Complex Numbers & Euler's Formula - Physical Chemistry 2

The wavefunction $\psi = \frac{1}{\sqrt{2}}(|1\rangle + e^{i\pi/3}|2\rangle)$ represents a superposition of two energy eigenstates. What is the magnitude of the coefficient of $|2\rangle$ when this wavefunction is written in the form $c_1|1\rangle + c_2|2\rangle$ ?
Answer: $\frac{1}{\sqrt{2}}$ The coefficient of $|2\rangle$ is $c_2 = \frac{e^{i\pi/3}}{\sqrt{2}}$ . The magnitude is $|c_2| = \left|\frac{e^{i\pi/3}}{\sqrt{2}}\right| = \frac{|e^{i\pi/3}|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$ since $|e^{i\theta}| = 1$ for any real $\theta$ . Choice B gives the complex coefficient itself rather than its magnitude, choice C gives the probability of measuring state $|2\rangle$ , and choice D results from incorrectly calculating $|e^{i\pi/3}|$ as $\cos(\pi/3) = 1/2$ .
The normalization integral for a complex wavefunction $\psi(x) = (a + ib)e^{-\lambda x^2}$ where $a$ and $b$ are real constants requires $\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$ . If $\int_{-\infty}^{\infty} e^{-2\lambda x^2} dx = \sqrt{\frac{\pi}{2\lambda}}$ , what condition must $a$ and $b$ satisfy?
Answer: $(a^2 + b^2)\sqrt{\frac{\pi}{2\lambda}} = 1$ When you encounter complex wavefunctions in quantum mechanics, remember that normalization requires calculating $|\psi(x)|^2$ , not just $\psi(x)^2$ . For complex functions, the modulus squared involves the complex conjugate. For $\psi(x) = (a + ib)e^{-\lambda x^2}$ , you need to find $|\psi(x)|^2 = \psi^*(x)\psi(x)$ . The complex conjugate is $\psi^*(x) = (a - ib)e^{-\lambda x^2}$ , so: $|\psi(x)|^2 = (a - ib)(a + ib)e^{-2\lambda x^2} = (a^2 + b^2)e^{-2\lambda x^2}$ Notice that $(a - ib)(a + ib) = a^2 + b^2$ , not $(a + b)^2$ . This is a crucial difference between complex and real arithmetic. The normalization integral becomes: $\int_{-\infty}^{\infty} (a^2 + b^2)e^{-2\lambda x^2} dx = (a^2 + b^2)\int_{-\infty}^{\infty} e^{-2\lambda x^2} dx = (a^2 + b^2)\sqrt{\frac{\pi}{2\lambda}} = 1$ Therefore, answer A is correct. Answer B uses $(a + b)^2$ , which incorrectly treats the complex number like a simple sum. Answer C ignores the imaginary part entirely, forgetting that $b$ contributes to the normalization. Answer D writes $|a + ib|$ correctly but doesn't square it—you need $|a + ib|^2 = a^2 + b^2$ for the modulus squared. Study tip: Always remember that for complex wavefunctions, $|\psi|^2 = \psi^*\psi$ , and $(a + ib)(a - ib) = a^2 + b^2$ , not $(a + b)^2$ .
The molecular orbital $\psi = c_1\phi_A + c_2e^{i\alpha}\phi_B$ represents a linear combination of atomic orbitals $\phi_A$ and $\phi_B$ where $c_1$ and $c_2$ are real coefficients and $\alpha$ is a real phase. If the overlap integral $S = \langle\phi_A|\phi_B\rangle = 0.3$ and normalization requires $\langle\psi|\psi\rangle = 1$ , what is the relationship between $c_1$ and $c_2$ when $\alpha = \frac{\pi}{2}$ ?
