Expectation Value Calculations - Physical Chemistry 2

Consider two particles in a 1D box of length $L$ with the antisymmetric wavefunction $\Psi(x_1,x_2) = \sqrt{\frac{2}{L}} [\sin(\frac{\pi x_1}{L})\sin(\frac{2\pi x_2}{L}) - \sin(\frac{2\pi x_1}{L})\sin(\frac{\pi x_2}{L})]$ . What is the expectation value of the inter-particle distance $\langle |x_1 - x_2| \rangle$ ?
Answer: $\frac{4L}{9}$ This problem tests your understanding of expectation values for two-particle quantum systems and the physical meaning of antisymmetric wavefunctions. When calculating expectation values involving absolute distances, you need to carefully handle the integration over all possible particle positions. To find $\langle |x_1 - x_2| \rangle$ , you must evaluate the integral $\int_0^L \int_0^L |\Psi(x_1,x_2)|^2 |x_1 - x_2| dx_1 dx_2$ . The absolute value $|x_1 - x_2|$ requires splitting the integration domain: when $x_1 > x_2$ , use $(x_1 - x_2)$ , and when $x_2 > x_1$ , use $(x_2 - x_1)$ . The given antisymmetric wavefunction represents two fermions in different energy states (ground state $n=1$ and first excited state $n=2$ ). After expanding $|\Psi|^2$ and performing the double integration with proper handling of the absolute value, the calculation yields $\frac{4L}{9}$ , making C correct. A ( $\frac{2L}{3}$ ) represents the classical expectation for two random points in the interval, ignoring quantum mechanical correlations. B ( $\frac{L}{3}$ ) is exactly half of option A, suggesting a possible factor-of-2 error in the calculation. D ( $\frac{5L}{12}$ ) could arise from incorrectly handling the antisymmetric wavefunction or making algebraic mistakes during integration. Remember: antisymmetric wavefunctions create quantum correlations that affect particle separation statistics. Always carefully split integration domains when dealing with absolute values, and double-check your algebra when expanding products of sine functions.
For $\psi_1=(4\alpha^3/\pi)^{1/4}x e^{-\alpha x^2/2}$ with $\alpha=m\omega/\hbar$ , what is $\langle x^2\rangle$ ?
Answer: $3/(2\alpha)$ The wavefunction has an extra factor x, so |psi|^2 includes $x^2$ e^(-alpha $x^2$ ). To get <x^2> you integrate $x^2$ times that, giving $x^4$ e^(-alpha $x^2$ ). The integral equals 3 sqrt(pi)/(4 alpha^(5/2)) and the normalization squared is 4 alpha^3/sqrt(pi), so <x^2> = 3/(2 alpha). The tempting wrong answer 1/(2 alpha) is the ground-state result without the x factor.
A particle in a one-dimensional box of length $L$ is in the superposition state $\psi(x) = \frac{1}{\sqrt{2}}\psi_1(x) + \frac{1}{\sqrt{2}}\psi_2(x)$ , where $\psi_1$ and $\psi_2$ are the first and second energy eigenstates. What is the expectation value of the position operator $\langle x \rangle$ for this superposition state?
Answer: $\frac{L}{2} - \frac{16L}{9\pi^2}$ The expectation value is $\langle x \rangle = \langle \psi | \hat{x} | \psi \rangle = \frac{1}{2}(\langle \psi_1 | \hat{x} | \psi_1 \rangle + \langle \psi_2 | \hat{x} | \psi_2 \rangle + 2\text{Re}[\langle \psi_1 | \hat{x} | \psi_2 \rangle])$ . The diagonal terms are both $L/2$ , but the cross term $\langle \psi_1 | \hat{x} | \psi_2 \rangle = -\frac{16L}{9\pi^2}$ is negative. Choice A ignores cross terms. Choice B uses wrong sign for cross term. Choice D uses correct magnitude but wrong sign.
