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Solve only 3rd1. Consider the following QM system. A particle in a QM system has wavefunction

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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:

Solve only 3rd1. Consider the following QM system. A particle in a QM system has wavefunction 't' that
is not an cigenfunction of the Energy Operator, H . The wave function '¥, however, can
be expanded as a superposition of Eigenfunctions of the Energy operator represented by
Wi, v2 and ys, and these Eigenfunctions have Eigenvalues Ey, Ey, and E; respectively of
the Energy operator. [This means H yi = Ejyi. and Hy: = Eaya, H ys = Eas]
Specifically, W can be written as ¥ = ciy) + cows + cays, where the Eigenfunctions yi,
wy and y; are Orthonormal (means?), and c;, ¢; and c; arc the Expansion Coefficients.
To reduce the complexity of the problem, assume ¥ is Normalized. Providing a brief
rationale for your answer, what is the probability of getting E1 as the energy value for
this QM state W? [Hint: A similar example was looked at in class. While this should be
a very straight forward question, do not just guess an answer!]. Section #5. Some
Details Required in providing a rationale.

2. For the particle in the system defined in Question 1, determine the average Ener

VIPSolutions · Accepted Answer

Certainly! Let's solve **Question 3** step-by-step as requested.

---

# 3. **Standard Deviation in the Measurement of Energy**

Given:
- The wavefunction $\Psi$ is a superposition of orthonormal eigenfunctions $\psi_1, \psi_2, \psi_3$ of the Hamiltonian $\hat{H}$ with eigenvalues $E_1, E_2, E_3$.
- Expansion: $\Psi = c_1\psi_1 + c_2\psi_2 + c_3\psi_3$ with $|c_1|^2 + |c_2|^2 + |c_3|^2 = 1$ (normalized).

We are to find the **standard deviation** $\sigma_E$ of energy measurements.

---

## **Step 1: Write the Formula**

The standard deviation is:
$$
\sigma_E = \sqrt{\langle E^2 \rangle - \langle E \rangle^2}
$$

---

## **Step 2: Compute $\langle E \rangle$ (Average Energy)**

$$
\langle E \rangle = \langle \Psi | \hat{H} | \Psi \rangle
$$

Expand using the basis:

$$
= \left(c_1^*\langle \psi_1| + c_2^*\langle \psi_2| + c_3^*\langle \psi_3|\right) \hat{H} \left(c_1\psi_1 + c_2\psi_2 + c_3\psi_3\right)
$$

Since the eigenfunctions are orthonormal and eigenstates:
$$
\hat{H}\psi_i = E_i\psi_i;\quad \langle \psi_i | \psi_j \rangle = \delta_{ij}
$$

Only diagonal terms survive:
$$
\langle E \rangle = |c_1|^2 E_1 + |c_2|^2 E_2 + |c_3|^2 E_3
$$

---

## **Step 3: Compute $\langle E^2 \rangle$**

$$
\langle E^2 \rangle = \langle \Psi | \hat{H}^2 | \Psi \rangle
$$

Since $\hat{H}^2 \psi_i = E_i^2 \psi_i$:
$$
\langle E^2 \rangle = |c_1|^2 E_1^2 + |c_2|^2 E_2^2 + |c_3|^2 E_3^2
$$

---

## **Step 4: Find the Standard Deviation**

$$
\sigma_E = \sqrt{\langle E^2 \rangle - \langle E \rangle^2}
$$

Substitute the results:

- $\langle E \rangle = |c_1|^2 E_1 + |c_2|^2 E_2 + |c_3|^2 E_3$
- $\langle E^2 \rangle = |c_1|^2 E_1^2 + |c_2|^2 E_2^2 + |c_3|^2 E_3^2$

So,

$$
\boxed{
\sigma_E = \sqrt{
\left(
|c_1|^2 E_1^2 + |c_2|^2 E_2^2 + |c_3|^2 E_3^2
\right)
-
\left(
|c_1|^2 E_1 + |c_2|^2 E_2 + |c_3|^2 E_3
\right)^2
}
}
$$

---

### **Summary**

1. **Compute the weighted average of energies** (expectation value): $\langle E \rangle$.
2. **Compute the weighted average of squared energies**: $\langle E^2 \rangle$.
3. **Plug into the standard deviation formula** above.

---

**This is the full step-by-step solution for the standard deviation of energy in the given quantum system!**

Let me know if you'd like a numerical example or further clarification on any step.

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