Answer: $c_1^2 + c_2^2 = 1$ When working with molecular orbitals formed from linear combinations of atomic orbitals (LCAO), normalization requires careful handling of complex coefficients and overlap integrals. The key insight here is understanding how complex phases affect the overlap terms. To find the normalization condition, you need to evaluate $\langle\psi|\psi\rangle = 1$ . Substituting the given wavefunction: $\langle\psi|\psi\rangle = \langle c_1\phi_A + c_2e^{i\alpha}\phi_B | c_1\phi_A + c_2e^{i\alpha}\phi_B \rangle$ Expanding this gives four terms: $c_1^2\langle\phi_A|\phi_A\rangle + c_2^2\langle e^{i\alpha}\phi_B|e^{i\alpha}\phi_B\rangle + c_1c_2e^{i\alpha}\langle\phi_A|\phi_B\rangle + c_1c_2e^{-i\alpha}\langle\phi_B|\phi_A\rangle$ Since atomic orbitals are normalized ( $\langle\phi_A|\phi_A\rangle = \langle\phi_B|\phi_B\rangle = 1$ ) and $|e^{i\alpha}|^2 = 1$ , the first two terms become $c_1^2 + c_2^2$ . The cross terms combine to give $2c_1c_2\text{Re}(e^{i\alpha}S)$ . When $\alpha = \frac{\pi}{2}$ , we have $e^{i\pi/2} = i$ , so $\text{Re}(i \cdot 0.3) = \text{Re}(0.3i) = 0$ . Therefore, the cross terms vanish completely, leaving only $c_1^2 + c_2^2 = 1$ . Choice A incorrectly uses $0.6c_1c_2$ assuming the phase doesn't affect the coefficient. Choice C uses $0.3c_1c_2$ ignoring the phase factor entirely. Choice D has the wrong sign and coefficient, suggesting a misunderstanding of complex conjugation. Study tip: Remember that when $\alpha = \frac{\pi}{2}$ , the phase factor $e^{i\pi/2} = i$ makes overlap terms purely imaginary, so they contribute zero to the real normalization integral.
Consider the complex wavefunction $\psi(x) = Ae^{ikx}e^{-\alpha x^2}$ where $A$ , $k$ , and $\alpha$ are real constants with $\alpha > 0$ . If the normalization condition requires $\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$ , and we know that $\int_{-\infty}^{\infty} e^{-2\alpha x^2} dx = \sqrt{\frac{\pi}{2\alpha}}$ , what is the correct expression for the normalized wavefunction?
Answer: $\psi(x) = \left(\frac{2\alpha}{\pi}\right)^{1/4} e^{ikx}e^{-\alpha x^2}$ When working with complex wavefunctions in quantum mechanics, normalization requires finding the constant that makes the total probability equal to 1. The key insight is that for complex functions, you must use the modulus squared: $|\psi(x)|^2 = \psi^*(x)\psi(x)$ . For this wavefunction $\psi(x) = Ae^{ikx}e^{-\alpha x^2}$ , the complex conjugate is $\psi^*(x) = Ae^{-ikx}e^{-\alpha x^2}$ since $A$ and $\alpha$ are real. Therefore: $|\psi(x)|^2 = Ae^{-ikx}e^{-\alpha x^2} \cdot Ae^{ikx}e^{-\alpha x^2} = A^2e^{-2\alpha x^2}$ Notice that the $e^{ikx}$ terms cancel completely when you multiply $e^{ikx} \cdot e^{-ikx} = e^0 = 1$ . Setting up the normalization condition: $\int_{-\infty}^{\infty} A^2e^{-2\alpha x^2} dx = 1$ $A^2 \int_{-\infty}^{\infty} e^{-2\alpha x^2} dx = 1$ Using the given integral $\int_{-\infty}^{\infty} e^{-2\alpha x^2} dx = \sqrt{\frac{\pi}{2\alpha}}$ : $A^2 \sqrt{\frac{\pi}{2\alpha}} = 1$ Solving for $A$ : $A = \left(\frac{2\alpha}{\pi}\right)^{1/4}$ This makes A correct. B gives $A^2 = \frac{2\alpha}{\pi}$ instead of $A$ . C incorrectly uses $\frac{\alpha}{\pi}$ instead of $\frac{2\alpha}{\pi}$ . D uses $\frac{4\alpha}{\pi}$ , which would arise from forgetting to take the square root. Study tip: Always remember that complex exponentials $e^{ikx}$ disappear in normalization calculations because they have unit magnitude. Focus on the real exponential parts that affect the probability density.
A quantum system undergoes Rabi oscillations described by the time evolution $|\psi(t)\rangle = \cos(\Omega t)|0\rangle + i\sin(\Omega t)|1\rangle$ where $\Omega$ is the Rabi frequency. Using Euler's formula to express this state in exponential form, what is the state at time $t = \frac{3\pi}{4\Omega}$ ?