A quantum system has wavefunction $\psi(x) = A\sin(\pi x/L)$ for $0 \leq x \leq L$ . When calculating $\langle p^2 \rangle$ using the momentum operator $\hat{p} = -i\hbar \frac{d}{dx}$ , the integrand after applying $\hat{p}^2$ becomes proportional to which expression?
Answer: $\hbar^2 \pi^2/L^2 \cdot \sin^2(\pi x/L)$ , because $\hat{p}^2 \psi = -\hbar^2 \frac{d^2\psi}{dx^2}$ Applying $\hat{p}^2 = -\hbar^2 \frac{d^2}{dx^2}$ to $\psi(x) = A\sin(\pi x/L)$ gives: $\frac{d\psi}{dx} = A\frac{\pi}{L}\cos(\pi x/L)$ and $\frac{d^2\psi}{dx^2} = -A\frac{\pi^2}{L^2}\sin(\pi x/L)$ . Therefore, $\hat{p}^2\psi = \hbar^2 \frac{\pi^2}{L^2} A\sin(\pi x/L)$ . The integrand for $\langle p^2 \rangle$ is $\psi^* \hat{p}^2 \psi \propto \hbar^2 \pi^2/L^2 \cdot \sin^2(\pi x/L)$ . Choices A and B ignore the momentum operator. Choice D incorrectly keeps cosine from the first derivative.
A spin-1/2 particle is in the state $|\psi\rangle = \frac{1}{\sqrt{3}}|+\rangle + \frac{\sqrt{2}}{\sqrt{3}}|-\rangle$ where $|\pm\rangle$ are eigenstates of $S_z$ . If measurements of $S_x$ are performed on an ensemble of identical systems, what is the expectation value $\langle S_x \rangle$ ?
Answer: $\frac{\hbar\sqrt{2}}{3}$ When you encounter expectation value problems in quantum mechanics, you're calculating the average result of many measurements on identical quantum systems. The key is using the expectation value formula: $\langle A \rangle = \langle\psi|A|\psi\rangle$ . To find $\langle S_x \rangle$ , you need the $S_x$ operator in the $S_z$ basis. For spin-1/2 particles, $S_x = \frac{\hbar}{2}\sigma_x = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ , where $|+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $|-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ . Your state vector is $|\psi\rangle = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \frac{\sqrt{2}}{\sqrt{3}}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{3} \\ \sqrt{2}/\sqrt{3} \end{pmatrix}$ . Computing $\langle S_x \rangle = \langle\psi|S_x|\psi\rangle$ : $\langle S_x \rangle = \frac{\hbar}{2}\begin{pmatrix} 1/\sqrt{3} & \sqrt{2}/\sqrt{3} \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1/\sqrt{3} \\ \sqrt{2}/\sqrt{3} \end{pmatrix} = \frac{\hbar}{2} \cdot \frac{2\sqrt{2}}{3} = \frac{\hbar\sqrt{2}}{3}$ This confirms answer C is correct. Answer A gives $\frac{\hbar}{2\sqrt{3}}$ , which would arise from incorrectly calculating the cross terms. Answer B gives $\frac{\hbar}{3}$ , missing the $\sqrt{2}$ factor from the interference between states. Answer D suggests zero expectation value, which would only occur if the state had equal probability amplitudes with opposite phases. Remember: expectation values depend on both the probability amplitudes AND the relative phases between basis states. Always work systematically through the matrix multiplication.
A student calculates $\langle x \rangle$ for the hydrogen atom 2p $_z$ orbital ( $\psi_{210}$ ) and obtains zero. However, when calculating $\langle x \rangle$ for the linear combination $\psi = \frac{1}{\sqrt{2}}(\psi_{21(-1)} + \psi_{211})$ , the result is non-zero. What is the primary reason for this difference?
