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By Ehrenfest Theorem, we know that $$\dfrac{\mathrm d \langle \Omega \rangle}{\mathrm dt} =\langle[H,\Omega]\rangle, $$ where $\Omega$ is an operator and $H$ is the Quantum Hamiltonian.

I would like to know what is wrong in the following steps:

\begin{equation} \frac{\mathrm d \langle \Omega \rangle}{\mathrm dt} := \frac{\mathrm d\langle \psi | \Omega|\psi \rangle}{\mathrm dt} = \left(\frac{\partial\langle\psi| }{\partial t} \right)\Omega|\psi\rangle + \langle \psi |\frac{\partial(\Omega|\psi\rangle|)}{\partial t} \tag{1} \end{equation} Using product rule, \begin{equation} \frac{\mathrm d\langle \psi | \Omega|\psi \rangle}{\mathrm dt} = \left(\frac{\partial\langle\psi| }{\partial t} \right)\Omega|\psi\rangle + \langle \psi |\frac{\partial(\Omega|\psi\rangle|)}{\partial t} \tag{2} \end{equation} Schroedinger's equation states:

\begin{equation} \frac{\partial|\psi \rangle}{\partial t} = \frac{-i H|\psi\rangle }{\hbar}\tag{3} \end{equation}

So taking the hermitian conjugation (and since the Hamiltonian is hermitian): \begin{equation} \frac{\partial\langle\psi| }{\partial t} = \frac{i \langle\psi|H }{\hbar} \tag{4} \end{equation} Now applying the time dependent Schrodinger equation to the state $\Omega|\psi\rangle$ (This can be done as $\Omega|\psi\rangle$ is also a state in the function space),

\begin{equation} \frac{\partial\Omega|\psi \rangle}{\partial t} = \frac{-i H\Omega|\psi\rangle }{\hbar}\tag{5} \end{equation} Thus, applying equations 4,5 to equation 2, \begin{equation} \frac{\mathrm d\langle \psi | \Omega|\psi \rangle}{\mathrm dt} = \left( \frac{i \langle\psi|H }{\hbar} \right)\Omega|\psi\rangle + \langle \psi| \frac{-i }{\hbar}H\Omega|\psi\rangle\tag{5} \end{equation} As we can observe this gives

\begin{equation} \frac{\mathrm d\langle \psi | \Omega|\psi \rangle}{\mathrm dt} = \frac{i}{\hbar}\langle\psi|H \Omega|\psi\rangle - \frac{i}{\hbar}\langle\psi|H \Omega|\psi\rangle = 0 \end{equation} This goes against Ehrefest's theorem as $\Omega$ was a general operator. I suspect that one of my steps has succinctly assumed something but I cant figure out what it is.

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Assume that $\Omega$ is not explicitly dependent on time. With this your (5) becomes $$ \frac{\partial}{\partial t}\Omega\vert\psi\rangle = \Omega\frac{\partial}{\partial t}\vert\psi\rangle=-\frac{i\Omega H}{\hbar} \vert\psi\rangle \tag{6} $$ Combining your (4) and my (6) will produce the required result.

More specifically, the error is that, if $\vert\psi\rangle$ is a solution then $\Omega\vert\psi\rangle$ is usually not a solution. To see this consider $$ \vert\Psi(t)\rangle= a_0e^{-i \omega t/2} \vert 0\rangle +a_1 e^{-i 3\omega t/2} \vert 1\rangle $$ with $\vert n\rangle$ the $n$'th harmonic oscillator ket. Then clearly \begin{align} i\hbar \frac{d}{dt}\vert\Psi(t)\rangle &= \frac{\hbar\omega}{2} a_0e^{-i \omega t/2} \vert 0\rangle +\frac{3\hbar\omega}{2} a_1e^{-i 3\omega t/2} \vert 1\rangle \, ,\\ &=\hat H\vert\Psi(t)\rangle \end{align} but if $\hat X=a+a^\dagger$ then \begin{align} \hat X\vert\Psi(t)\rangle&= e^{-i 3\omega t/2} a_1\vert 0\rangle +a_0e^{-i \omega t/2} \vert 1\rangle +\sqrt{2}a_1e^{-i 3\omega t/2} \vert 2\rangle\, ,\\ \hat H\hat X\vert\Psi(t)\rangle&= e^{-i 3\omega t/2} a_1\textstyle\frac{1}{2}\hbar\omega\vert 0\rangle + \frac{3}{2}\hbar\omega a_0e^{-i \omega t/2} \vert 1\rangle + \frac{5}{2}\hbar\omega\sqrt{2}a_1e^{-i 3\omega t/2} \vert 2\rangle\\ \end{align} whereas \begin{align} i\hbar\frac{d}{dt}\hat X\vert\Psi(t)\rangle&= \textstyle\frac{3}{2}\hbar\omega a_1e^{-i 3\omega t/2}\vert 0\rangle + \frac{1}{2}\hbar a_0e^{-i \omega t/2} \vert 1\rangle +\frac{3\hbar\omega}{2}\sqrt{2}a_1e^{-i 3\omega t/2} \vert 2\rangle\, . \end{align}


Edit: In general start with \begin{align} \vert\Psi(t)\rangle&=\sum_n \vert n\rangle c_n e^{-iE_nt/\hbar} \, ,\\ \Omega\vert\Psi(t)\rangle&= \sum_{nm} \vert m\rangle c_n \Omega_{mn} e^{-iE_nt/\hbar}\, , \tag{1} \end{align} where $\Omega_{mn}=\langle m\vert \Omega \vert n\rangle$.

Now, start from (1) and compare \begin{align} i\hbar\frac{d}{dt}\Omega \vert\Psi(t)\rangle &= \sum_{nm} \vert m\rangle c_n \Omega_{mn} E_n e^{-iE_n t/\hbar} \, , \tag{2}\\ H\Omega\vert\Psi(t)\rangle&=\sum_{mn} \vert m\rangle c_n \Omega_{mn} E_m e^{-iE_nt/\hbar} \tag{3} \end{align} and you can see the difference between (2) and (3) comes because $H$ “sees” $\vert m\rangle$ and returns a factor of $E_m$ but the derivative w/r to $t$ “sees” the factor $e^{-iE_nt/\hbar}$ and so returns a factor of $E_n$ instead.

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  • $\begingroup$ Thank you! I understand the example, But why is it that $$\Omega|\psi\rangle$$ is not a solution to the time dependent Schrodinger's equation? could you provide a reference please? $\endgroup$ – Sparsh Mishra Dec 30 '17 at 4:42
  • $\begingroup$ @SparshMishra There is no reason to believe it would be a solution. Of course if $\Omega$ and $H$ commute it’s another matter. $\endgroup$ – ZeroTheHero Dec 30 '17 at 5:06
  • $\begingroup$ @SparshMishra I’ve added a bit so you can see why in general it doesn’t work out. $\endgroup$ – ZeroTheHero Dec 30 '17 at 5:25
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The problem is Eq. (5). The TDSE's solutions don't include all vectors in the Hilbert space. If you write $|\psi\rangle$ as a superposition of the Hamiltonian's eigenstates, you can prove (5) provided the coefficients are time-independent, which isn't true in general.

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You have an error in (2) and (5). In (2), you are taking the time derivative of the operator acting on the state ket. When taking this time derivative, think of the product rule for differentiation. You'll end up with the partial time derivative of the operator and the operator acting on the partial time derivative of the state ket.

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  • $\begingroup$ Yeah I understand that, but what is wrong with the method that I posted? $\endgroup$ – Sparsh Mishra Dec 30 '17 at 4:20

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