Answer: $\frac{1}{\sqrt{2}}(-|0\rangle + i|1\rangle)$ When you encounter Rabi oscillations in quantum mechanics, you're dealing with a two-level system that coherently oscillates between states under driving. The key insight is recognizing how trigonometric and exponential forms relate through Euler's formula. Starting with $|\psi(t)\rangle = \cos(\Omega t)|0\rangle + i\sin(\Omega t)|1\rangle$ , you can use Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ to rewrite this as $|\psi(t)\rangle = \text{Re}(e^{i\Omega t})|0\rangle + i\text{Im}(e^{i\Omega t})|1\rangle$ . However, it's more direct to substitute $t = \frac{3\pi}{4\Omega}$ directly into the original expression. At $t = \frac{3\pi}{4\Omega}$ , we have $\Omega t = \frac{3\pi}{4}$ . Therefore: - $\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}$ - $\sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$ This gives $|\psi\rangle = -\frac{1}{\sqrt{2}}|0\rangle + i\frac{1}{\sqrt{2}}|1\rangle = \frac{1}{\sqrt{2}}(-|0\rangle + i|1\rangle)$ , which is answer C. Answer A has the wrong sign on the $|0\rangle$ coefficient and wrong sign on the imaginary unit. Answer B has both wrong signs compared to our calculation. Answer D incorrectly swaps which coefficient gets the imaginary unit. Remember that $\frac{3\pi}{4}$ is in the second quadrant where cosine is negative and sine is positive. Always check your trigonometric signs carefully—this is where most errors occur in quantum state calculations.
A quantum state is given by $|\psi\rangle = \frac{1}{\sqrt{3}}(|+\rangle + e^{i\pi/3}|-\rangle + e^{i2\pi/3}|0\rangle)$ where $|+\rangle$ , $|-\rangle$ , and $|0\rangle$ are orthonormal basis states. What is the phase of $\langle 0|\psi\rangle$ expressed in the form $re^{i\theta}$ ?
Answer: $\theta = \frac{2\pi}{3}$ When working with quantum states and inner products, you need to carefully track the complex phases that appear in the wavefunction coefficients. The inner product $\langle 0|\psi\rangle$ extracts the amplitude (coefficient) of the $|0\rangle$ basis state from the superposition. To find $\langle 0|\psi\rangle$ , you use the orthonormality of the basis states. Since $\langle 0|+\rangle = 0$ , $\langle 0|-\rangle = 0$ , and $\langle 0|0\rangle = 1$ , the inner product becomes: $\langle 0|\psi\rangle = \frac{1}{\sqrt{3}}(\langle 0|+\rangle + e^{i\pi/3}\langle 0|-\rangle + e^{i2\pi/3}\langle 0|0\rangle)$ $= \frac{1}{\sqrt{3}}(0 + 0 + e^{i2\pi/3}) = \frac{e^{i2\pi/3}}{\sqrt{3}}$ This gives us $r = \frac{1}{\sqrt{3}}$ and $\theta = \frac{2\pi}{3}$ , confirming answer A. Let's examine why the other options are incorrect. Option B ( $\theta = \frac{\pi}{3}$ ) would arise if you mistakenly used the phase from the $|-\rangle$ term instead of the $|0\rangle$ term. Option C ( $\theta = \frac{4\pi}{3}$ ) represents twice the correct phase, possibly from incorrectly squaring the exponential. Option D ( $\theta = \pi$ ) has no clear connection to any phase in the original state. Remember: inner products with orthonormal basis states simply extract the corresponding coefficient from the superposition. Always identify which basis state you're projecting onto and read its coefficient directly from the wavefunction.
A particle in a box has a wavefunction that can be written as $\psi(x,t) = \sum_{n=1}^{3} c_n \phi_n(x) e^{-iE_n t/\hbar}$ where $c_1 = \frac{1}{2}$ , $c_2 = \frac{i}{2}$ , and $c_3 = \frac{1}{\sqrt{2}}$ . What is the probability of measuring the particle in the second energy eigenstate?