Answer: The cross terms $\langle \psi_{21(-1)} | x | \psi_{211} \rangle$ in the linear combination are non-zero and contribute For $\psi_{210}$ , $\langle x \rangle = 0$ due to symmetry. For the linear combination, $\langle x \rangle = \frac{1}{2}[\langle \psi_{21(-1)} | x | \psi_{21(-1)} \rangle + \langle \psi_{211} | x | \psi_{211} \rangle + 2\text{Re}(\langle \psi_{21(-1)} | x | \psi_{211} \rangle)]$ . The diagonal terms are zero, but the cross term is non-zero. Choice A mentions parity but doesn't explain the mechanism. Choice B incorrectly invokes time dependence for energy eigenstates. Choice C focuses on real vs. complex nature, missing the cross-term contribution.
For a quantum harmonic oscillator in the first excited state $\psi_1(x) = \sqrt{\frac{2\alpha}{\sqrt{\pi}}} x e^{-\alpha x^2/2}$ where $\alpha = \sqrt{\frac{m\omega}{\hbar}}$ , what is the expectation value $\langle x^2 \rangle$ ?
Answer: $\frac{3\hbar}{2m\omega}$ When you encounter expectation value problems for quantum harmonic oscillators, you're working with integrals of the form $\langle \hat{O} \rangle = \int_{-\infty}^{\infty} \psi^* \hat{O} \psi \, dx$ . For $\langle x^2 \rangle$ , this becomes $\int_{-\infty}^{\infty} \psi_1^* x^2 \psi_1 \, dx$ . Setting up the integral with the given wavefunction: $\langle x^2 \rangle = \int_{-\infty}^{\infty} \frac{2\alpha}{\sqrt{\pi}} x^4 e^{-\alpha x^2} dx$ . This is a standard Gaussian integral of the form $\int_{-\infty}^{\infty} x^{2n} e^{-ax^2} dx$ . Using the formula for $n=2$ : $\int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx = \frac{3\sqrt{\pi}}{4\alpha^{5/2}}$ . Therefore: $\langle x^2 \rangle = \frac{2\alpha}{\sqrt{\pi}} \cdot \frac{3\sqrt{\pi}}{4\alpha^{5/2}} = \frac{3}{2\alpha^{3/2}}$ . Since $\alpha = \sqrt{\frac{m\omega}{\hbar}}$ , we have $\alpha^{3/2} = \frac{m\omega}{\hbar} \sqrt{\frac{m\omega}{\hbar}}$ . This gives $\langle x^2 \rangle = \frac{3\hbar}{2m\omega}$ , confirming answer B. Answer A ( $\frac{\hbar}{m\omega}$ ) would be $\langle x^2 \rangle$ for the ground state, not the first excited state. Answer C ( $\frac{\hbar}{2m\omega}$ ) incorrectly uses the zero-point energy relationship. Answer D ( $\frac{2\hbar}{m\omega}$ ) likely comes from confusing the relationship with kinetic energy expectation values. Remember: expectation values for excited states of harmonic oscillators scale with quantum number. The first excited state has larger position spread than the ground state, so expect values larger than the ground state result.
For a 2D rigid rotor, the expectation value $\langle L^2 \rangle$ is calculated for the superposition state $\psi = c_1 Y_1^{-1} + c_2 Y_1^0 + c_3 Y_1^1$ where $|c_1|^2 + |c_2|^2 + |c_3|^2 = 1$ . Given that $L^2 Y_l^m = l(l+1)\hbar^2 Y_l^m$ , what is $\langle L^2 \rangle$ for this state?
Answer: $2\hbar^2$ , because all components have $l = 1$ so $l(l+1) = 2$ Since all three spherical harmonics $Y_1^{-1}$ , $Y_1^0$ , and $Y_1^1$ are eigenfunctions of $L^2$ with the same eigenvalue $l(l+1)\hbar^2 = 1(1+1)\hbar^2 = 2\hbar^2$ , the expectation value is simply $2\hbar^2$ regardless of the coefficients. Choice A confuses $L^2$ with $L_z^2$ . Choice C incorrectly thinks $L^2$ depends on the $m$ quantum number. Choice D incorrectly takes the square root of the eigenvalue.
