What does ## \psi^{*}\hat{H}\psi ## mean?

[ConjugatedClanger]

I was reading a section of The Physics off Quantum Mechanics by James Binney and David Skinner. On page 45, when discussing the probability current (in the wave mechanics formalism) in calculating it they state:

"We multiply the TDSE by $\psi^{*}$ and subtract it from the conjugate of of the TDSE multiplied with $\psi$"

I.e.$$ i \hbar \left( \psi^{*} \frac{\partial}{\partial t} \psi + \psi \frac{\partial}{\partial t} \psi^{*} \right) = \psi^{*} \hat{H} \psi - \psi \hat{H}^{*} \psi^{*} \tag{1}$$The intent is obviously to derive a continuity equation for the TDSE and I've seen this approach in a few other places, but I've always wondered: what is the intuition, or motivation, behind calculating $(1)$, straight off the bat? What does the quantity $\psi^{*}\hat{H}\psi$ mean -- an "expectation density"?

To me the more intuitive approach (and one shown in David Tong's lecture notes (page 11)) is to say
$$ \frac{\partial}{\partial t} \rho = \psi^{*} \frac{\partial}{\partial t} \psi + \psi \frac{\partial}{\partial t} \psi^{*} \tag{2}$$ where $\rho = |\psi|^{2} = \psi^{*} \psi$ is the probability density (as defined by the postulates of wave mechanics), and then rearrange the TDSE and its conjugate for $\partial/\partial t \psi$ and $\partial/\partial t \psi^{*}$ respectively and substitute back into $(2)$, which will give us $(1)$.

It might be that I'm looking for significance where there is none, and the approach shown in $(1)$ is from the insight we gained by first doing $(2)$ and lecturers and authors just choose this way to jump to an answer and then work backwards. But if anyone has some insight please let me know!

 

[Haborix]

If I had to conjure up a pedagogical reason, I'd say you might prefer technique (1) because it starts with known variables and arrives at the new one $\rho$. As someone familiar with this material, I think (2) seems less convoluted.

 

[ConjugatedClanger]

If I had to conjure up a pedagogical reason, I'd say you might prefer technique (1) because it starts with known variables and arrives at the new one $\rho$. As someone familiar with this material, I think (2) seems less convoluted.

I couldn't think of a physics/mathematical motivation for $(1)$, hence I am wondering if others have a better insight.

I agree, I think $(2)$ is the far more physically intuitive approach. If $(1)$, as you say, is a story telling/pedagogical approach, I don't think its a good one. I think that we should motivate why we do operations, especially when teaching it to students.

 

[vanhees71]

I was reading a section of The Physics off Quantum Mechanics by James Binney and David Skinner. On page 45, when discussing the probability current (in the wave mechanics formalism) in calculating it they state:

I.e.$$ i \hbar \left( \psi^{*} \frac{\partial}{\partial t} \psi + \psi \frac{\partial}{\partial t} \psi^{*} \right) = \psi^{*} \hat{H} \psi - \psi \hat{H}^{*} \psi^{*} \tag{1}$$The intent is obviously to derive a continuity equation for the TDSE and I've seen this approach in a few other places, but I've always wondered: what is the intuition, or motivation, behind calculating $(1)$, straight off the bat? What does the quantity $\psi^{*}\hat{H}\psi$ mean -- an "expectation density"?

To me the more intuitive approach (and one shown in David Tong's lecture notes (page 11)) is to say
$$ \frac{\partial}{\partial t} \rho = \psi^{*} \frac{\partial}{\partial t} \psi + \psi \frac{\partial}{\partial t} \psi^{*} \tag{2}$$ where $\rho = |\psi|^{2} = \psi^{*} \psi$ is the probability density (as defined by the postulates of wave mechanics), and then rearrange the TDSE and its conjugate for $\partial/\partial t \psi$ and $\partial/\partial t \psi^{*}$ respectively and substitute back into $(2)$, which will give us $(1)$.

It might be that I'm looking for significance where there is none, and the approach shown in $(1)$ is from the insight we gained by first doing $(2)$ and lecturers and authors just choose this way to jump to an answer and then work backwards. But if anyone has some insight please let me know!

Of course, Tong's approach is more to the physicist's taste. The idea is simple: The probability to find the particle somewhere must be always 1,
$$\int_{\mathbb{R}^3} \mathrm{d}^3 |\psi(t,\vec{x})|^2=1,$$
i.e., you have a "conservation law" for a "charge" with the density of this charge given by $\rho=|\psi|^2$.

On the other hand you have the wave equation, which should obey this conservation law. That's the case, because $\hat{H}$ is self-adjoint, and the formal solution of the Schrödinger equation is
$$\psi(t,\vec{x})=\exp(-\mathrm{i} \hat{H} t/\hbar) \psi(0,\vec{x}).$$
Since $\hat{H}$ is self-adjoint the exp is unitary and thus the "conservation of probability" holds.

On the other hand from field theories, e.g., classical electrodynamics, a conservation of this kind should also have a "local version", i.e., for the given charge density $\rho$ there should also be a current density $\vec{j}$, and the local version of the conservation law is the continuity equation
$$\partial_t \rho +\vec{\nabla} \cdot \vec{j}=0.$$
So the goal in Tong's approach is to rewrite $\partial_t \rho$ as a divergence of a vector field, using the "field equation" (i.e., in this case the Schrödinger equation), and indeed that's what Tong does.

