##H=i\hbar\frac{d}{dt}## is weird

[Foracle]

TL;DR Summary

I tried calculating the time derivative of the probability density of a wavefunction but end up getting $$\frac{d}{dt}P(x,t)=0$$ for any wavefunction.

$\frac{d}{dt}P(x,t)=\frac{d}{dt}<\Psi|\Psi>$
$=<\frac{d}{dt}\Psi|\Psi>+<\Psi|\frac{d}{dt}\Psi>$

By using $\frac{d}{dt}=\frac{-i}{\hbar}H$ ,

$\frac{d}{dt}P(x,t)=<\frac{-i}{\hbar}H\Psi|\Psi>+<\Psi|\frac{-i}{\hbar}H\Psi>$
$=\frac{i}{\hbar}<\Psi|H\Psi> - \frac{i}{\hbar}<\Psi|H\Psi> = 0$
The above equation shows that the probability density is not changing in time for any state, though clearly this situation only applies to energy eigenstate. Can someone tell me where I went wrong?

Edit : Another problem that I have that resembles this is :
For a time independent operator Q, a quick (and probably naive) calculation shows that [H,Q] = 0. But take position operator for example $\hat{x}$. It's clear that $\frac{d}{dt}<\hat{x}>$ which can be written as $\frac{d}{dt}<\hat{x}> = \frac{i}{\hbar}<[H,\hat{x}]>$ is not zero for most states.

 

[Demystifier]

$P(x,t)\neq \langle\Psi|\Psi\rangle$. The correct equality is
$$P(x,t)=\langle\Psi(t)|x\rangle\langle x|\Psi(t)\rangle$$

 

[Foracle]

$P(x,t)\neq \langle\Psi|\Psi\rangle$. The correct equality is
$$P(x,t)=\langle\Psi(t)|x\rangle\langle x|\Psi(t)\rangle$$

Thanks for pointing out!
Following the previous calculation, I end up with $\frac{d}{dt}P(x,t) = \frac{i}{\hbar}[<H\Psi|x><x|\Psi> - <\Psi|x><x|H\Psi>]$
If I assume $<\Psi|x> = \Psi^{*}(x)>$ , I get
$$\frac{d}{dt}P(x,t) = \frac{i}{\hbar}[H\Psi^{*}\Psi - \Psi^{*}H\Psi]$$
If the wavefunction is real, than this is 0 again. Did I do some mistake?

 

[PeroK]

Thanks for pointing out!
Following the previous calculation, I end up with $\frac{d}{dt}P(x,t) = \frac{i}{\hbar}[<H\Psi|x><x|\Psi> - <\Psi|x><x|H\Psi>]$
If I assume $<\Psi|x> = \Psi^{*}(x)>$ , I get
$$\frac{d}{dt}P(x,t) = \frac{i}{\hbar}[H\Psi^{*}\Psi - \Psi^{*}H\Psi]$$
If the wavefunction is real, than this is 0 again. Did I do some mistake?

How can a wavefunction be real if it satisfies the Schrodinger equation?

 

[Foracle]

How can a wavefunction be real if it satisfies the Schrodinger equation?

Thanks for replying!
Can you please give me a quick explanation on why this is the case?
What about energy eigenfunction that can be chosen to be real?

 

[PeroK]

Thanks for replying!
Can you please give me a quick explanation on why this is the case?
What about energy eigenfunction that can be chosen to be real?

That's only the spatial function. There's a time-dependent function that is complex.

 

[Foracle]

That's only the spatial function. There's a time-dependent function that is complex.

Ah I see..

Schrodinger equation : $$i\hbar\frac{d}{dt}\Psi = H\Psi$$
Since everything in this equation but i is real, nothing can counter the imaginary part hence $\Psi$ must be complex. Is this why wavefunction satisfying Schrodinger's equation must be complex?

 

[PeroK]

Ah I see..

Schrodinger equation : $$i\hbar\frac{d}{dt}\Psi = H\Psi$$
Since everything in this equation but i is real, nothing can counter the imaginary part hence $\Psi$ must be complex. Is this why wavefunction satisfying Schrodinger's equation must be complex?

Yes, the full wavefunction cannot be real.

 

[Demystifier]

If the wavefunction is real, than this is 0 again.

The full wave function cannot be real, as @PeroK said. But if the wave function is an energy eigenstate, then $P(x,t)=P(x)$ does not depend on time. There is nothing inconsistent with it, that's indeed why such states are called stationary.

 

[PeterDonis]

if the wave function is an energy eigenstate, then $P(x,t)=P(x)$ does not depend on time.

Mathematically, this is because for an eigenstate, $H \Psi^* = E \Psi^*$ and $H \Psi = E \Psi$, so the RHS of the equation in post #3 does vanish for this special case.

 

[Demystifier]

Mathematically, this is because for an eigenstate, $H \Psi^* = \Psi^*$ and $H \Psi = \Psi$, so the RHS of the equation in post #3 does vanish for this special case.

An $E$ is missing on your right hand sides, but yes.

 

[PeterDonis]

An $E$ is missing on your right hand sides, but yes.

Ah, yes. Fixed now. Thanks!

 

[vanhees71]

Thanks for replying!
Can you please give me a quick explanation on why this is the case?
What about energy eigenfunction that can be chosen to be real?

