetry, exactly half the probability density lies in each half of the box. Therefore, each integral equals $1/2$ , giving $P = (1/2)(1/2) = 1/4$ . Option B incorrectly assumes you need to adjust $A$ to meet the probability condition, but this probability emerges naturally from the wavefunction symmetry. Option C wrongly claims the individual functions aren't normalized. Option D suggests an arbitrary normalization factor without justification. Study tip: When you see separable wavefunctions, always check if the individual components are already normalized first—this often determines the overall normalization constant immediately.
A wavefunction $\psi(x,t) = \frac{1}{\sqrt{2}}[\psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}]$ represents a superposition of two energy eigenstates where $\psi_1$ and $\psi_2$ are orthonormal. The time-averaged probability density $\langle|\psi(x,t)|^2\rangle_t$ over one complete period is:
Answer: $\frac{1}{2}[|\psi_1(x)|^2 + |\psi_2(x)|^2]$ When you encounter a superposition of energy eigenstates, you're dealing with quantum interference effects that oscillate in time. The key insight is understanding what happens when you average these oscillations over a complete period. To find the time-averaged probability density, start by calculating $|\psi(x,t)|^2$ : $|\psi(x,t)|^2 = \frac{1}{2}[|\psi_1(x)|^2 + |\psi_2(x)|^2 + \psi_1^*(x)\psi_2(x)e^{i(E_2-E_1)t/\hbar} + \psi_1(x)\psi_2^*(x)e^{-i(E_2-E_1)t/\hbar}]$ The last two terms combine to give $2\text{Re}(\psi_1^*(x)\psi_2(x)\cos((E_2-E_1)t/\hbar))$ . These are the interference terms that oscillate with frequency $(E_2-E_1)/\hbar$ . When you time-average over one complete period, the oscillating cosine terms average to zero, leaving only the time-independent probability densities from each eigenstate. This gives answer B: $\frac{1}{2}[|\psi_1(x)|^2 + |\psi_2(x)|^2]$ . Answer A incorrectly keeps the interference term without time-averaging. Answer C shows the full time-dependent expression before averaging. Answer D uses the imaginary part instead of the real part and fails to time-average. Study tip: Remember that time-averaging superposition states always eliminates oscillating cross-terms between different energy eigenstates. The time-averaged probability density becomes just the sum of individual state probabilities weighted by their coefficients—quantum interference washes out over time.
Two students attempt to normalize the wavefunction $\psi(x) = xe^{-x/a}$ over the domain $[0, \infty)$ . Student A computes $\int_0^\infty |\psi(x)|^2 dx = 2a^3$ and concludes the normalized function is $\psi_N(x) = \frac{1}{\sqrt{2a^3}}xe^{-x/a}$ . Student B argues that since $x$ can be negative in principle, the domain should be $(-\infty, \infty)$ and uses $\int_{-\infty}^\infty |\psi(x)|^2 dx$ for normalization. Who is correct?
Answer: Student A is correct; the domain $[0, \infty)$ is appropriate and the normalization constant is $\frac{1}{\sqrt{2a^3}}$ The given wavefunction $\psi(x) = xe^{-x/a}$ is only defined for $x \geq 0$ as written. The integral $\int_0^\infty x^2 e^{-2x/a} dx = 2a^3$ is correct, making the normalization constant $1/\sqrt{2a^3}$ . Choice B incorrectly assumes all wavefunctions must be defined over all space. Choice C suggests both are valid when only the specified domain matters. Choice D is wrong since $x^2 e^{-2x/a}$ is clearly square-integrable over $[0,\infty)$ .
A particle's wavefunction is $\psi(x) = Ne^{-\alpha x^2}$ where $\alpha > 0$ . If the probability density at $x = 0$ is four times the probability density at $x = \sigma$ , what is the relationship between $\alpha$ and $\sigma$ ?