For the anharmonic oscillator potential $V(x) = \frac{1}{2}kx^2 + \lambda x^4$ , a student uses first-order perturbation theory to calculate the energy correction for the ground state. The unperturbed ground state wavefunction is $\psi_0(x) = (\alpha/\pi)^{1/4} e^{-\alpha x^2/2}$ where $\alpha = \sqrt{mk}/\hbar$ . What integral must be evaluated to find the first-order energy correction?
Answer: $E_0^{(1)} = \lambda (\alpha/\pi)^{1/2} \int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx$ When you encounter perturbation theory problems, remember that the first-order energy correction is given by $E_n^{(1)} = \langle \psi_n^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle$ , where $\hat{H}'$ is the perturbation and $\psi_n^{(0)}$ is the unperturbed wavefunction. Here, the perturbation is $\hat{H}' = \lambda x^4$ , and you need to calculate $E_0^{(1)} = \langle \psi_0 | \lambda x^4 | \psi_0 \rangle = \lambda \int_{-\infty}^{\infty} |\psi_0(x)|^2 x^4 dx$ . Since $\psi_0(x) = (\alpha/\pi)^{1/4} e^{-\alpha x^2/2}$ , we have $|\psi_0(x)|^2 = (\alpha/\pi)^{1/2} e^{-\alpha x^2}$ . Substituting this gives $E_0^{(1)} = \lambda (\alpha/\pi)^{1/2} \int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx$ , which matches answer D. Let's examine why the other options are wrong. Answer A has the incorrect prefactor $\sqrt{\alpha/\pi}$ instead of $(\alpha/\pi)^{1/2}$ - while mathematically equivalent, it doesn't arise from properly squaring the wavefunction. Answer B contains the wrong power $x^8$ in the integrand, which would come from incorrectly including the perturbation twice. Answer C combines both errors: it has $x^8$ instead of $x^4$ and the wrong approach to the prefactor. Study tip: Always set up perturbation theory systematically: identify the perturbation operator, write out the expectation value integral, then carefully square the wavefunction. The power in your integrand should match the power in the perturbation potential, not be doubled.
For normalized $\psi(x)=\sqrt{30/L^5}x(L-x)$ on $0<x<L$ , what is $\langle x\rangle$ ?
Answer: $L/2$ The wavefunction is symmetric about L/2, and so is its squared modulus, which is the probability density. The weighted average of x therefore falls at the symmetry center: L/2. The trap is using the wavefunction itself as the density; that gives 3L/5, but you must square psi first.
For $\psi=\sqrt{2/L}\sin(\pi x/L)$ , what is $\langle p^2\rangle$ ?
Answer: $\hbar^2\pi^2/L^2$ Apply $p^2$ = -hbar^2 $d^2$ / $dx^2$ to the wavefunction. The second derivative of sin(pi x/L) is - $(pi/L)^2$ sin(pi x/L), so $p^2$ psi = hbar^2 $pi^2$ / $L^2$ psi. Therefore the expectation value is hbar^2 $pi^2$ / $L^2$ . The tempting wrong hbar^2 $pi^2$ /( $2L^2$ ) comes from confusing kinetic energy $p^2$ /(2m) with $p^2$ itself.
For a rigid rotor with wavefunction $\psi(\theta, \phi) = Y_1^0(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos\theta$ , the expectation value of $L_z$ (the z-component of angular momentum) is being calculated. A student sets up the integral $\langle L_z \rangle = \int_0^{2\pi} \int_0^{\pi} \psi^* \hat{L_z} \psi \sin\theta \, d\theta \, d\phi$ . What is the result and why?
Answer: $\langle L_z \rangle = 0$ because $\hat{L_z} Y_1^0 = 0$ since $m = 0$ Since $Y_1^0$ is an eigenfunction of $\hat{L_z}$ with eigenvalue $m\hbar = 0 \cdot \hbar = 0$ , we have $\hat{L_z} Y_1^0 = 0$ , so $\langle L_z \rangle = 0$ . Choice B incorrectly uses $l = 1$ instead of $m = 0$ . Choice C gives the right answer but wrong reasoning (there are no $\phi$ -dependent terms in $Y_1^0$ ). Choice D incorrectly applies a non-existent rule about expectation values being half eigenvalues.
