"Inverted" quantum oscillator

I'm trying to understand the problem of the "inverted" oscillator, which has the following Hamiltonian:

$$\hat{H}=\frac{\hat{p}^{2}}{2m}-\frac{k\hat{x}^{2}}{2}$$

Suppose that a particle at the initial time is in a state

$$\Psi(x)=\frac{1}{(\pi b^{2})^{\frac{1}{4}}}\exp^{-x^{2}/2b^{2}}$$

We want to find $\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle$ and $\langle \Psi \mid \hat{x}^{2}(t) \mid \Psi \rangle$ and their asymptotic when $t \rightarrow +\infty$.

Heisenberg equation for operator $\hat{x}$ (we use that $\hat{x}$ does not depend explicitly on time, $[\hat{x},\hat{x}]=0$, $[\hat{x},\hat{p}]=i\hbar$ and Leibniz rule:$[A,BC]=[A,B]C+B[A,C]$):

$$i\hbar \dot{\hat{x}}=\left[\hat{x},\hat{H}\right]=\left[\hat{x},\frac{\hat{p}^{2}}{2m}-\frac{k\hat{x}^{2}}{2}\right]=\left[\hat{x},\frac{\hat{p}^{2}}{2m}\right]=\frac{\hat{p}}{m}\left[\hat{x},\hat{p}\right]=i\hbar\frac{\hat{p}}{m}$$

$$\dot{\hat{x}}=\hat{p}/m$$

Similarly, Heisenberg equation for operator $\hat{p}$:

$$i\hbar\dot{\hat{p}}=\left[\hat{p},\hat{H}\right]=\left[\hat{p},\frac{\hat{p}^{2}}{2m}-\frac{k\hat{x}^{2}}{2}\right]=-\left[\hat{p},\frac{k\hat{x}^{2}}{2}\right]=-k\hat{x}\left[\hat{p},\hat{x}\right]=i\hbar k\hat{x}$$

$$\dot{\hat{p}}=k\hat{x}$$

So, we have following set of equations: $\dot{\hat{x}}=\hat{p}/m$; $\dot{\hat{p}}=k\hat{x}$.

Using them, we can write:

$$m\ddot{\hat{x}}=k\hat{x}$$

Solution of this differential equation is $\hat{x}=A\cosh(\omega t)+B\sinh(\omega t)$, where $\omega=\sqrt{\frac{k}{m}}$.

Initial conditions: $A=\hat{x}(0)$; $B=\frac{\dot{\hat{x}}(0)}{\omega}$. Remember that $\dot{\hat{x}}=\hat{p}/m$. Then $B=\frac{\hat{p}(0)}{m\omega}$.

So we have

$$\hat{x}=\hat{x}(0)\cosh(\omega t)+\frac{\hat{p}(0)}{m\omega}\sinh(\omega t)$$

Then

$$\hat{x}^{2}=\hat{x}^{2}(0)\cosh^{2}(0)+\frac{\hat{p}^{2}(0)}{(m\omega)^{2}}\sinh^{2}(0)+\frac{1}{m\omega}(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\sinh(\omega t)\cosh(\omega t)$$

Now we want to calculate $\langle\Psi\mid\hat{x}\mid\Psi\rangle$ and $\langle \Psi \mid \hat{x}^{2} \mid \Psi \rangle$:

$$\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=\langle\Psi\mid\hat{x}(0)\cosh(\omega t)+\frac{\hat{p}(0)}{m\omega}\sinh(\omega t)\mid\Psi\rangle=\langle\Psi\mid\hat{x}(0)\mid\Psi\rangle\cosh(\omega t)+\langle\Psi\mid\hat{p}(0)\mid\Psi\rangle\frac{\sinh(\omega t)}{m\omega}$$

Similarly

$$\langle \Psi \mid \hat{x}^{2}(t) \mid \Psi \rangle=\langle\Psi\mid\hat{x}^{2}(0)\mid\Psi\rangle\cosh^{2}(\omega t)+\langle\Psi\mid\hat{p}^2(0)\mid\Psi\rangle\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}}+\frac{1}{m\omega}\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle\sinh(\omega t)\cosh(\omega t)$$