Answer: $\alpha = \frac{2\ln(2)}{\sigma^2}$ The probability density is $|\psi(x)|^2 = N^2e^{-2\alpha x^2}$ . At $x = 0$ : $N^2$ . At $x = \sigma$ : $N^2e^{-2\alpha\sigma^2}$ . Setting up the ratio: $\frac{N^2}{N^2e^{-2\alpha\sigma^2}} = 4$ , so $e^{2\alpha\sigma^2} = 4$ . Taking the natural log: $2\alpha\sigma^2 = \ln(4) = 2\ln(2)$ , giving $\alpha = \frac{2\ln(2)}{\sigma^2}$ . Choice A uses $\alpha$ instead of $2\alpha$ . Choice B forgets the factor of 2. Choice C uses $\ln(4)$ directly without the factor of 2 in the exponent.
Two wavefunctions $\psi_1(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$ and $\psi_2(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{2\pi x}{L}\right)$ are each normalized over $[0, L]$ . A linear combination $\Psi(x) = c_1\psi_1(x) + c_2\psi_2(x)$ is formed where $|c_1|^2 + |c_2|^2 = 1$ . Which statement about the normalization of $\Psi(x)$ is correct?
Answer: $\Psi(x)$ is normalized only if $\psi_1$ and $\psi_2$ are orthogonal, which they are in this case The norm squared of $\Psi$ is $|c_1|^2 + |c_2|^2 + 2\text{Re}(c_1^*c_2\langle\psi_1|\psi_2\rangle)$ . Since $\psi_1$ and $\psi_2$ are orthogonal (different sine functions), $\langle\psi_1|\psi_2\rangle = 0$ , so $\Psi$ is normalized. Choice A ignores the cross terms. Choice C incorrectly suggests the zero cross terms still affect normalization. Choice D makes an overly broad false statement about linear combinations.
Two particles with identical mass are described by the symmetric wavefunction $\psi(x_1,x_2) = A[\phi(x_1)\phi(x_2) + \phi(x_2)\phi(x_1)]$ where $\phi(x) = e^{-\alpha x^2}$ and $\alpha > 0$ . The probability density $|\psi(x_1,x_2)|^2$ along the line $x_1 = x_2$ compared to that at the origin $(0,0)$ is:
Answer: Maximum at the origin and decreases as $x_1 = x_2$ moves away from zero When you encounter symmetric wavefunctions for identical particles, focus on how the mathematical form affects probability density in different regions of space. Let's analyze this step by step. First, notice that $\psi(x_1,x_2) = A[\phi(x_1)\phi(x_2) + \phi(x_2)\phi(x_1)]$ . Since $\phi(x) = e^{-\alpha x^2}$ is the same function for both terms, we have $\phi(x_1)\phi(x_2) = \phi(x_2)\phi(x_1)$ . This means $\psi(x_1,x_2) = 2A\phi(x_1)\phi(x_2) = 2Ae^{-\alpha(x_1^2 + x_2^2)}$ . The probability density is $|\psi|^2 = 4A^2e^{-2\alpha(x_1^2 + x_2^2)}$ . Along the line $x_1 = x_2$ , this becomes $|\psi|^2 = 4A^2e^{-4\alpha x_1^2}$ . At the origin $(0,0)$ , $|\psi|^2 = 4A^2$ . As we move along $x_1 = x_2$ away from zero, the exponential term $e^{-4\alpha x_1^2}$ decreases, making the probability density smaller. Option A incorrectly assumes the interference effect varies with position—the "constructive interference" is constant everywhere. Option B misunderstands what symmetry means; symmetric doesn't mean uniform probability. Option D confuses this with antisymmetric wavefunctions (like for fermions), where the wavefunction itself would be zero along $x_1 = x_2$ . The correct answer is C: the probability density is maximum at the origin and decreases exponentially as you move away from zero along the diagonal. Study tip: For Gaussian-type wavefunctions, probability density always peaks where the exponent is least negative (usually at the origin) and decreases as variables increase in magnitude.
