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Chapter 2: Problem 25

Evaluate the matrix elements (a) \(\langle v+1|x| v\rangle\) and (b) \(\left\langle v+2\left|x^{2}\right| v\right\rangle\) of a harmonic oscillator by using the relations given at the bottom of Table 2.1 for the Hermite polynomials.

Short Answer

Expert verified
The matrix elements for the harmonic oscillator are (a) \(\langle v+1|x| v\rangle = 0\) and (b) \(\langle v+2|x^2|v\rangle = 0\). These are derived from the Hermite polynomial relations, taking into account that different energy states are orthogonal to each other.

Step by step solution

01

Apply the relation for Hermite polynomials

For part (a), start by noting the expression for the \(v+1\) term derived from the Hermite polynomial: \(H_{v+1} = 2vH_v -2(v-1)H_{v-1}\) where \(v\) is the index of the Hermite polynomial. In terms of the state of a harmonic oscillator, the term \(x|v\rangle\) is related to Hermite polynomials. Substitute \(x|v\rangle\) into the above expression.
02

Simplify and evaluate the matrix element

Rewrite \(2vH_v -2(v-1)H_{v-1}\) as \(2v|x|v\rangle -2(v-1)|x|v-1\rangle\). Assuming that the wave functions are normalized, the inner product \(\langle v+1|v\rangle\) is 0 for \(v+1 \neq v\). Thus, \(\langle v+1|x| v\rangle = 0\). Simplify to solve for part (a).
03

Apply the relation for Hermite polynomials for part (b)

For part (b), note that the expression for the \(x^2\) term is related to Hermite polynomials as \(H_{v+1} + H_{v-1} = 2vH_v\). Substitute \(x^2|v\rangle\) instead of \(|x|v\rangle\) into this expression.
04

Simplify and evaluate the matrix element for part (b)

Rewrite as \(\langle v+2|x^2|v \rangle = 2v|\langle v+1|x|v\rangle + |\langle v-1|x|v\rangle\rangle\). From Step 2, it was found that \(\langle v+1|x|v\rangle = 0\), so simplify the expression, which results in the term \(\langle v+2|x^2|v\rangle = |\langle v-1|x|v\rangle\rangle\). The inner product \(\langle v+2|v\rangle\) is 0 for \(v+2 \neq v\). So, \(\langle v+2|x^2|v \rangle = 0\). This is the simplified answer for part (b).

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