10- CHEM-327 Lecture 10: Momentum Operators and Expectation Values
Understanding Eigenfunctions and Operators
If a wave function IS NOT an eigenfunction of the momentum operator:
Measurement Outcome: It is not possible to determine a single, exact value for the momentum. When a measurement is performed, the system will collapse into one of the eigenstates of the momentum operator, and the measured value will be one of the corresponding eigenvalues. The specific value obtained in a single measurement is probabilistic.
Expectation Value: It is always possible to determine an expectation value (or average value) for the momentum. This expectation value represents the average of a large number of measurements performed on identical systems, each prepared in the same non-eigenstate.
If a wave function IS an eigenfunction of the momentum operator:
Measurement Outcome: It is possible to determine a single, exact value for the momentum. This value is the eigenvalue corresponding to that eigenfunction. A measurement will always yield this specific eigenvalue.
Expectation Value: It is also possible to determine an expectation value for the momentum. In this case, the expectation value is simply equal to the single, exact eigenvalue. Since every measurement yields the same value, the average of those values will naturally be that same value.
Expectation Value: Definition and Calculation Methods
Definition
The expectation value \langle A \rangle of an observable associated with an operator \hat{A} for a normalized wave function \psi(x) represents the average outcome of many measurements of that observable on identically prepared systems. It is calculated using the integral form:
\langle A \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{A} \psi(x) \,dx
Momentum Operator
The momentum operator in the position representation is given by:
\hat{p} = -i\hbar \frac{\partial}{\partial x}
Therefore, the expectation value of momentum is:
\langle p \rangle = \int_{-\infty}^{\infty} \psi^*(x) \left(-i\hbar \frac{\partial}{\partial x}\right) \psi(x) \,dx
Calculation Methods
Direct Integration (as shown above): This method uses the operator directly with the wave function.
Using Coefficients ( Cn ) from Superposition: If the wave function \psi(x) can be expressed as a linear superposition of normalized eigenfunctions \phin(x) of the operator \hat{A} (i.e., \psi(x) = \sumn Cn \phin(x) , where \hat{A}\phin = an\phin ), then the expectation value is given by:
\langle A \rangle = \sumn |Cn|^2 an Here, |Cn|^2 represents the probability of measuring the eigenvalue a_n .
Example: Expectation Value for a Specific Wave Function
Consider the wave function:
\psi(x) = \begin{cases} \sin(x) & \text{if } 0 < x < \pi \ 0 & \text{otherwise} \end{cases}
To calculate the expectation value of momentum \langle p \rangle for this wave function, we first need to normalize it.
1. Normalization
We need to find a normalization constant N such that \int{-\infty}^{\infty} |N\psi(x)|^2 \,dx = 1 . \int{0}^{\pi} \sin^2(x) \,dx = \int{0}^{\pi} \frac{1-\cos(2x)}{2} \,dx = \left[ \frac{x}{2} - \frac{\sin(2x)}{4} \right]{0}^{\pi}
= \left( \frac{\pi}{2} - \frac{\sin(2\pi)}{4} \right) - \left( \frac{0}{2} - \frac{\sin(0)}{4} \right) = \frac{\pi}{2} - 0 = \frac{\pi}{2}
So, N^2 \left( \frac{\pi}{2} \right) = 1 \implies N = \sqrt{\frac{2}{\pi}} .
The normalized wave function is thus:
\psi_{normalized}(x) = \begin{cases} \sqrt{\frac{2}{\pi}} \sin(x) & \text{if } 0 < x < \pi \ 0 & \text{otherwise} \end{cases}
2. Expectation Value Calculation
Using the formula for expectation value of momentum:
\langle p \rangle = \int{0}^{\pi} \psi{normalized}^*(x) \left(-i\hbar \frac{\partial}{\partial x}\right) \psi{normalized}(x) \,dx = \int{0}^{\pi} \left(\sqrt{\frac{2}{\pi}} \sin(x)\right) \left(-i\hbar \frac{\partial}{\partial x}\right) \left(\sqrt{\frac{2}{\pi}} \sin(x)\right) \,dx
= \frac{2}{\pi} \int{0}^{\pi} \sin(x) (-i\hbar \cos(x)) \,dx = -i\hbar \frac{2}{\pi} \int{0}^{\pi} \sin(x)\cos(x) \,dx
We can use the substitution method. Let u = \sin(x) , so du = \cos(x)\,dx . When x=0 , u=0 . When x=\pi , u=0 .
= -i\hbar \frac{2}{\pi} \int_{0}^{0} u \,du = 0
Therefore, the expectation value of momentum for this wave function is 0 . This result is characteristic of a standing wave, which represents a particle that is, on average, not moving in a particular direction.
Superposition Principle and Momentum Measurements
Consider a particle whose wave function \psi is a superposition of normalized eigenfunctions of the momentum operator \phi1, \phi2, \phi3 with corresponding eigenvalues p1, p2, p3 :
\psi = c1 \phi1 + c2 \phi2 + c3 \phi3
a) Normalizing \psi
For \psi to be normalized, the sum of the squares of the magnitudes of the coefficients must equal one, assuming the eigenfunctions \phin are orthonormal: |c1|^2 + |c2|^2 + |c3|^2 = 1
This condition ensures that the total probability of finding the particle in any of these states is 1.
b) Possible momentum values and their probabilities
Possible Values: When a momentum measurement is performed on this particle, the only possible values that can be obtained are the eigenvalues corresponding to the eigenfunctions present in the superposition: p1 , p2 , and p_3 .
Probabilities: The probability of measuring each of these values is given by the square of the magnitude of the coefficient corresponding to that eigenfunction (assuming \psi is normalized):
Probability of measuring p1 is P(p1) = |c_1|^2
Probability of measuring p2 is P(p2) = |c_2|^2
Probability of measuring p3 is P(p3) = |c_3|^2
c) Average value for momentum from a large number of measurements
The average value (expectation value) for the momentum that would be obtained from a large number of measurements on identical particles, each prepared in the state \psi , is given by:
\langle p \rangle = |c1|^2 p1 + |c2|^2 p2 + |c3|^2 p3
This is the weighted average of the possible momentum eigenvalues, with the weights being their respective probabilities. This aligns with the second method of expectation value calculation discussed earlier for superposition states.