A particle's wavefunction is given by $\psi(x) = \begin{cases} A(x + L) & \text{for } -L \leq x \leq 0 \\ A(L - x) & \text{for } 0 \leq x \leq L \\ 0 & \text{elsewhere} \end{cases}$ where $A$ is a normalization constant. The probability current density $J(x) = \frac{\hbar}{2mi}[\psi^*\nabla\psi - \psi\nabla\psi^*]$ at $x = 0$ is:
Answer: Zero because the wavefunction is real and continuous at $x = 0$ When you encounter probability current density problems, remember that this quantity tells you about particle flow and requires careful handling of derivatives, especially at points where the wavefunction changes form. The probability current density formula $J(x) = \frac{\hbar}{2mi}[\psi^*\nabla\psi - \psi\nabla\psi^*]$ involves the gradient (derivative) of the wavefunction. Since this wavefunction is real everywhere, $\psi^* = \psi$ , so the formula becomes $J(x) = \frac{\hbar}{2mi}[\psi\frac{d\psi}{dx} - \psi\frac{d\psi}{dx}] = 0$ . For any real wavefunction, the two terms in the brackets are identical and cancel out completely. Additionally, at $x = 0$ , both pieces of the wavefunction give $\psi(0) = AL$ , confirming continuity. The current density is zero because there's no net particle flow when the wavefunction is purely real. Looking at the wrong answers: B incorrectly suggests that derivative discontinuity makes the current undefined, but even with discontinuous derivatives, you can still evaluate limits and find that both terms in the current formula remain equal. C and D both propose non-zero currents, but these ignore the fundamental fact that real wavefunctions always produce zero current density due to the mathematical structure of the formula. Study tip: Remember that probability current density is always zero for real wavefunctions. Only complex wavefunctions (with imaginary components) can produce non-zero particle flow. This is a key quantum mechanics principle that appears frequently on physical chemistry exams.
Two unnormalized wavefunctions are given: $\psi_1(x) = Ae^{-\alpha x^2}$ and $\psi_2(x) = Bxe^{-\beta x^2}$ where $A$ , $B$ , $\alpha$ , and $\beta$ are positive real constants. After normalization, which statement about their probability densities at $x = 0$ is correct?
Answer: The normalized $|\psi_1|^2$ is non-zero at $x = 0$ , while normalized $|\psi_2|^2$ is zero at $x = 0$ When evaluating probability densities of wavefunctions, you need to examine the behavior of $|\psi|^2$ at specific points, remembering that normalization changes the amplitude but not the fundamental shape or zeros of the function. Let's analyze each wavefunction at $x = 0$ . For $\psi_1(x) = Ae^{-\alpha x^2}$ , when $x = 0$ : $\psi_1(0) = Ae^{0} = A$ . Since $A$ is a positive constant, $|\psi_1(0)|^2 = A^2 > 0$ . After normalization, this becomes some positive value. For $\psi_2(x) = Bxe^{-\beta x^2}$ , when $x = 0$ : $\psi_2(0) = B(0)e^{0} = 0$ . The factor of $x$ in front makes the entire function zero at the origin, so $|\psi_2(0)|^2 = 0$ . Normalization cannot change a zero into a non-zero value. Looking at the wrong answers: (A) incorrectly claims both are zero— $\psi_1$ clearly isn't zero at $x = 0$ . (C) incorrectly states both have maxima at $x = 0$ —while $\psi_1$ does have its maximum there, $\psi_2$ equals zero. (D) reverses the correct relationship entirely. The correct answer is (B): normalized $|\psi_1|^2$ is non-zero at $x = 0$ , while normalized $|\psi_2|^2$ is zero there. Study tip: When analyzing wavefunctions, always substitute the specific coordinate values directly into the original functions first. Normalization affects the overall magnitude but preserves zeros and the relative shape of the probability density.
A complex wavefunction $\psi(x) = (a + ib)e^{ikx}$ where $a$ , $b$ , and $k$ are real constants is defined over a finite interval and normalized. If the probability current density $J \propto \text{Im}[\psi^* \frac{d\psi}{dx}]$ is measured, what determines its magnitude?
Answer: The product $k(a^2 + b^2)$ , combining both the momentum and the total probability amplitude The probability current is $J \propto \text{Im}[(a-ib)ik(a+ib)e^{ikx}e^{-ikx}] = \text{Im}[ik(a^2+b^2)] = k(a^2+b^2)$ . The magnitude depends on both the wave vector $k$ (momentum) and the total probability amplitude $(a^2+b^2)$ . Choice A incorrectly excludes the imaginary part. Choice B ignores the amplitude dependence. Choice D gives the ratio rather than recognizing that both $a$ and $b$ contribute to the total amplitude.