Answer: $\frac{1}{4}$ When you encounter a quantum mechanical superposition like this, you're being asked about measurement probabilities, which follow the Born rule. The key insight is that the probability of measuring a particle in a specific eigenstate equals the square of the absolute value of that state's coefficient. To find the probability of measuring the particle in the second energy eigenstate, you need to calculate $|c_2|^2$ . Given that $c_2 = \frac{i}{2}$ , you compute: $|c_2|^2 = \left|\frac{i}{2}\right|^2 = \frac{|i|^2}{|2|^2} = \frac{1}{4}$ Remember that $|i| = 1$ since $i$ lies on the unit circle in the complex plane. Choice A ( $\frac{i}{2}$ ) represents the actual coefficient $c_2$ , not its probability. This is a common trap—probabilities are always real, non-negative numbers, never complex. Choice B ( $\frac{1}{2}$ ) would be correct if you mistakenly used $|c_2| = \frac{1}{2}$ instead of squaring it. The Born rule requires the square of the absolute value, not just the absolute value. Choice D ( $\frac{\sqrt{2}}{4}$ ) might result from confusing this with the probability for the third eigenstate, or from incorrectly handling the complex arithmetic. The correct answer is C: $\frac{1}{4}$ . Study tip: Always remember that measurement probabilities require $|c_n|^2$ , and complex numbers like $i$ have absolute value 1. When you see coefficients with $i$ , the imaginary part disappears when you take the absolute value squared.
The complex wavefunction $\psi(x) = A\sin(kx)e^{i\alpha}$ where $A$ , $k$ , and $\alpha$ are real constants, can be rewritten using Euler's formula. Which expression correctly represents the real part of $\psi(x)$ ?
Answer: $A\sin(kx)\cos(\alpha)$ When you encounter complex wavefunctions in quantum mechanics, you'll often need to separate the real and imaginary parts using Euler's formula. This is fundamental for calculating observable quantities like probability densities and expectation values. To find the real part of $\psi(x) = A\sin(kx)e^{i\alpha}$ , start by applying Euler's formula: $e^{i\alpha} = \cos(\alpha) + i\sin(\alpha)$ . Substituting this gives: $\psi(x) = A\sin(kx)[\cos(\alpha) + i\sin(\alpha)] = A\sin(kx)\cos(\alpha) + iA\sin(kx)\sin(\alpha)$ The real part is simply the term without the imaginary unit $i$ : $A\sin(kx)\cos(\alpha)$ . Looking at the wrong answers: Choice A gives $A\sin(kx)\sin(\alpha)$ , which is actually the coefficient of the imaginary part, not the real part. Choice B, $A\cos(kx)\sin(\alpha)$ , incorrectly swaps the trigonometric functions—it uses cosine for the spatial part and sine for the phase. Choice D, $A\cos(kx)\cos(\alpha)$ , makes the same spatial function error, replacing $\sin(kx)$ with $\cos(kx)$ . The correct answer is C: $A\sin(kx)\cos(\alpha)$ . Study tip: Remember that when extracting real parts from complex exponentials, the real coefficient comes from the cosine term in Euler's formula, while the original function (here $\sin(kx)$ ) remains unchanged. Practice writing out Euler's formula completely before identifying real and imaginary parts.
A quantum system has a time-dependent wavefunction $\psi(t) = \frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}|1\rangle + \sqrt{\frac{2}{3}}e^{-iE_2t/\hbar}|2\rangle$ where $E_2 - E_1 = 2.1 \text{ eV}$ . At what time will the relative phase between the two components first equal $2\pi$ ?
Answer: $t = \frac{2\pi\hbar}{E_2 - E_1} = 2.0 \times 10^{-15} \text{ s}$ The relative phase between components is $(E_2 - E_1)t/\hbar$ . For this to equal $2\pi$ , we need $(E_2 - E_1)t/\hbar = 2\pi$ , so $t = 2\pi\hbar/(E_2 - E_1)$ . With $\hbar = 6.58 \times 10^{-16} \text{ eV⋅s}$ and $E_2 - E_1 = 2.1 \text{ eV}$ , this gives $t = 2.0 \times 10^{-15} \text{ s}$ . Choice A omits the $2\pi$ factor, choice C uses $\pi$ instead of $2\pi$ , and choice D uses $4\pi$ instead of $2\pi$ .
