Question

A quantum simple harmonic oscillator (SHO) of mass $m$ and angular frequency $omega$ has been prepared in the following state: $vert Psi angle = frac{1}{sqrt{2}}(vert 1 angle + ivert 2 angle)$, (1) where $vert n angle$ is the energy eigenvector of the SHO for $n = 0, 1, 2, dots$. Compute the expectation value of the position, $langle x angle$, for this state and express it in terms of $m$, $omega$, and $hbar$.

          A quantum simple harmonic oscillator (SHO) of mass $m$ and angular frequency $omega$ has been prepared in the following state:
$vert Psi 
angle = frac{1}{sqrt{2}}(vert 1 
angle + ivert 2 
angle)$, (1)
where $vert n 
angle$ is the energy eigenvector of the SHO for $n = 0, 1, 2, dots$. Compute the expectation value of the position, $langle x 
angle$, for this state and express it in terms of $m$, $omega$, and $hbar$.

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$A quantum simple harmonic oscillator (SHO) of mass m and angular frequency omega has been prepared in the following state: vert Psi angle = frac1sqrt2(vert 1 angle + ivert 2 angle), (1) where vert n angle is the energy eigenvector of the SHO for n = 0, 1, 2, dots. Compute the expectation value of the position, langle x angle, for this state and express it in terms of m, omega, and hbar.$

Added by Jacob T.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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05/29/2023

Step 1

The state must be normalized, so we have: N^2 (|0⟩⟨0| + |1⟩⟨1| + |2⟩⟨2| + 1/4 |1⟩⟨1|) = 1 N^2 (1 + 1 + 1 + 1/4) = 1 N^2 (11/4) = 1 N = 2/√11 Now, we can write the normalized state as: |ψ⟩ = (2/√11) (|0⟩ + |1⟩ + |2⟩ + i/2 |1⟩) Show more…

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00:01 In this problem, we are given the following quantum mechanical state.

00:05 Ψ equal to 1 over root 2 state 1 plus i times state 2.

00:14 And these states belong to the simple harmonic oscillator.

00:18 So these are the number states.

00:20 And with this ψ state, we are going to compute the expectation value of the position operator.

00:28 Since we are dealing with the number states like 1, 2, 3, etc.

00:33 It would be really beneficial if we use the creation and annihilation operators.

00:40 So a and a dagger.

00:42 And we remember that x can be written, can be expressed in terms of a and a dagger in this manner.

00:52 A plus a dagger times some constant.

00:56 And we also remember how a and a dagger work.

01:02 A acting on n will give us root n, n minus 1.

01:05 And a dagger acting on n will give us root n plus 1 and the state n plus 1.

01:16 So let us carry out this calculation.

01:21 We have this expectation value.

01:23 So on the left, we take the conjugate of the given state.

01:27 We have 1 minus i 2 over root 2 times x.

01:34 Omega times a plus a dagger.

01:42 And we have the state itself over root 2.

01:46 Here we have 1 over 2 square root of h bar over 2m.

01:54 Omega.

01:56 Then 1 minus i 2 times...

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