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Let's go through the given statement step by step and construct a proof.

### Problem:

We are given that \( V(x) \) is an even function, meaning:
\[
V(-x) = V(x)
\]
We are asked to prove that if \( \psi(x) \) is a solution to a certain equation (e.g., a Schrödinger-like equation) for a given energy \( E \), then we can take \( \psi(x) \) to be either even or odd. The hint provided states that if \( \psi(x) \) satisfies the equation for \( E \), then \( \psi(-x) \) also satisfies the same equation. This leads to the conclusion that the linear combinations \( \psi(x) \pm \psi(-x) \) are even and odd functions, respectively.

### Proof Outline:

1. **Even Potential \( V(x) \):**
   \[
   V(-x) = V(x)
   \]
   This means the potential is symmetric around \( x = 0 \).

2. **Schrödinger Equation (or a similar equation):**
   Let \( \psi(x) \) be the solution of an equation like:
   \[
   H \psi(x) = E \psi(x)
   \]
   where \( H \) is some Hamiltonian operator that typically involves the potential \( V(x) \). The explicit form might be:
   \[
   -\frac{d^2}{dx^2} \psi(x) + V(x) \psi(x) = E \psi(x)
   \]
   Since \( V(x) \) is even, this equation holds for both \( \psi(x) \) and \( \psi(-x) \).

3. **Symmetry of the Equation:**
   Let’s evaluate what happens if we replace \( x \) with \( -x \). Applying this transformation, we get:
   \[
   -\frac{d^2}{dx^2} \psi(-x) + V(-x) \psi(-x) = E \psi(-x)
   \]
   Since \( V(-x) = V(x) \), this simplifies to:
   \[
   -\frac{d^2}{dx^2} \psi(-x) + V(x) \psi(-x) = E \psi(-x)
   \]
   This is the same equation as for \( \psi(x) \). Therefore, \( \psi(-x) \) is also a solution for the same energy \( E \).

4. **Constructing Even and Odd Solutions:**
   Now, we define two new functions:
   - The **even** linear combination:
     \[
     \psi_{\text{even}}(x) = \frac{1}{2} \left( \psi(x) + \psi(-x) \right)
     \]
     - The **odd** linear combination:
     \[
     \psi_{\text{odd}}(x) = \frac{1}{2} \left( \psi(x) - \psi(-x) \right)
     \]

   These functions satisfy the following properties:
   - \( \psi_{\text{even}}(-x) = \psi_{\text{even}}(x) \), so it is an even function.
   - \( \psi_{\text{odd}}(-x) = -\psi_{\text{odd}}(x) \), so it is an odd function.

5. **Final Step:**
   Since both \( \psi_{\text{even}}(x) \) and \( \psi_{\text{odd}}(x) \) are valid solutions to the original equation with the same energy \( E \), the solution \( \psi(x) \) can always be decomposed into even and odd parts. Therefore, we can always take the solution to be either even or odd, as required.

### Conclusion:
We have shown that for an even potential \( V(x) \), if \( \psi(x) \) is a solution, then both \( \psi(x) \) and \( \psi(-x) \) are solutions for the same energy. Therefore, the linear combinations \( \psi(x) \pm \psi(-x) \) give even and odd solutions, proving that \( \psi(x) \) can always be taken to be either even or odd.

Would you like more details on any step?

### 5 related questions:
1. How do even and odd functions differ in terms of their symmetry properties?
2. What types of physical problems often involve potentials that are even functions?
3. Can the decomposition into even and odd functions be extended to higher dimensions?
4. How does the parity of the potential function influence the selection of solutions in quantum mechanics?
5. Why is it useful to consider both even and odd solutions in physical problems?

### Tip:
When working with differential equations that involve symmetry, always check if you can exploit even-odd symmetry to simplify the problem or reduce the number of solutions you need to consider.

If V(x) is an even function (that is, V(-x) = V(x)), prove that ψ(x) can be either even or odd.

Learn how to prove that the wavefunction ψ(x) can be either even or odd if V(x) is an even function. This solution explores the symmetry in the Schrödinger equation, using even and odd function combinations. Suitable for undergraduate students studying quantum mechanics or mathematical physics.

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