Consider a wavefunction $\psi(x) = A[\phi_1(x) + i\phi_2(x)]$ where $\phi_1(x)$ and $\phi_2(x)$ are real, normalized, and orthogonal functions. If $\phi_1(x) = \sqrt{\frac{2}{L}}\cos\left(\frac{\pi x}{L}\right)$ and $\phi_2(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$ for $x \in [0,L]$ , what is the phase of $\psi(x)$ at $x = L/4$ ?
Answer: $\pi/4$ When you encounter a complex wavefunction written as a linear combination with an imaginary component, you're dealing with quantum mechanical phase analysis. The key insight is that any complex number can be expressed in polar form, where the phase (argument) tells you the angle in the complex plane. To find the phase of $\psi(x) = A[\phi_1(x) + i\phi_2(x)]$ , you need to evaluate both component functions at $x = L/4$ . First, calculate $\phi_1(L/4) = \sqrt{\frac{2}{L}}\cos\left(\frac{\pi}{4}\right) = \sqrt{\frac{2}{L}} \cdot \frac{1}{\sqrt{2}} = \sqrt{\frac{1}{L}}$ . Similarly, $\phi_2(L/4) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi}{4}\right) = \sqrt{\frac{1}{L}}$ . The wavefunction becomes $\psi(L/4) = A\sqrt{\frac{1}{L}}[1 + i]$ . For a complex number $z = a + bi$ , the phase is $\theta = \arctan(b/a)$ . Here, $a = b = 1$ , so the phase is $\arctan(1/1) = \arctan(1) = \pi/4$ . Answer A ( $\pi/4$ ) is correct. Answer B ( $\pi/2$ ) would occur if the real part were zero. Answer C ( $3\pi/4$ ) would arise if the real part were negative and equal in magnitude to the imaginary part. Answer D ( $\pi/6$ ) corresponds to $\arctan(1/\sqrt{3})$ , which has no relevance here. Remember: for phase calculations, always convert to rectangular form first, then use $\arctan(imaginary/real)$ to find the argument.
A wavefunction in spherical coordinates has the form $\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$ where $\Phi(\phi) = \frac{1}{\sqrt{2\pi}}e^{im\phi}$ with $m = 2$ . The probability density $|\psi|^2$ is measured at two points: $(r_0, \pi/3, 0)$ and $(r_0, \pi/3, \pi/2)$ . The ratio of these probability densities is:
Answer: $1$ because $|\Phi(\phi)|^2$ is independent of $\phi$ When you encounter wavefunctions in spherical coordinates, remember that probability densities involve the squared magnitude of complex functions, which eliminates phase information. Let's examine what happens when we calculate $|\psi|^2$ at both points. Since $\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$ , we have: $|\psi|^2 = |R(r)|^2|\Theta(\theta)|^2|\Phi(\phi)|^2$ For $\Phi(\phi) = \frac{1}{\sqrt{2\pi}}e^{i2\phi}$ , the magnitude squared is: $|\Phi(\phi)|^2 = \left|\frac{1}{\sqrt{2\pi}}e^{i2\phi}\right|^2 = \frac{1}{2\pi}|e^{i2\phi}|^2$ Since $|e^{i\alpha}| = 1$ for any real $\alpha$ , we get $|\Phi(\phi)|^2 = \frac{1}{2\pi}$ , which is completely independent of $\phi$ . At both points $(r_0, \pi/3, 0)$ and $(r_0, \pi/3, \pi/2)$ , the $r$ and $\theta$ values are identical, so the ratio of probability densities equals 1. Answer A correctly identifies this independence. Answer B incorrectly suggests the phase $e^{i\pi}$ matters, but phases disappear when taking magnitudes. Answer C makes the error of thinking you square the exponential factor rather than taking its magnitude first. Answer D wrongly applies interference concepts—individual probability measurements don't involve interference between different spatial points. Study tip: Remember that probability densities always involve $|\psi|^2$ , so any pure phase factors (like $e^{im\phi}$ ) will always contribute a magnitude of 1, making the angular probability distribution uniform for single eigenstates.