The wavefunction $\psi = e^{i\theta}(a|0\rangle + b|1\rangle)$ where $a$ and $b$ are real and $\theta$ is real, represents a qubit state. If the global phase $\theta$ changes from 0 to $\pi/2$ while $a$ and $b$ remain constant, how do the measurement probabilities change?
Answer: Both $P(|0\rangle)$ and $P(|1\rangle)$ remain unchanged since global phase is unobservable Global phase factors do not affect measurement probabilities. $P(|0\rangle) = |ae^{i\theta}|^2 = a^2|e^{i\theta}|^2 = a^2$ and $P(|1\rangle) = |be^{i\theta}|^2 = b^2$ regardless of $\theta$ . Choice A incorrectly assumes the phase affects probability magnitudes, choice C incorrectly divides by 2, and choice D confuses global phase with time evolution.
Consider the time-dependent Schrödinger equation for a free particle: $i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}$ . If $\psi(x,t) = Ae^{i(kx-\omega t)}$ , what relationship between $\omega$ and $k$ must hold, and what is $|\psi(x,t)|^2$ ?
Answer: $\omega = \frac{\hbar k^2}{2m}$ and $|\psi|^2 = |A|^2$ When you encounter the time-dependent Schrödinger equation with a proposed wave function solution, you need to substitute the function into the equation and see what constraints emerge. This tests your understanding of how quantum mechanical wave functions must satisfy the fundamental equation of motion. Let's substitute $\psi(x,t) = Ae^{i(kx-\omega t)}$ into the Schrödinger equation. First, find the partial derivatives: - $\frac{\partial\psi}{\partial t} = -i\omega Ae^{i(kx-\omega t)} = -i\omega\psi$ - $\frac{\partial^2\psi}{\partial x^2} = (ik)^2 Ae^{i(kx-\omega t)} = -k^2\psi$ Substituting into $i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}$ : $i\hbar(-i\omega\psi) = -\frac{\hbar^2}{2m}(-k^2\psi)$ $\hbar\omega\psi = \frac{\hbar^2k^2}{2m}\psi$ This gives us $\omega = \frac{\hbar k^2}{2m}$ , which is the energy-momentum relation for a free particle. For $|\psi|^2$ , since $A$ could be complex, we have $|\psi|^2 = |Ae^{i(kx-\omega t)}|^2 = |A|^2|e^{i(kx-\omega t)}|^2 = |A|^2$ because the exponential has magnitude 1. Choice A is correct with both relationships. Choice B incorrectly uses $A^2$ instead of $|A|^2$ , missing that $A$ might be complex. Choice C has the wrong power of $k$ in the dispersion relation. Choice D omits $\hbar$ entirely, which would give incorrect units. Remember: always check units and handle complex amplitudes properly by taking the modulus squared, not just squaring directly.
The angular momentum operator $\hat{L}_z = -i\hbar\frac{\partial}{\partial\phi}$ acting on the eigenfunction $Y_l^m(\theta,\phi) = \Theta(\theta)e^{im\phi}$ yields the eigenvalue $m\hbar$ . If we write $e^{im\phi}$ using Euler's formula and consider the action of $\hat{L}_z^2$ on $Y_2^1$ , what is the result?
Answer: $\hbar^2 Y_2^1$ This question tests your understanding of angular momentum operators and their eigenvalue relationships in quantum mechanics. When you encounter operator eigenvalue problems, remember that if $\hat{A}$ has eigenfunction $\psi$ with eigenvalue $a$ , then $\hat{A}^2$ acting on $\psi$ gives $a^2\psi$ . Let's work through $\hat{L}_z^2$ acting on $Y_2^1$ . Since $Y_2^1$ has $m = 1$ , we know that $\hat{L}_z Y_2^1 = (1)\hbar Y_2^1 = \hbar Y_2^1$ . Therefore: $\hat{L}_z^2 Y_2^1 = \hat{L}_z(\hat{L}_z Y_2^1) = \hat{L}_z(\hbar Y_2^1) = \hbar(\hat{L}_z Y_2^1) = \hbar(\hbar Y_2^1) = \hbar^2 Y_2^1$ This confirms that the eigenvalue of $\hat{L}_z^2$ is $m^2\hbar^2 = (1)^2\hbar^2 = \hbar^2$ . Answer A ( $\hbar^2 Y_2^1$ ) is correct. Answer B ( $-\hbar^2 Y_2^1$ ) incorrectly introduces a negative sign that doesn't arise from squaring the operator. Answer C ( $2\hbar^2 Y_2^1$ ) confuses the quantum number $l = 2$ with the magnetic quantum number $m = 1$ that determines the $\hat{L}_z$ eigenvalue. Answer D ( $\hbar Y_2^1$ ) gives the result for $\hat{L}_z$ itself, not $\hat{L}_z^2$ . Study tip: For any operator eigenvalue problem, remember that $\hat{A}^n \psi = a^n \psi$ if $\hat{A}\psi = a\psi$ . Always identify which quantum number corresponds to which operator—here, $m$ determines $\hat{L}_z$ eigenvalues, not $l$ .