The usual energy eigenfunction is calculated as a solution of the time-indpendent Schrödinger equation, but here you have to use the time-dependent equation. If $u_E(x)$ is the solution of the time-independent Schrödinger equation i.e., simply an eigenfunction of $\hat{H}$) then $\psi_E(t,x)=u_E(x) \exp(-\mathrm{i} E t/\hbar)$ is a solution of the time-dependent equation. Nevertheless, and this is important to understand the probability distribution for position in this case
$$P(t,x)=|\psi(t,x)|^2=|u_E(x)|^2$$
is indeed time-independent. That shows that the energy-eigenfunctions lead to the stationary states of the system. The time-dependent phase factor doesn't change the state, because the state is defined by the wave function up to a phase factor, because the physics is in $P=|\psi|^2$ only.

For an arbitrary solution of the time-dependent Schrödinger equation you get
$$\partial_t P(t,x)=\partial_t [\psi^*(t,x) \psi(t,x)]=(\partial_t \psi)^* \psi + \psi^* \partial_t \psi.$$
Now from the time-dependent Schrödinger equation you have
$$\partial_t \psi=-\mathrm{i}/\hbar \hat{H} \psi = -\frac{\mathrm{i}}{\hbar} \left [-\frac{\hbar^2 \partial_x^2}{2m} + V(x) \right] \psi$$
and taking the conjugate complex of this
$$\partial_t \psi^* = +\frac{\mathrm{i}}{\hbar} \left [-\frac{\hbar^2 \partial_x^2}{2m} + V(x) \right] \psi^*=+\frac{\mathrm{i}}{\hbar} \hat{H} \psi^*.$$
So you finally get
$$\partial_t P(t,x)=\frac{\mathrm{i}}{\hbar} [\psi \hat{H} \psi^*-\psi^* \hat{H} \psi]=-\frac{\mathrm{i} \hbar}{2m} [\psi \partial_x^2 \psi^*-\psi^* \partial_x^2 \psi].$$
From this you also get the very important continuity equation
$$\partial_t P(t,x)+\partial_x j(t,x)=0,$$
where the "probability current" is defined by
$$j=\frac{\mathrm{i} \hbar}{2m} [\psi \partial_x \psi^*-\psi^* \partial_x \psi].$$
Integrating over $x$ along the real axis you get (for a proper wave function which sufficiently quickly goes to 0 for $x \rightarrow \pm \infty$)
$$\mathrm{d}_t \int_{R} \mathrm{d} x P(t,x)=0.$$
This means that, if $P(t,x)$ is properly normalized at $t=0$ it stays normalized forever, as it should be.

 

[Foracle]

The usual energy eigenfunction is calculated as a solution of the time-indpendent Schrödinger equation, but here you have to use the time-dependent equation. If $u_E(x)$ is the solution of the time-independent Schrödinger equation i.e., simply an eigenfunction of $\hat{H}$) then $\psi_E(t,x)=u_E(x) \exp(-\mathrm{i} E t/\hbar)$ is a solution of the time-dependent equation. Nevertheless, and this is important to understand the probability distribution for position in this case
$$P(t,x)=|\psi(t,x)|^2=|u_E(x)|^2$$
is indeed time-independent. That shows that the energy-eigenfunctions lead to the stationary states of the system. The time-dependent phase factor doesn't change the state, because the state is defined by the wave function up to a phase factor, because the physics is in $P=|\psi|^2$ only.

For an arbitrary solution of the time-dependent Schrödinger equation you get
$$\partial_t P(t,x)=\partial_t [\psi^*(t,x) \psi(t,x)]=(\partial_t \psi)^* \psi + \psi^* \partial_t \psi.$$
Now from the time-dependent Schrödinger equation you have
$$\partial_t \psi=-\mathrm{i}/\hbar \hat{H} \psi = -\frac{\mathrm{i}}{\hbar} \left [-\frac{\hbar^2 \partial_x^2}{2m} + V(x) \right] \psi$$
and taking the conjugate complex of this
$$\partial_t \psi^* = +\frac{\mathrm{i}}{\hbar} \left [-\frac{\hbar^2 \partial_x^2}{2m} + V(x) \right] \psi^*=+\frac{\mathrm{i}}{\hbar} \hat{H} \psi^*.$$
So you finally get
$$\partial_t P(t,x)=\frac{\mathrm{i}}{\hbar} [\psi \hat{H} \psi^*-\psi^* \hat{H} \psi]=-\frac{\mathrm{i} \hbar}{2m} [\psi \partial_x^2 \psi^*-\psi^* \partial_x^2 \psi].$$
From this you also get the very important continuity equation
$$\partial_t P(t,x)+\partial_x j(t,x)=0,$$
where the "probability current" is defined by
$$j=\frac{\mathrm{i} \hbar}{2m} [\psi \partial_x \psi^*-\psi^* \partial_x \psi].$$
Integrating over $x$ along the real axis you get (for a proper wave function which sufficiently quickly goes to 0 for $x \rightarrow \pm \infty$)
$$\mathrm{d}_t \int_{R} \mathrm{d} x P(t,x)=0.$$
This means that, if $P(t,x)$ is properly normalized at $t=0$ it stays normalized forever, as it should be.

Thank you for taking your time to write this answer!