Okay, let's see. I need to find the eigenvalues of the Hamiltonian operator H, which is given by H = ε σ · n, where ε is a constant with energy dimensions, σ are the Pauli spin matrices, and n is a unit vector. Hmm, the Pauli matrices are 2x2 matrices, right? So σ · n would be the dot product of the vector σ and the unit vector n. That would result in a 2x2 matrix. Then multiplying by ε gives the Hamiltonian matrix.

First, I remember that the Pauli matrices have eigenvalues ±1. But here, we're dealing with a linear combination of the Pauli matrices. So maybe the eigenvalues of H will depend on the magnitude of the vector σ · n. Wait, but σ · n is a vector of Pauli matrices dotted with n. Let me think. The Pauli matrices are σ_x, σ_y, σ_z, and they each have eigenvalues ±1. But when you take a linear combination like σ · n = n_x σ_x + n_y σ_y + n_z σ_z, this operator will have eigenvalues equal to the magnitude of the vector scaled by something. Wait, no. Wait, the eigenvalues of σ · n should be the magnitude of n times something. Since n is a unit vector, the magnitude of σ · n is... Wait, maybe I should recall that for any unit vector n, the operator σ · n has eigenvalues ±1. Because each Pauli matrix has eigenvalues ±1, and when you take a linear combination with coefficients summing to 1 (since n is a unit vector), the eigenvalues would be ±1. Wait, but actually, the eigenvalues of σ · n should be ±1 regardless of the direction of n. Let me check that. For example, take n along the x-axis: σ · n = σ_x. Its eigenvalues are ±1. Same for y and z. So regardless of the direction of n, σ · n is a Hermitian matrix with eigenvalues ±1. Therefore, multiplying by ε, the Hamiltonian H = ε σ · n would have eigenvalues ±ε. So the eigenvalues should be ε and -ε. Looking at the options, that's option B: +ε, -ε. But wait, let me make sure I'm not missing anything here.

Wait, another way to think about it: The Pauli matrices are all traceless and have determinant -1. The operator σ · n is a 2x2 matrix with eigenvalues ±1 because for any unit vector, σ · n squared is the identity matrix. Because (σ · n)^2 = (σ · n)(σ · n) = σ · σ + cross terms. Wait, no. Let me compute (σ · n)^2. Since σ is a vector of matrices, (σ · n)^2 = (n_x σ_x + n_y σ_y + n_z σ_z)^2. Expanding this, we get n_x² σ_x² + n_y² σ_y² + n_z² σ_z² + 2n_x n_y σ_x σ_y + 2n_x n_z σ_x σ_z + 2n_y n_z σ_y σ_z. Now, σ_i² for i=x,y,z are all identity matrices (since σ_x² = I, same for y and z). So the first three terms sum to (n_x² + n_y² + n_z²) I = I, since n is a unit vector. Now the cross terms: σ_i σ_j for i ≠ j. Remember that σ_i σ_j = i ε_ijk σ_k, where ε is the Levi-Civita symbol. So σ_x σ_y = i σ_z, σ_y σ_z = i σ_x, etc. So the cross terms would be 2n_x n_y (i σ_z) + 2n_x n_z (i σ_y) + 2n_y n_z (i σ_x). Wait, but then these terms are imaginary. However, when you square (σ · n), the imaginary terms would combine in such a way that when you take the trace, they might vanish. Wait, but (σ · n)^2 should be I, right? Because σ · n is a Hermitian matrix with eigenvalues ±1, so squaring it gives the identity. Let me check. Let's take n along the x-axis: σ · n = σ_x. Then (σ_x)^2 = I. Similarly for any direction, (σ · n)^2 = I. Therefore, the eigenvalues of σ · n must satisfy λ² = 1, so λ = ±1. Therefore, H = ε σ · n would have eigenvalues ±ε. So the answer should be B. But wait, the options are given in terms of constants with possible factors of ℏ/2. Wait, the Pauli matrices are defined with factors of ℏ/2 in some conventions. Let me check. Wait, the Pauli matrices are typically defined as σ_x, σ_y, σ_z, which are 2x2 matrices with entries ±1 and 0, and trace 0. Their eigenvalues are ±1. However, in some contexts, especially when dealing with spin-1/2 particles, the spin operators are often written as (ℏ/2) σ. For example, the spin angular momentum operator S = (ℏ/2) σ. So if the problem had used S · n, the eigenvalues would be ±ℏ/2. But in this problem, the Hamiltonian is given as H = ε σ · n. Therefore, if σ · n has eigenvalues ±1, then H has eigenvalues ±ε. But in the options, option B is ±ε, option C is ±ε ℏ/2, etc. So according to this, the answer should be B. However, I need to make sure that the Pauli matrices here are indeed with eigenvalues ±1. Let me confirm. The Pauli matrices are:

