Question

Are the total energy eigenfunctions for the free particle in one dimension, \(\psi^{+}(x)=A_{+} e^{+i \sqrt{\left(2 m E / \hbar^{2}\right)} x}\) and \(\psi^{-}(x)=A_{-} e^{-i \sqrt{\left(2 m E / \hbar^{2}\right)} x},\) eigenfunctions of the one-dimensional linear momentum operator? If so, what are the eigenvalues?

Short answer

Both wave functions, \(\psi^{+}(x)\) and \(\psi^{-}(x)\), are eigenfunctions of the one-dimensional linear momentum operator, with eigenvalues of \(\sqrt{2mE}\) and \(-\sqrt{2mE}\), respectively.

Step-by-step solution

Wave Function Definitions

$$\psi^{+}(x)=A_{+} e^{+i \sqrt{\left(2 m E / \hbar^{2}\right)} x}$$ $$\psi^{-}(x)=A_{-} e^{-i \sqrt{\left(2 m E / \hbar^{2}\right)} x}$$ Now, we will apply the linear momentum operator on both wave functions.

Applying the Linear Momentum Operator on the First Wave Function

Apply the linear momentum operator on the first wave function, \(\psi^{+}(x)\). $$\hat{P}_x \psi^{+}(x) = -i\hbar\frac{d}{dx}\left(A_{+} e^{+i \sqrt{\left(2 m E / \hbar^{2}\right)} x}\right)$$

Solving the Differential of the First Wave Function

Take the derivative and solve for the expression. $$-i\hbar \frac{d}{dx}\left(A_{+} e^{+i \sqrt{\left(2 m E / \hbar^{2}\right)} x}\right) = -i\hbar A_{+}\left(+i \sqrt{\left(2 m E / \hbar^{2}\right)}\right)e^{+i \sqrt{\left(2 m E / \hbar^{2}\right)} x}$$

Simplify the Expression for the First Wave Function

After simplifying, we get the eigenvalue for the first wave function, \(\psi^{+}(x)\): $$\hat{P}_x \psi^{+}(x) = \sqrt{2mE}A_{+}e^{+i \sqrt{\left(2 m E / \hbar^{2}\right)} x} = \sqrt{2mE}\psi^{+}(x)$$

Applying the Linear Momentum Operator on the Second Wave Function

Now, apply the linear momentum operator on the second wave function, \(\psi^{-}(x)\). $$\hat{P}_x \psi^{-}(x) = -i\hbar\frac{d}{dx}\left(A_{-} e^{-i \sqrt{\left(2 m E / \hbar^{2}\right)} x}\right)$$

Solving the Differential of the Second Wave Function

Take the derivative and solve for the expression. $$-i\hbar \frac{d}{dx}\left(A_{-} e^{-i \sqrt{\left(2 m E / \hbar^{2}\right)} x}\right) = -i\hbar A_{-}\left(-i \sqrt{\left(2 m E / \hbar^{2}\right)}\right)e^{-i \sqrt{\left(2 m E / \hbar^{2}\right)} x}$$

Simplify the Expression for the Second Wave Function

After simplifying, we get the eigenvalue for the second wave function, \(\psi^{-}(x)\): $$\hat{P}_x \psi^{-}(x) = -\sqrt{2mE}A_{-}e^{-i \sqrt{\left(2 m E / \hbar^{2}\right)} x} = -\sqrt{2mE}\psi^{-}(x)$$ Now, we can state our conclusions.

Conclusion

Both wave functions, \(\psi^{+}(x)\) and \(\psi^{-}(x)\), are eigenfunctions of the one-dimensional linear momentum operator, with eigenvalues of \(\sqrt{2mE}\) and \(-\sqrt{2mE}\), respectively.

Key concepts

Eigenfunctions

In quantum mechanics, eigenfunctions are crucial as they are solutions to the Schrödinger equation, which describe the states of a quantum system. An eigenfunction, when operated on by a particular operator, yields a constant (the eigenvalue) multiplied by the original function.

• In the context of this problem, the wave functions \(\psi^{+}(x)\) and \(\psi^{-}(x)\) are being tested to see if they are eigenfunctions of the linear momentum operator.
• If indeed they are, applying the operator will not change their form except for a multiplicative constant (eigenvalue).

Understanding eigenfunctions helps us interpret physical properties of a system, like energy and momentum, in a more manageable way.

Linear Momentum Operator

The linear momentum operator in quantum mechanics represents momentum, a fundamental property of particles. It is denoted as \(\hat{P}_x\) for the one-dimensional case and is defined as:

\[ \hat{P}_x = -i\hbar \frac{d}{dx} \]

• This operator acts on the wave function to extract information about the momentum of the particle described by the wave function.
• In our exercise, applying this operator to the wave functions \(\psi^{+}(x)\) and \(\psi^{-}(x)\) confirms if they are eigenfunctions and reveals the eigenvalues, which are related to the momentum of the particle.

The linear momentum operator is central in quantum mechanics to explore how particles move and behave.

Wave Functions

Wave functions \(\psi(x)\) are key to understanding the quantum state of a particle, containing all possible information about the system.

• They are generally complex-valued functions and can be used to calculate probabilities of a particle's position or momentum.
• In this scenario, the wave functions \(\psi^{+}(x)\) and \(\psi^{-}(x)\) describe a particle's state in a free particle model, with the exponentials indicating oscillatory behavior.

Wave functions bridge the gap between the conceptual and measurable properties of quantum systems, enabling detailed predictions about physical phenomena.

Free Particle Model

The free particle model describes a quantum particle that is not influenced by external forces or potential fields. This model is fundamental to understanding how particles behave when not bound by external constraints.

• In one dimension, the solution to the Schrödinger equation in this model is an exponential function, reflecting the particle’s wave nature.
• The functions \(\psi^{+}(x)\) and \(\psi^{-}(x)\) seen here are examples of such wave functions, showing the particle propagating in space.

This model is essential for building intuition about particle behavior and is a stepping stone to more complex systems in quantum mechanics.