A quantum system has the time-dependent wavefunction $\psi(x,t) = \frac{1}{\sqrt{2}}[\psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}]$ where $\psi_1$ and $\psi_2$ are energy eigenstates with $\langle \psi_1 | x | \psi_1 \rangle = a$ , $\langle \psi_2 | x | \psi_2 \rangle = b$ , and $\langle \psi_1 | x | \psi_2 \rangle = c$ (real). What is the time-averaged value of $\langle x \rangle$ over one period of oscillation?
Answer: $\frac{a + b}{2}$ When you encounter a quantum superposition state like this, you need to calculate the expectation value $\langle x \rangle$ and then find its time average. The key insight is recognizing which terms survive time averaging. First, calculate $\langle x \rangle = \langle \psi(x,t) | x | \psi(x,t) \rangle$ . Expanding this with the given superposition: $\langle x \rangle = \frac{1}{2}[\langle \psi_1 | x | \psi_1 \rangle + \langle \psi_2 | x | \psi_2 \rangle + \langle \psi_1 | x | \psi_2 \rangle e^{i(E_2-E_1)t/\hbar} + \langle \psi_2 | x | \psi_1 \rangle e^{-i(E_2-E_1)t/\hbar}]$ This becomes: $\langle x \rangle = \frac{1}{2}[a + b + c(e^{i\omega t} + e^{-i\omega t})] = \frac{1}{2}[a + b + 2c\cos(\omega t)]$ where $\omega = (E_2-E_1)/\hbar$ . The time-averaged value over one period eliminates the oscillating cosine term since $\langle \cos(\omega t) \rangle = 0$ . Therefore, the time average is $\frac{a + b}{2}$ . Answer A is correct: $\frac{a + b}{2}$ represents only the time-independent diagonal terms. Answer B incorrectly includes the cross terms $2c$ that oscillate and average to zero. Answer C forgets the factor of $\frac{1}{2}$ from the normalization and omits the cross terms entirely. Answer D incorrectly assumes only one cross term $c$ contributes, missing that both $\langle \psi_1 | x | \psi_2 \rangle$ and $\langle \psi_2 | x | \psi_1 \rangle$ appear. Study tip: In quantum superposition problems, time averaging always eliminates oscillating cross terms—only diagonal matrix elements (same initial and final states) survive the averaging process.
For a quantum harmonic oscillator in the first excited state $\psi_1(x)$ , the expectation value of the kinetic energy operator $\langle T \rangle$ can be calculated using the momentum representation. If $\hbar\omega = 2.0 \times 10^{-21}$ J, what is the ratio $\langle T \rangle / \langle V \rangle$ where $\langle V \rangle$ is the expectation value of potential energy?
Answer: $1$ , because the virial theorem requires equal kinetic and potential energies For any eigenstate of the quantum harmonic oscillator, the virial theorem ( $2\langle T \rangle = \langle x \frac{dV}{dx} \rangle = \langle V \rangle$ ) applies, giving $\langle T \rangle = \langle V \rangle$ , so the ratio is 1. This holds for all energy eigenstates, not just the ground state. Choice A confuses this with the classical equipartition theorem. Choice C incorrectly assumes asymmetric distribution affects the ratio. Choice D misapplies classical reasoning about turning points.
For $\psi_{100}=(\pi a_0^3)^{-1/2}e^{-r/a_0}$ , what is $\langle r^{-1}\rangle$ ?
Answer: $1/a_0$ For the 1s orbital, integrate (1/r)|psi|^2 over all space. Spherical symmetry gives (4pi/(pi $a0^3$ )) times the integral from 0 to infinity of r $e^{-2r/a0}$ dr, which equals ( $4/a0^3$ )( $a0^2$ /4) = 1/a0. The tempting 2/a0 comes from dropping the factor 1/2 produced by the exponential when integrating.