σ_x = [[0, 1], [1, 0]]
σ_y = [[0, -i], [i, 0]]
σ_z = [[1, 0], [0, -1]]

Each of these matrices has eigenvalues ±1. For example, σ_z has eigenvalues 1 and -1. σ_x and σ_y also have eigenvalues ±1. So yes, their eigenvalues are ±1. Therefore, σ · n is a linear combination of these matrices, but since n is a unit vector, the sum of squares of the components of n is 1. So σ · n is a Hermitian matrix with eigenvalues ±1. Therefore, multiplying by ε (which has units of energy), the eigenvalues become ±ε. So the answer is B. But wait, let me check the options again. The options are:

A. +1, -1

B. +ε, -ε

C. +εℏ/2, -εℏ/2

D. +ℏ/2, -ℏ/2

So the correct answer is B. But wait, in some textbooks, the Pauli matrices are defined with a factor of 1/2. For instance, when dealing with spin-1/2 systems, the spin matrices are (ℏ/2) σ. So perhaps in this problem, σ · n is actually (ℏ/2) times some other vector. Wait, but the problem states that the components of σ are the Pauli spin matrices. So σ here is the vector of Pauli matrices. Therefore, σ · n is a sum of Pauli matrices scaled by the components of n. Therefore, σ · n is a 2x2 matrix with eigenvalues ±1. Therefore, H = ε σ · n would have eigenvalues ±ε. So the answer is B. But let me check another way. Let's compute the eigenvalues of σ · n. Let me take a general unit vector n = (n_x, n_y, n_z). Then σ · n = n_x σ_x + n_y σ_y + n_z σ_z. Let's compute the matrix:

σ · n = [[n_z, n_x - i n_y], [n_x + i n_y, -n_z]]

The trace of this matrix is 0, and the determinant is (n_z)^2 + (n_x - i n_y)(n_x + i n_y) - (-n_z)(n_z) = n_z² + (n_x² + n_y²) - (-n_z²) = n_x² + n_y² + n_z² + n_z² = (n_x² + n_y² + n_z²) + n_z². Wait, no. Wait, determinant is (n_z)(-n_z) - (n_x - i n_y)(n_x + i n_y). Let's compute that. The determinant is (-n_z²) - [ (n_x)^2 + (n_y)^2 ] because (n_x - i n_y)(n_x + i n_y) = n_x² + n_y². So determinant is -n_z² - n_x² - n_y² = - (n_x² + n_y² + n_z²) = -1, since n is a unit vector. So determinant is -1. The trace is 0. The eigenvalues squared would be trace squared minus determinant, but for a 2x2 matrix, the eigenvalues λ satisfy λ² = trace² - 4 determinant. Wait, no. For a 2x2 matrix, the characteristic equation is λ² - trace λ + determinant = 0. Since trace is 0, it's λ² + determinant = 0. So determinant is -1, so λ² = 1, so λ = ±1. Therefore, the eigenvalues of σ · n are indeed ±1. Therefore, H = ε σ · n has eigenvalues ±ε. So the correct answer is B. But wait, looking at the options again, option B is "+ε, -ε", which matches. But let me check if there's a possibility that the eigenvalues are scaled by ℏ/2. Wait, maybe the problem is using a different convention where the Pauli matrices are scaled by ℏ/2. Let me see. The question states that the components of σ are the Pauli spin matrices. The standard Pauli matrices have entries ±1 and 0, with trace 0 and determinant -1. Therefore, σ · n has eigenvalues ±1. Therefore, multiplying by ε gives eigenvalues ±ε. Therefore, answer B. However, sometimes in physics, especially when dealing with spin, the spin angular momentum is (ℏ/2) σ. So if the problem had written (ℏ/2) σ · n, then the eigenvalues would be ±ℏ/2. But here, the Hamiltonian is given as ε σ · n. So unless ε already includes a factor of ℏ/2, the eigenvalues are ±ε. Therefore, the answer is B.
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