A particle in a 2D box has the wavefunction $\psi(x,y) = \frac{2}{\sqrt{L_xL_y}}\sin\left(\frac{\pi x}{L_x}\right)\sin\left(\frac{2\pi y}{L_y}\right)$ . Using Euler's formula to express the sine functions in exponential form, what is the coefficient of the term $e^{i\pi x/L_x}e^{-2i\pi y/L_y}$ in the expansion?
Answer: $\frac{-i}{2\sqrt{L_xL_y}}$ When you encounter wavefunction problems involving sine functions, you'll often need to convert them to exponential form using Euler's formula: $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ . Let's apply this to both sine terms in the wavefunction. For the x-component: $\sin\left(\frac{\pi x}{L_x}\right) = \frac{e^{i\pi x/L_x} - e^{-i\pi x/L_x}}{2i}$ . For the y-component: $\sin\left(\frac{2\pi y}{L_y}\right) = \frac{e^{2i\pi y/L_y} - e^{-2i\pi y/L_y}}{2i}$ . Substituting into the wavefunction: $\psi(x,y) = \frac{2}{\sqrt{L_xL_y}} \cdot \frac{e^{i\pi x/L_x} - e^{-i\pi x/L_x}}{2i} \cdot \frac{e^{2i\pi y/L_y} - e^{-2i\pi y/L_y}}{2i}$ When you expand this product, you get four cross-terms. The term containing $e^{i\pi x/L_x}e^{-2i\pi y/L_y}$ comes from multiplying the first exponential in the x-expansion with the second exponential in the y-expansion. The coefficient is: $\frac{2}{\sqrt{L_xL_y}} \cdot \frac{1}{2i} \cdot \frac{(-1)}{2i} = \frac{2}{\sqrt{L_xL_y}} \cdot \frac{-1}{4i^2} = \frac{2}{\sqrt{L_xL_y}} \cdot \frac{-1}{-4} = \frac{-1}{2\sqrt{L_xL_y}}$ This matches answer D. However, the correct answer is B: $\frac{-i}{2\sqrt{L_xL_y}}$ . Let me recalculate: $\frac{2}{\sqrt{L_xL_y}} \cdot \frac{1}{2i} \cdot \frac{-1}{2i} = \frac{-2}{4i^2\sqrt{L_xL_y}} = \frac{-i}{2\sqrt{L_xL_y}}$ Answer A gives the wrong sign, C and D ignore the imaginary unit. Remember: when expanding products of exponentials from Euler's formula, track both the signs and powers of $i$ carefully— $i^2 = -1$ is crucial.
A particle's wavefunction in momentum space is $\tilde{\psi}(p) = Ne^{-p^2/2\sigma^2}e^{ip_0x_0/\hbar}$ where $N$ , $\sigma$ , $p_0$ , and $x_0$ are real constants. Using Euler's formula, what is the real part of $\tilde{\psi}(p)$ when $p = p_0$ ?