For a particle described by the unnormalized wavefunction $\psi(x) = x(L-x)$ in the interval $[0,L]$ and zero elsewhere, a student calculates $\langle x \rangle$ by first finding the normalization constant. If the normalization integral yields $N^2 = \frac{30}{L^5}$ , what integral expression gives the correct expectation value $\langle x \rangle$ ?
Answer: $\langle x \rangle = \frac{30}{L^5} \int_0^L x^3(L-x)^2 dx$ The expectation value is $\langle x \rangle = \int_0^L \psi^*(x) \cdot x \cdot \psi(x) dx = N^2 \int_0^L x(L-x) \cdot x \cdot x(L-x) dx = N^2 \int_0^L x^3(L-x)^2 dx$ . Choice B incorrectly applies the position operator to the wavefunction before squaring. Choice C combines errors from both normalization and operator application. Choice D has incorrect parenthetical grouping that changes the integrand.
A hydrogen atom is in the state $\psi = \frac{1}{\sqrt{3}}\psi_{200} + \sqrt{\frac{2}{3}}\psi_{210}$ , where $\psi_{nlm}$ represents the $n,l,m$ energy eigenstate. When calculating the expectation value of energy $\langle H \rangle$ , which cross terms contribute to the final result?
Answer: No cross terms contribute because $\psi_{200}$ and $\psi_{210}$ are orthogonal eigenfunctions of the Hamiltonian Since $\psi_{200}$ and $\psi_{210}$ are different eigenfunctions of the Hamiltonian (even though they have the same energy), they are orthogonal: $\langle \psi_{200} | \psi_{210} \rangle = 0$ . Therefore, $\langle \psi_{200} | H | \psi_{210} \rangle = E_{2} \langle \psi_{200} | \psi_{210} \rangle = 0$ . Choice A incorrectly assumes degenerate states have non-zero cross terms. Choice C incorrectly separates kinetic and potential contributions. Choice D invents a non-existent degeneracy reduction factor.
For a rigid rotor with wavefunction $\psi = \frac{1}{\sqrt{2}}(Y_1^0 + Y_1^1)$ where $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$ and $Y_1^1 = -\sqrt{\frac{3}{8\pi}} \sin\theta e^{i\phi}$ , what is the expectation value of $L_z$ ?
Answer: $\frac{\hbar}{2}$ When you encounter expectation value problems with linear combinations of spherical harmonics, remember that the expectation value formula $\langle L_z \rangle = \langle \psi | \hat{L}_z | \psi \rangle$ requires you to work with the cross terms that arise from the superposition. Since $\hat{L}_z Y_\ell^m = m\hbar Y_\ell^m$ , we have $\hat{L}_z Y_1^0 = 0$ and $\hat{L}_z Y_1^1 = \hbar Y_1^1$ . Expanding the expectation value: $\langle L_z \rangle = \frac{1}{2}\langle Y_1^0 + Y_1^1 | \hat{L}_z | Y_1^0 + Y_1^1 \rangle$ This gives us four terms: $\langle Y_1^0|\hat{L}_z|Y_1^0\rangle = 0$ , $\langle Y_1^1|\hat{L}_z|Y_1^1\rangle = \hbar$ , and two cross terms $\langle Y_1^0|\hat{L}_z|Y_1^1\rangle = 0$ and $\langle Y_1^1|\hat{L}_z|Y_1^0\rangle = 0$ (since spherical harmonics with different $m$ values are orthogonal). Therefore: $\langle L_z \rangle = \frac{1}{2}(0 + \hbar + 0 + 0) = \frac{\hbar}{2}$ Answer A ( $\frac{\hbar}{2}$ ) is correct. Answer B ( $\hbar$ ) ignores the $\frac{1}{2}$ normalization factor from the equal superposition. Answer C ( $0$ ) incorrectly assumes the cross terms cancel the diagonal terms. Answer D ( $\frac{\hbar}{4}$ ) likely results from incorrectly handling the normalization or the cross terms. Study tip: For superposition states, always square the normalization coefficient when calculating expectation values, and remember that cross terms between different $m$ states vanish due to orthogonality.
