- #1

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- Homework Statement
- Show that the Hamiltonian operator $$\hat H = -\frac{\hbar}{2m} \frac{d^2}{dx^2} + V(x)$$ is hermitian

- Relevant Equations
- ##<f|\hat H g> = <\hat H f|g>##

$$<f|\hat H g> = \int_{-\infty}^{\infty} f^*\Big(-\frac{\hbar}{2m} \frac{d^2}{dx^2} + V(x) \Big) g dx$$

Integrating (twice) by parts and assuming the potential term is real (AKA ##V(x) = V^*(x)##) we get

$$<f|\hat H g> = -\frac{\hbar}{2m} \Big( f^* \frac{dg}{dx}|_{-\infty}^{\infty} - \frac{df^*}{dx}|_{-\infty}^{\infty} g + \int_{-\infty}^{\infty} \frac{d^2 f}{dx^2}g dx \Big) + \int_{-\infty}^{\infty} V^*(x) f^* g dx $$

In order to get the desired I had to assume that

$$f^* \frac{dg}{dx}|_{-\infty}^{\infty} = 0$$

$$\frac{df^*}{dx}|_{-\infty}^{\infty} g = 0$$

Then we get

$$<f|\hat H g> = -\frac{\hbar}{2m} \int_{-\infty}^{\infty} \frac{d^2 f}{dx^2}g dx + \int_{-\infty}^{\infty} V^*(x) f^* g dx = <\hat H f|g>$$

Checking the solution, they say that these terms indeed vanish 'because both f and g live on Hilbert space'.

But what property of Hilbert space makes this true?

Thanks.

Integrating (twice) by parts and assuming the potential term is real (AKA ##V(x) = V^*(x)##) we get

$$<f|\hat H g> = -\frac{\hbar}{2m} \Big( f^* \frac{dg}{dx}|_{-\infty}^{\infty} - \frac{df^*}{dx}|_{-\infty}^{\infty} g + \int_{-\infty}^{\infty} \frac{d^2 f}{dx^2}g dx \Big) + \int_{-\infty}^{\infty} V^*(x) f^* g dx $$

In order to get the desired I had to assume that

$$f^* \frac{dg}{dx}|_{-\infty}^{\infty} = 0$$

$$\frac{df^*}{dx}|_{-\infty}^{\infty} g = 0$$

Then we get

$$<f|\hat H g> = -\frac{\hbar}{2m} \int_{-\infty}^{\infty} \frac{d^2 f}{dx^2}g dx + \int_{-\infty}^{\infty} V^*(x) f^* g dx = <\hat H f|g>$$

Checking the solution, they say that these terms indeed vanish 'because both f and g live on Hilbert space'.

But what property of Hilbert space makes this true?

Thanks.