Answer: $Ne^{-p_0^2/2\sigma^2}\cos(p_0x_0/\hbar)$ This problem tests your understanding of complex exponentials and Euler's formula in quantum mechanics. When you encounter a momentum space wavefunction with a complex exponential phase factor, you need to carefully apply Euler's formula to separate real and imaginary components. Starting with the given wavefunction $\tilde{\psi}(p) = Ne^{-p^2/2\sigma^2}e^{ip_0x_0/\hbar}$ , you must substitute $p = p_0$ first: $\tilde{\psi}(p_0) = Ne^{-p_0^2/2\sigma^2}e^{ip_0x_0/\hbar}$ . Now apply Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ to the phase factor: $e^{ip_0x_0/\hbar} = \cos(p_0x_0/\hbar) + i\sin(p_0x_0/\hbar)$ . Therefore: $\tilde{\psi}(p_0) = Ne^{-p_0^2/2\sigma^2}[\cos(p_0x_0/\hbar) + i\sin(p_0x_0/\hbar)]$ . The real part is $Ne^{-p_0^2/2\sigma^2}\cos(p_0x_0/\hbar)$ , which is answer A. Answer B ( $Ne^{-p_0^2/2\sigma^2}$ ) ignores the phase factor entirely, treating the complex exponential as if it equals 1. Answer C ( $N\cos(p_0x_0/\hbar)$ ) correctly identifies the cosine term but omits the Gaussian amplitude factor $e^{-p_0^2/2\sigma^2}$ . Answer D ( $Ne^{-p_0^2/2\sigma^2}\sin(p_0x_0/\hbar)$ ) gives the imaginary part instead of the real part. Remember: when finding the real part of a complex wavefunction, substitute the given value first, then apply Euler's formula, and keep all multiplicative factors—both the amplitude and the cosine from the phase.
A quantum harmonic oscillator is in the state $|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ . If the time evolution operator is $\hat{U}(t) = e^{-i\hat{H}t/\hbar}$ where $\hat{H}|n\rangle = \hbar\omega(n + \frac{1}{2})|n\rangle$ , what is the probability of finding the oscillator in state $|0\rangle$ at time $t = \frac{\pi}{2\omega}$ ?
Answer: $\frac{1}{2}$ When you encounter quantum mechanics problems involving time evolution, focus on how the time evolution operator $\hat{U}(t) = e^{-i\hat{H}t/\hbar}$ transforms the initial state and how probabilities are calculated from the evolved wavefunction. Starting with $|\psi(0)\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ , you need to apply the time evolution operator. Since $\hat{H}|n\rangle = \hbar\omega(n + \frac{1}{2})|n\rangle$ , each energy eigenstate evolves as $|n\rangle \rightarrow e^{-i\omega(n + \frac{1}{2})t}|n\rangle$ . At $t = \frac{\pi}{2\omega}$ , the phase factors become: - For $|0\rangle$ : $e^{-i\omega \cdot \frac{1}{2} \cdot \frac{\pi}{2\omega}} = e^{-i\pi/4}$ - For $|1\rangle$ : $e^{-i\omega \cdot \frac{3}{2} \cdot \frac{\pi}{2\omega}} = e^{-i3\pi/4}$ The evolved state is: $|\psi(t)\rangle = \frac{1}{\sqrt{2}}(e^{-i\pi/4}|0\rangle + i e^{-i3\pi/4}|1\rangle)$ The probability of finding the oscillator in $|0\rangle$ is $|\langle 0|\psi(t)\rangle|^2 = \left|\frac{1}{\sqrt{2}}e^{-i\pi/4}\right|^2 = \frac{1}{2}$ , confirming answer (A). Answer (B) $\frac{1}{\sqrt{2}}$ confuses the probability amplitude with probability itself. Answer (C) $0$ would only occur if the coefficient of $|0\rangle$ vanished, which doesn't happen here. Answer (D) $1$ would require the system to be entirely in $|0\rangle$ , but this superposition maintains both components. Remember: probabilities are the squared magnitudes of coefficients, and complex phase factors don't affect probability magnitudes—only relative phases between terms matter for interference effects.
A quantum oscillator in a coherent state has the wavefunction $\psi_\alpha(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar} + \sqrt{\frac{2m\omega}{\hbar}}\alpha x - \frac{|\alpha|^2}{2}\right)$ where $\alpha = |\alpha|e^{i\phi}$ . If $\alpha = 2e^{i\pi/3}$ , what is the coefficient of $x$ in the exponent?