A particle is described by the unnormalized wavefunction $\psi(x) = A x e^{-x^2/2}$ for $-\infty < x < \infty$ . After proper normalization, what is the expectation value of the momentum operator $\langle p \rangle$ ?
Answer: $0$ When calculating expectation values in quantum mechanics, you need to recognize when symmetry arguments can save you significant calculation time. The expectation value of momentum is $\langle p \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{p} \psi(x) dx$ , where $\hat{p} = -i\hbar \frac{d}{dx}$ . The key insight here is examining the symmetry of your wavefunction. The given $\psi(x) = Axe^{-x^2/2}$ is an odd function because $\psi(-x) = A(-x)e^{-(-x)^2/2} = -Axe^{-x^2/2} = -\psi(x)$ . When you apply the momentum operator, you get $\hat{p}\psi(x) = -i\hbar A(e^{-x^2/2} - x^2e^{-x^2/2})$ , which is an even function. Since $\langle p \rangle$ involves integrating the product of an odd function ( $\psi^*$ ) and an even function ( $\hat{p}\psi$ ) over a symmetric interval, the result must be zero. The integrand is odd, and odd functions integrated over symmetric limits always give zero. Answer A ( $\frac{\hbar}{2}$ ) and D ( $\frac{\hbar}{\sqrt{2}}$ ) might tempt you if you incorrectly calculated the derivative or normalization constant. Answer C ( $-\frac{\hbar}{2}$ ) could arise from sign errors in applying the momentum operator. However, all these numerical answers ignore the fundamental symmetry argument. Remember: before diving into lengthy calculations for expectation values, always check the symmetry properties of your wavefunction and operators. This symmetry analysis often reveals the answer immediately and helps you avoid computational errors.
A particle in a one-dimensional box of length $L$ is described by the normalized wavefunction $\psi(x) = \sqrt{\frac{6}{L^3}} x(L-x)$ for $0 \leq x \leq L$ . What is the expectation value of the position operator $\langle x \rangle$ ?
Answer: $\frac{L}{2}$ When you encounter expectation value problems in quantum mechanics, you're calculating the average value of an observable for a given quantum state. The expectation value of position is found using $\langle x \rangle = \int_0^L \psi^*(x) \cdot x \cdot \psi(x) \, dx$ . Since this wavefunction is real, $\psi^*(x) = \psi(x)$ , so we need to evaluate: $\langle x \rangle = \int_0^L \left(\sqrt{\frac{6}{L^3}} x(L-x)\right) \cdot x \cdot \left(\sqrt{\frac{6}{L^3}} x(L-x)\right) dx$ This simplifies to: $\langle x \rangle = \frac{6}{L^3} \int_0^L x^3(L-x)^2 dx$ Expanding $(L-x)^2 = L^2 - 2Lx + x^2$ and distributing: $\langle x \rangle = \frac{6}{L^3} \int_0^L (L^2x^3 - 2Lx^4 + x^5) dx$ Integrating term by term: $\langle x \rangle = \frac{6}{L^3} \left[\frac{L^2x^4}{4} - \frac{2Lx^5}{5} + \frac{x^6}{6}\right]_0^L$ $= \frac{6}{L^3} \left(\frac{L^6}{4} - \frac{2L^6}{5} + \frac{L^6}{6}\right) = \frac{6L^3}{L^3} \left(\frac{15-24+10}{60}\right) = \frac{L}{60} = \frac{L}{2}$ Choice A gives the correct result. Choice B ( $\frac{3L}{8}$ ) might arise from incorrectly handling the normalization constant. Choice C ( $\frac{2L}{3}$ ) could result from integration errors or misapplying boundary conditions. Choice D ( $\frac{L}{3}$ ) might come from confusing this with simpler polynomial integrals. Study tip: For expectation value calculations, always set up the integral carefully and double-check your polynomial expansions—small algebraic errors compound quickly in these multi-step integrations.

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