Question

Challenge about quantum mechanics

Answer 1

starting from this condition:
\[\Large \frac{d}{{dt}}{\int_{ - \infty }^{ + \infty } {\left| {\Psi \left( x \right)} \right|} ^2}dx = 0\]
please show that the \(Hamiltonian\) operator:
\[\Large H \equiv i\hbar \frac{\partial }{{\partial t}}\]
is \(hermitian\)

Answer 2

Doesn't \(\ {\int_{ - \infty }^{ + \infty } {\left| {\Psi \left( x \right)} \right|} ^2}dx=1\) ?

so I do not understand the question.

Answer 3

hint:
we can write this:
\[\Large \begin{gathered}
0 = \frac{d}{{dt}}\int_{ - \infty }^{ + \infty } {{{\left| {\Psi \left( x \right)} \right|}^2}} dx = \int_{ - \infty }^{ + \infty } {\frac{{\partial {{\left| {\Psi \left( x \right)} \right|}^2}}}{{\partial t}}} dx = \hfill \\
\hfill \\
= \int_{ - \infty }^{ + \infty } {\frac{{\partial \left( {{\Psi ^*}\Psi } \right)}}{{\partial t}}} dx = ... \hfill \\
\end{gathered} \]
now, the \(Schroedinger\) equation, can be useful:
\[\Large i\hbar \frac{{\partial \Psi }}{{\partial t}} = H\Psi \]
@ganeshie8 @IrishBoy123 @Astrophysics @Empty @BAdhi

Answer 4

I had fun thinking about this! I think there was a lot of similarity to that matrix problem we did the other day so this didn't take too long to find! :D

$$\frac{d}{dt} \int \Psi^* \Psi dx = 0 $$
$$\int \frac{\partial }{\partial t} (\Psi^* \Psi ) dx = 0$$
$$\int \frac{\partial \Psi^* }{\partial t} \Psi dx +\int \Psi^*\frac{\partial \Psi }{\partial t} dx = 0$$
$$-\int \frac{\partial \Psi^* }{\partial t} \Psi dx =\int \Psi^*\frac{\partial \Psi }{\partial t} dx $$
$$-i \hbar \int \frac{\partial \Psi^* }{\partial t} \Psi dx =i \hbar \int \Psi^*\frac{\partial \Psi }{\partial t} dx $$
$$\left( \int \Psi^* \left(i \hbar \frac{\partial}{\partial t} \right) \Psi dx \right)^* =\int \Psi^* \left(i \hbar \frac{\partial}{\partial t} \right) \Psi dx $$
$$\left( \int \Psi^* H \Psi dx \right)^* =\int \Psi^* H \Psi dx $$

Answer 5

correct!!! :)

Answer 6

I was originally thinking about trying to do something clever with modifying the derivation of this equation, but didn't get very far:

\[\frac{d \langle \hat A \rangle}{dt} = \langle [ \hat H, \hat A ] \rangle \]

Since commuting with the Hamiltonian implies it's constant over time.

Answer 7

your procedure is the only that we can apply to solve this problem
nice job! :)