In time dependent Schrodinger's equation as given in Schrodinger's lecture (Four Lectures on Wave mechanics, Blackie & Son, 1949, pg22) he arrives at

$$\nabla^2\psi-\frac{4 \pi m i}{h}\dot\psi-\frac{8\pi^2 m V}{h^2}\psi=0.\tag{22}$$

Here note the sign before Hamiltonian! How shoould one intrepret this? Again in Mott's book(An outline of wave mechanics, 1934, pg 44) the same sign problem

$$i \hbar \frac{\partial \psi}{\partial t}=\frac{h^2}{8\pi^2 m}\nabla^2\psi-V\psi.\tag{6}$$

Only when Krammers (Quantentheorie,1938) started with

$$\psi=A \exp\{2\pi i[(\vec{\sigma}\cdot \vec{x})-\nu t] \}$$

ends up with

$$\nabla^2\psi+(\frac{2 m i}{\hbar})\frac{\partial \psi}{\partial t}=0.\tag{62}$$

This is the way all books write now. I wish to understand what is problem if we start with $e^{i \omega t}$ as in the first two cases then I end up with Schrodinger equation with a minus sign. Can anyone give some possible reference where this issue has been clearly discussed? Another point is whether momentum, Hamiltonian would be Hermitian ?


The important thing is the relative sign between the potential and the Laplacian.

Otherwise there are two square roots of $-1$ namely $\pm i.$ You could write $j=-i$ and then you have two perfectly good square roots of $-1$ you can use $i$ or you can use $j$.

For square roots of positive numbers you get things like $\pm\sqrt 2$ and there is a way to pick one consistently, pick the positive one. For $\pm\sqrt {-1}$ there is no way to prefer one over the other.

And in fact engineers like to use $j$ for $\sqrt{-1}$ and they like to write $e^{j\omega t}$ and physicists like to write $i$ for $\sqrt{-1}$ and use $e^{-i\omega t}$ but there isn't really a difference.

In real life you will meet people (usually engineers) that write oscillating waves like $E^{j\omega t}$ and you will meet other people that write oscillating waves like $E^{-i\omega t}.$ And they both are referring to a real wave that oscillates like $\cos (\omega t).$ The difference is one of convention, not of physics. It's like if you drew the x axis going to the right, drawing the y axis as going up (rather then down) is mere convention and means nothing. But you do need to stick to a convention and not mix and match.

In fact, selecting one of those two direction up versus down is all that is going on when you pick one of the two square roots. If you can't see this easily, it is because your favorite meaningless convention is so ingrained in your mind that you can't even see what you are doing. There are two square roots, and no one can tell them apart. You can draw them of the lien from 0 to 1 goes rightwards and then you pick the upwards direction and call it $+i$ and call downwards $-i$ but that is just a convention about how you like to draw things, in reality there are simply those two directions.

If you met an alien there would be literally no way to explain the difference between $i$ and $-i$ they would be different, like clockwise and counterclockwise, but the reason to have one of them have a $-$ or to have the word $counter$ is mere convention and history. It doesn't mean anything.

And I can't explain something that has no meaning. How would I explain clockwise other than people made sundials in the northern hemisphere during the northern summer and the sun rises in the east and people liked to keep doing that? Whoops, that used the words east and north, how is an alien going to no that other than from context and history, they are mere conventions. We called the direction the sun rises east and the north pole and the south pole are equally good poles, we put the north pole up on our globes by convention, not because that means anything.

What is correct is to have a sign difference between the Laplacian and the potential. What is correct is to make sure your momentum and Hamiltonian are related to your spatial and temporal derivatives correctly. If you like writing your waves like $e^{i(\vec k \cdot \vec x -\omega t)}$ then one choice will seem more natural to you. If you prefer to write your waves like $e^{j(\omega t -\vec k \cdot \vec x)}$ then other choices might seem more natural. But those aren't really different, they are both cosines as real parts and there is no way to distinguish between the two imaginary directions. Absolutely no way whatsoever. Some people even do quaternion quantum mechanics where the imaginary unit is a unit imaginary quaternion that can change from place to place, so it might point different ways in different places because then $+i$ and $-i$ are continuously connected to each other by rotating through the $i,k$ plane. That is different physics since it changes from place to place, but it shows that there is no meaning to picking one direction over the other.

