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- Homework Statement
- (d) Show that ##\hat H## Hermitian for a particle in one dimension.

- Relevant Equations
- ##\hat H##= ##\frac{\hat p^2}{2m} +V(x)##

##\hat p^\dagger=\hat p##

I need help with part d of this problem. I believe I completed the rest correctly, but am including them for context

(a)Show that the hermitian conjugate of the hermitian conjugate of any operator ##\hat A## is itself, i.e. ##(\hat A^\dagger)^\dagger##

(b)Consider an arbitrary operator ##\hat A##. Show that ##\hat A^\dagger \hat A##, a Hermitian operator, and its expectation value satisfies ##\langle \hat A^\dagger \hat A \rangle \geq 0##

(c) Show that ˆx is Hermitian for a particle in one dimension.

Attempt

Using ##\hat H##= ##\frac{\hat p^2}{2m} +V(x)##

##\langle f~|\frac{\hat p^2}{2m} +V(x)~g \rangle##

##=\langle f~|\frac{\hat p^2}{2m} ~g \rangle~+\langle f~|V(x)~g \rangle##

##=\frac{1}{2m}~\langle f~|\hat p^2 ~g \rangle~+\langle f~|V(x)~g \rangle##

##=\frac{1}{2m}~\langle (\hat p^2)^\dagger f~|~g \rangle~+\langle\mathbf {V^*(x)} f~|~g \rangle## Is it wrong to write V(x) like this? Since it isn't an operator, I assumed it might act like a constant

##=\frac{1}{2m}~\langle \hat p^2 f~|~g \rangle~+\langle V^*(x) f~|~g \rangle## Since ##\hat p## is hermitian ##(\hat p^2)^\dagger=\hat p^2##

##=\frac{1}{2m}~\langle g~|~\hat p^2 f \rangle^*~+\langle g~|~V^*(x) f \rangle^*##

##=(\langle g|\frac{\hat p^2}{2m}+ V^*(x) f \rangle^*)^*##

##=\langle \frac{\hat p^2}{2m}+ V^*(x) f|g \rangle##

Is this the correct approach? If it is, what do I do with my ##V^*(x)##

(a)Show that the hermitian conjugate of the hermitian conjugate of any operator ##\hat A## is itself, i.e. ##(\hat A^\dagger)^\dagger##

(b)Consider an arbitrary operator ##\hat A##. Show that ##\hat A^\dagger \hat A##, a Hermitian operator, and its expectation value satisfies ##\langle \hat A^\dagger \hat A \rangle \geq 0##

(c) Show that ˆx is Hermitian for a particle in one dimension.

**(d) Show that ##\hat H## is Hermitian for a particle in one dimension.**Attempt

Using ##\hat H##= ##\frac{\hat p^2}{2m} +V(x)##

##\langle f~|\frac{\hat p^2}{2m} +V(x)~g \rangle##

##=\langle f~|\frac{\hat p^2}{2m} ~g \rangle~+\langle f~|V(x)~g \rangle##

##=\frac{1}{2m}~\langle f~|\hat p^2 ~g \rangle~+\langle f~|V(x)~g \rangle##

##=\frac{1}{2m}~\langle (\hat p^2)^\dagger f~|~g \rangle~+\langle\mathbf {V^*(x)} f~|~g \rangle## Is it wrong to write V(x) like this? Since it isn't an operator, I assumed it might act like a constant

##=\frac{1}{2m}~\langle \hat p^2 f~|~g \rangle~+\langle V^*(x) f~|~g \rangle## Since ##\hat p## is hermitian ##(\hat p^2)^\dagger=\hat p^2##

##=\frac{1}{2m}~\langle g~|~\hat p^2 f \rangle^*~+\langle g~|~V^*(x) f \rangle^*##

##=(\langle g|\frac{\hat p^2}{2m}+ V^*(x) f \rangle^*)^*##

##=\langle \frac{\hat p^2}{2m}+ V^*(x) f|g \rangle##

Is this the correct approach? If it is, what do I do with my ##V^*(x)##