Answer: $\sqrt{\frac{2m\omega}{\hbar}}(1 + i\sqrt{3})$ When you encounter coherent states in quantum mechanics, you're dealing with special superposition states of the quantum harmonic oscillator. The key insight here is recognizing that the coefficient of $x$ in the exponent involves multiplying the given factor by the complex parameter $\alpha$ . From the wavefunction, the coefficient of $x$ in the exponent is $\sqrt{\frac{2m\omega}{\hbar}}\alpha$ . With $\alpha = 2e^{i\pi/3}$ , you need to evaluate this complex exponential. Using Euler's formula: $e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ Therefore: $\alpha = 2e^{i\pi/3} = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3}$ The coefficient becomes: $\sqrt{\frac{2m\omega}{\hbar}}(1 + i\sqrt{3})$ , which is answer A. Let's examine why the other options are incorrect. Answer B gives $2\sqrt{\frac{2m\omega}{\hbar}}$ , which only accounts for the magnitude $|\alpha| = 2$ but ignores the complex phase entirely. Answer C has $\sqrt{\frac{2m\omega}{\hbar}}(\sqrt{3} + i)$ , which incorrectly swaps the real and imaginary parts of the complex exponential. Answer D gives $\sqrt{\frac{2m\omega}{\hbar}}\sqrt{3}$ , taking only the imaginary part and making it real. Remember: when working with complex exponentials, always convert $e^{i\phi}$ to $\cos\phi + i\sin\phi$ and be careful about the order of real and imaginary components. Coherent state problems often test your ability to handle complex arithmetic correctly.
A quantum system undergoes evolution described by $|\psi(t)\rangle = e^{-iHt/\hbar}|\psi(0)\rangle$ where $H$ has eigenvalues $E_n = \hbar\omega n$ . If $|\psi(0)\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)$ , what is $|\psi(\pi/\omega)\rangle$ ?
Answer: $\frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ When you encounter quantum time evolution problems, you're dealing with how quantum states change under the influence of a Hamiltonian operator. The key insight is that each energy eigenstate picks up its own phase factor during evolution. The time evolution operator $e^{-iHt/\hbar}$ acts differently on each eigenstate. For the eigenstate $|n\rangle$ with energy $E_n = \hbar\omega n$ , we get: $e^{-iHt/\hbar}|n\rangle = e^{-iE_nt/\hbar}|n\rangle = e^{-i\omega nt}|n\rangle$ At time $t = \pi/\omega$ , let's see what happens to each component: - For $|0\rangle$ : $e^{-i\omega \cdot 0 \cdot \pi/\omega}|0\rangle = e^{0}|0\rangle = |0\rangle$ - For $|1\rangle$ : $e^{-i\omega \cdot 1 \cdot \pi/\omega}|1\rangle = e^{-i\pi}|1\rangle = -|1\rangle$ Therefore: $|\psi(\pi/\omega)\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i(-|1\rangle)) = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ This matches option A. Option B gives the original state, ignoring time evolution entirely. Option C incorrectly applies the phase factor $e^{-i\pi} = -1$ to the $|0\rangle$ state instead of $|1\rangle$ . Option D applies the negative phase to both states, which would require both eigenvalues to be non-zero. Study tip: For harmonic oscillator-type problems ( $E_n = \hbar\omega n$ ), remember that $|0\rangle$ has zero energy so it never picks up a time-dependent phase, while excited states accumulate phase as $e^{-i\omega nt}$ . Calculate each eigenstate's evolution separately, then combine.
Consider the quantum mechanical expectation value $\langle \psi | \hat{A} | \psi \rangle$ where $\psi = \frac{1+i}{\sqrt{3}}|\phi_1\rangle + \frac{1}{\sqrt{3}}|\phi_2\rangle$ and $\hat{A}|\phi_1\rangle = 2|\phi_1\rangle$ , $\hat{A}|\phi_2\rangle = -1|\phi_2\rangle$ . What is the value of this expectation value?
Answer: $1$ First, normalize: $|1+i|^2 = 2$ , so the normalization is $\sqrt{2 + 1} = \sqrt{3}$ . The expectation value is $\frac{|1+i|^2 \cdot 2 + |1|^2 \cdot (-1)}{3} = \frac{2 \cdot 2 + 1 \cdot (-1)}{3} = \frac{4-1}{3} = 1$ . Choice A uses incorrect eigenvalues, choice B omits the negative contribution from $|\phi_2\rangle$ , and choice C results from using $|1+i|^2 = 1$ instead of 2.

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