A wavefunction $\psi(r, \theta, \phi) = R(r)Y(\theta, \phi)$ is separable in spherical coordinates. If the radial part is normalized such that $\int_0^\infty |R(r)|^2 r^2 dr = 1$ and the angular part satisfies $\int_0^{2\pi}\int_0^\pi |Y(\theta, \phi)|^2 \sin\theta \, d\theta \, d\phi = 1$ , what additional condition must be verified to confirm that $\psi$ is properly normalized?
Answer: No additional condition is needed; the separate normalization of radial and angular parts guarantees total normalization For separable functions $\psi = R(r)Y(\theta,\phi)$ , if both parts are individually normalized, then $\int |\psi|^2 d\tau = \int |R(r)|^2 r^2 dr \cdot \int |Y(\theta,\phi)|^2 \sin\theta \, d\theta \, d\phi = 1 \times 1 = 1$ . Choice B misapplies orthogonality concepts. Choice C suggests unnecessary verification when the mathematical proof is already complete. Choice D incorrectly invokes wave interference concepts that don't apply to normalization of separable functions.
A particle has the normalized wavefunction $\psi(x) = \sqrt{\frac{15}{16a^5}}x^2e^{-x/(2a)}$ for $x \geq 0$ and $\psi(x) = 0$ for $x < 0$ . The most probable position (where $|\psi(x)|^2$ is maximum) occurs at:
Answer: $x = 4a$ When you encounter a wavefunction and need to find the most probable position, you're looking for where the probability density $|\psi(x)|^2$ reaches its maximum value. This requires finding the critical points by taking the derivative and setting it equal to zero. Since the wavefunction is already real, $|\psi(x)|^2 = \psi(x)^2 = \frac{15}{16a^5}x^4e^{-x/a}$ . To find the maximum, take the derivative with respect to $x$ : $\frac{d}{dx}|\psi(x)|^2 = \frac{15}{16a^5}\frac{d}{dx}(x^4e^{-x/a})$ Using the product rule: $\frac{d}{dx}(x^4e^{-x/a}) = 4x^3e^{-x/a} + x^4(-\frac{1}{a})e^{-x/a} = x^3e^{-x/a}(4 - \frac{x}{a})$ Setting this equal to zero: $x^3e^{-x/a}(4 - \frac{x}{a}) = 0$ Since $x^3e^{-x/a} > 0$ for $x > 0$ , we need $4 - \frac{x}{a} = 0$ , giving us $x = 4a$ . Choice A ( $x = 2a$ ) would give $4 - 2 = 2 > 0$ , so this is where the probability is still increasing. Choice C ( $x = 6a$ ) gives $4 - 6 = -2 < 0$ , meaning the probability is decreasing past the maximum. Choice D ( $x = 8a$ ) is even further past the maximum where the exponential decay dominates. The answer is B: $x = 4a$ . Study tip: For probability density maxima, always differentiate $|\psi|^2$ , not just $\psi$ . Look for the balance point where polynomial growth meets exponential decay.
A time-dependent wavefunction $\Psi(x,t) = \psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}$ is formed from two normalized energy eigenstates with $\langle\psi_1|\psi_2\rangle = 0$ . At $t = 0$ , the probability density $|\Psi(x,0)|^2$ shows an interference pattern. What happens to this interference pattern as time evolves?
Answer: The interference pattern oscillates with frequency $\omega = (E_2 - E_1)/\hbar$ while the individual probability densities remain constant The probability density is $|\Psi|^2 = |\psi_1|^2 + |\psi_2|^2 + 2\text{Re}[\psi_1^*\psi_2 e^{i(E_2-E_1)t/\hbar}]$ . The cross term oscillates with frequency $(E_2-E_1)/\hbar$ while $|\psi_1|^2$ and $|\psi_2|^2$ are time-independent. Choice A ignores the different energies. Choice C invokes irrelevant thermodynamics. Choice D misunderstands that the oscillation is coherent, not random.

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