Saturday, 19 September 2020

"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.




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...