Lets introduce the notation:$x_{0}=\langle\Psi\mid\hat{x}(0)\mid\Psi\rangle$, $p_{0}=\langle\Psi\mid\hat{p}(0)\mid\Psi\rangle$,$x_{0}^{2}=\langle\Psi\mid\hat{x}^{2}(0)\mid\Psi\rangle$, $p_{0}^{2}=\langle\Psi\mid\hat{p}^{2}(0)\mid\Psi\rangle$. Then:

$$\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=x_{0}\cosh(\omega t)+\frac{p_{0}}{m\omega}\sinh(\omega t)$$

$$\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=x_{0}^{2}\cosh^{2}(\omega t)+p_{0}^{2}\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}}+\frac{1}{m\omega}\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle\sinh(\omega t)\cosh(\omega t)$$

Here is the first big trouble: i've found some lecture notes in the internet, where they say that

$\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle=0$ for real $\Psi(x)$. I tried to prove this statement, but i can't. Can you explain why it is equal to zero?

Well, suppose that this statement is true. Then:

$$\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=x_{0}\cosh(\omega t)+\frac{p_{0}}{m\omega}\sinh(\omega t)$$

$$\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=x_{0}^{2}\cosh^{2}(\omega t)+p_{0}^{2}\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}}$$

Thus, we expressed values through their meanings at the initial time. So, lets find $x_{0}$, $p_{0}$, $x_{0}^{2}$, $p_{0}^{2}$. From this point we will work in the coordinate representation.

$$x_{0}=\langle\Psi\mid\hat{x}\mid\Psi\rangle=\int\Psi^{\ast}x\Psi dx=\int\Psi x\Psi dx=\int\Psi^{2}xdx$$

Remember that $\Psi(x)=\frac{1}{(\pi b^{2})^{\frac{1}{4}}}\exp^{-x^{2}/2b^{2}}$ :

$$x_{0}=\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\int x\exp^{-x^{2}/b^{2}}dx$$ We integrate in domains from minus infinity to plus infinity. But this integral is equal to zero! So $$x_{0}=0$$ Similarly: $$p_{0}=\langle\Psi\mid\hat{p}\mid\Psi\rangle=-i\hbar\int\Psi^{\ast}\frac{\partial}{\partial x}\Psi dx=\frac{2}{b^{2}}i\hbar\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\int x\exp^{-x^{2}/b^{2}}dx$$ The domains of integration are the same:from minus infinity to plus infinity. But this is also is equal to zero! So, we have : $$\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=0\cosh(\omega t)+\frac{0}{m\omega}\sinh(\omega t)=0$$ I'm confused by this result! Is there a mistake? If not, can you explain me this result?

Now we are going to calculate $x_{0}^{2}$ and $p_{0}^{2}$:

$$x_{0}^{2}=\langle\Psi\mid\hat{x}^{2}\mid\Psi\rangle=\int\Psi^{\ast}x\Psi dx=\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\int x^{2}\exp^{-x^{2}/b^{2}}dx=\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\frac{\sqrt{\pi} (b^{2})^{\frac{3}{2}}}{2}=\frac{b^2}{2}$$

$$p_{0}^{2}=\langle\Psi\mid\hat{p}^{2}\mid\Psi\rangle=-\hbar^{2}\int\Psi^{\ast}\frac{\partial^{2}}{\partial x^{2}}\Psi dx=\frac{\hbar^{2}}{2b^{2}}$$

Finally:

$$\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=\frac{b^2}{2}\cosh^{2}(\omega t)+\frac{\hbar^2}{2b^{2}}\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}}$$

Using the equation $\cosh^{2}(x)-\sinh^{2}(x)=1$ we get:

$$\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=\frac{1}{2}\left(b^{2}+\frac{\hbar^{2}}{b^{2}(m\omega)^{2}}\right)\cosh^{2}(\omega t)-\frac{\hbar^{2}}{2b^{2}(m\omega)^{2}}$$

Thus we see that when $t \rightarrow +\infty$ dispersion of coordinate $\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle$ tends to infinity.

To sum up, I have these several difficulties:

1. why $\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle=0$ for real $\Psi(x)$?


2. why $\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=0$?



3. why when $t \rightarrow +\infty$ $\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle$ tends to infinity? Also i would be happy if you provide some articles(or books) where i can read about it.