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 $\langle \Psi \mid \hat{x}^{2} \mid \Psi \rangle$:

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


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

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.
  • $\begingroup$ Hey Dima, please, please, please do me a big favour and write ‘Heisenberg’ rather than ‘Heizenberg’ (even though I don’t have the time to provide a proper answer right now), as the latter really hurts :\ $\endgroup$
    – Claudius
    Dec 8, 2012 at 15:49

2 Answers 2


I did not follow your calculations in detail, but I might be able to address your summed-up questions (excuse the sloppy notation of leaving the hats off):

Firstly, $\langle \Psi | x(0)p(0) + p(0)x(0) | \Psi \rangle$:

The calculation I originally put here had a sign error ($xp = i\hbar - px$ rather than the correct $xp = i\hbar +px$), sorry. I fear I cannot help with this problem.

Secondly, $\langle \Psi | x | \Psi \rangle = 0$:

Recall the original Hamiltonian, which is symmetric in $x$. Furthermore, your initial state $\Psi(0)$ is also symmetric. The expectation value of the position operator being $0$ is hence no surprise, as there is an equal chance that the system is in state $x_0$ as that it is in state $-x_0$. If you write the expectation value of the position operator as

$$ \langle x \rangle = \sum_x x\cdot p(x) \quad,$$

for discrete $x$ or similarly as an integral for continuous $x$, you can see that if $p(x) = p(-x)$, the summand $-x_0 p(-x_0)$ cancels with $x_0 p(x_0)$, leaving $\langle x \rangle \equiv 0$.

Thirdly, $\langle \Psi | x^2(t \to \infty) | \Psi \rangle \to \infty$:

The variance $\langle x^2 \rangle$ gives you a measure of the uncertainty in the expectation value of $x$, that is, how much the actual position will differ from the expectation value. If you look at the Hamiltonian, you can see that the term containing $x^2$ has a negative sign, i.e. energy is minimised for $|x| \to \infty$.

Classically speaking, the Hamiltonian describes a potential with a maximum at $x = 0$, the particle then "rolls off" either in the $+x$ or $-x$ direction. If you give the particle an arbitrarily long time to roll down this potential, it will get arbitrarily far away from the initial position at $x = 0$, either in the positive or negative direction. That means that (mathematicians, please stop reading :)) for $t = \infty$, the particle is either at $x = \infty$ or $x = -\infty$. Both positions are inifitely far away from the average ($x=0$), the variance is hence infinite, too.

This argument carries over to the quantum mechanical case and hopefully clears up your confusion. :)

  • $\begingroup$ You’re more than welcome, answering questions makes me revise everything I’ve learnt so far :-) (Just don’t ever write Heizenberg again :P) $\endgroup$
    – Claudius
    Dec 8, 2012 at 21:13

Claudius has appropriately addressed your second and third questions. I'll deal with the first. For strictly real $\langle x|\Psi\rangle$, you have $$ \langle \Psi|xp+px|\Psi\rangle =\langle \Psi|xp|\Psi\rangle+\langle \Psi|px|\Psi\rangle =\mathrm{Re}\left[\int \Psi^\ast x p \Psi \mathrm{d}x\right] =\mathrm{Re}\left[-i\hbar\int \Psi x \frac{\mathrm{d} \Psi}{\mathrm{d}x} \mathrm{d}x\right]. $$ Since $\Psi$ is purely real its derivative is purely real and the whole integral is a real number. The correlation $\langle \Psi|xp+px|\Psi\rangle$ is then the real part of a purely imaginary number and therefore zero.

  • 2
    $\begingroup$ It might be helpful to add that this relies on $x$ and $p$ being hermitian $\left(x^\dagger = x \; ;\; p^\dagger = p\right)$, as otherwise $\left( \langle \Psi | x p | \Psi \rangle \right)^\star = \langle \Psi | p^\dagger x^\dagger | \Psi \rangle$. $\endgroup$
    – Claudius
    Dec 8, 2012 at 17:52

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