You can write $i\hbar \frac{\partial}{\partial t} \Psi = -\frac{\hbar^2}{2m}\vec \nabla^2 \Psi + V \Psi$ and $\hat p = -i\hbar \nabla$ or else you can hit the first with a minus sign (no change in physics) and get $-i\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi - V \Psi$ and $\hat p = -i\hbar \nabla$ and then say hey, $\pm i$ are equally good and I dont' like writing so many minus signs, so I'll use $j=-i$ as my favorite square root and get $j\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi - V \Psi$ and $\hat p = j\hbar \nabla$ and all I did was write it in terms of the other square root which didn't change the physics.

No one is changing any physics. And both square roots are indeed equally good, and this is not deep. There is a minus sign difference between the Laplacian and the potential and there is no way around that. And you can call a square root of minus one anything you want because there are two and there is no way to know which is which. But like any convention you need to stick to it.

What about the case of free particle when V=0?

Now you don't have to worry about the relative sign between the two. Energy eigenstates will evolve like $e^{j\omega t}$ which is great if that is traditional in your field and annoying if $e^{-i\omega t}$ is traditional for your field, but deal with conventions or else get everyone to change (and make translations of old works as needed).

What is the correct Schrödinger equation then?

This is confusing, they aren't even really different equations. The equation $i\hbar \frac{\partial}{\partial t} \Psi = -\frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ and the equation $j\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ aren't saying different things, they both say that if you differentiate Psi and multiple by $\hbar$ then multiply by one (of the two equally good) square roots of minus one it is equal to $\frac{\hbar^2}{2m}$ times the Laplacian. They might disagree about where you want to draw that square root, if you should go clockwise or counterclockwise, but this is not physics, the equations are not saying different things.

What will be the momentum operator?

A wave travels in the $+x$ direction if a point of constant phase travels in the $+x$ direction as time evolves. So a wave like $e^{j(\omega t-kx)}$ or a wave like $e^{i(kx-\omega t)}$ (they are the same wave) works fine to describe something with a fixed energy and momentum with positive momentum in the x direction. If you use the equation $i\hbar \frac{\partial}{\partial t} \Psi = -\frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ then you get $i\hbar(-i\omega) = -\frac{\hbar^2}{2m}(-k^2)$ and everything is fine, and you get $\hat p_x=\frac{\hbar}{i}\frac{partial}{\partial x}$ as usual for the reason that it gives $\hbar k =p >0$ for the state with positive $x$ momentum above.

If instead you use the equation $j\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ then you get $j\hbar(j\omega) = \frac{\hbar^2}{2m}(-k^2)$ and everything is fine, and you get $\hat p_x=j\hbar\frac{\partial}{\partial x}$ for the reason that it gives $\hbar k =p >0$ for the state with positive $x$ momentum above.

Is this any different? No, you are just preferring to use the other root of $-1$.

What is the Hamiltonian operator?

That is a better question. It should be an operator that gives a positive number for a positive energy eigenstate if you want it to be an energy operator. So it should have the negative sign in front of the Laplacian.

Are these operators Hermitian?

I don't understand your concern. Your feelings (or your field's feelings) about which root of $-1$ you like best doesn't change whether something is Hermitian. And whether you slap a minus sign on it doesn't change whether it is Hermitian either. They are Hermitian.

But there is a similar convention that maybe you should address now. Why now? Because then you will less stuff to go recheck if your doubts are raised later in life. The question is about whether the inner product is $\int\overline\Psi\Psi$ or whether it is $\int\Psi\overline\Psi.$ Mathematicians prefer the latter when they are young mathematicians or are teaching young mathematicians. As far as I can tell it is because they like to say the word sesquilinear so they want it it be linear in the first argument. At first it doesn't make too much problem (well, it is nicer to write $\int\overline\Psi\frac{\hbar}{i}\frac{\partial}{\partial x}\Psi$ with each thing more clearly where it is because of what it is doing) but deep into operator algebra theory, the wrong choice turns out to create actual real headaches, so it would be better if everyone switched.

But there isn't really anything wrong, the two inner products are equally good (they are just complex conjugates of each other), and they are both there, making the same metric and everything. But the choice about which to be your favorite, that is similar to the $i$ versus $j$ issue, and I don't want it to create problems for you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.