Finally, if there's a conservation law, then in almost all cases Noether is around, i.e., there should be some Lie symmetry of the equations of motion that explains why there's a conserved quantity, and why the conserved quantity is given by a corresponding density and current density and how these look the way they look.

Indeed, in this case you can first find a Lagrangian for the Schrödinger equation, from which it is derived by the principle of least action (setting $\hbar=1$ for convenience):
$$\mathcal{L}=\frac{1}{2m} (\vec{\nabla} \psi^*) \cdot (\vec{\nabla} \psi) + \mathrm{i} \psi^* \partial_t \psi-V \psi^* \psi.$$
Here you can vary $\psi^*$ and $\psi$ independently, because $\psi$ represents two real field-degrees of freedom. Stationarity of the action under variations of $\psi^*$ leads to the Schrödinger equation,
$$\mathrm{i} \partial_t \psi=-\frac{1}{2m} \Delta \psi + V \psi.$$
Then you can use Noether's approach to analyze the symmetries of the action. As it turns out there are of course the 10 conservation laws from Galilei symmetry, which has to be true, because the Schrödinger equation should be compatible with the very fundamentals of Newtonian physics, which is Galilei invariance.

Another symmetry of the entire edifice of quantum theory (wave mechanics) is invariance under multiplication of the wave function with a constant phase factor, $\exp(-\mathrm{i} \phi)$ with $\phi \in \mathbb{R}$ constant. Indeed, the action is invariant under this transformation (since already the Lagrangian is invariant). Then using Noether's theory indeed leads to the probality density and current density,
$$\rho=|\psi^2|, \quad \vec{j}=\frac{1}{2m \mathrm{i}} (\psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi^*).$$

 

[ConjugatedClanger]

Of course, Tong's approach is more to the physicist's taste. The idea is simple: The probability to find the particle somewhere must be always 1,
$$\int_{\mathbb{R}^3} \mathrm{d}^3 |\psi(t,\vec{x})|^2=1,$$
i.e., you have a "conservation law" for a "charge" with the density of this charge given by $\rho=|\psi|^2$.

On the other hand you have the wave equation, which should obey this conservation law. That's the case, because $\hat{H}$ is self-adjoint, and the formal solution of the Schrödinger equation is
$$\psi(t,\vec{x})=\exp(-\mathrm{i} \hat{H} t/\hbar) \psi(0,\vec{x}).$$
Since $\hat{H}$ is self-adjoint the exp is unitary and thus the "conservation of probability" holds.

On the other hand from field theories, e.g., classical electrodynamics, a conservation of this kind should also have a "local version", i.e., for the given charge density $\rho$ there should also be a current density $\vec{j}$, and the local version of the conservation law is the continuity equation
$$\partial_t \rho +\vec{\nabla} \cdot \vec{j}=0.$$
So the goal in Tong's approach is to rewrite $\partial_t \rho$ as a divergence of a vector field, using the "field equation" (i.e., in this case the Schrödinger equation), and indeed that's what Tong does.

Finally, if there's a conservation law, then in almost all cases Noether is around, i.e., there should be some Lie symmetry of the equations of motion that explains why there's a conserved quantity, and why the conserved quantity is given by a corresponding density and current density and how these look the way they look.

Indeed, in this case you can first find a Lagrangian for the Schrödinger equation, from which it is derived by the principle of least action (setting $\hbar=1$ for convenience):
$$\mathcal{L}=\frac{1}{2m} (\vec{\nabla} \psi^*) \cdot (\vec{\nabla} \psi) + \mathrm{i} \psi^* \partial_t \psi-V \psi^* \psi.$$
Here you can vary $\psi^*$ and $\psi$ independently, because $\psi$ represents two real field-degrees of freedom. Stationarity of the action under variations of $\psi^*$ leads to the Schrödinger equation,
$$\mathrm{i} \partial_t \psi=-\frac{1}{2m} \Delta \psi + V \psi.$$
Then you can use Noether's approach to analyze the symmetries of the action. As it turns out there are of course the 10 conservation laws from Galilei symmetry, which has to be true, because the Schrödinger equation should be compatible with the very fundamentals of Newtonian physics, which is Galilei invariance.

Another symmetry of the entire edifice of quantum theory (wave mechanics) is invariance under multiplication of the wave function with a constant phase factor, $\exp(-\mathrm{i} \phi)$ with $\phi \in \mathbb{R}$ constant. Indeed, the action is invariant under this transformation (since already the Lagrangian is invariant). Then using Noether's theory indeed leads to the probality density and current density,
$$\rho=|\psi^2|, \quad \vec{j}=\frac{1}{2m \mathrm{i}} (\psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi^*).$$

Thank you very much for the detailed reply.

Indeed, calculating $\partial_{t} \rho$ is by far the most intuitive route to a continuity equation for me -- it just makes physical sense.

The problem for me is that many authours (such as Binney) simply state $$i(\psi^{*} \partial_{t} \psi + \psi \partial_{t} \psi^{*}) = \psi^{*} \hat{H} \psi - \psi\hat{H}^{*}\psi^{*} \tag{1}$$as the starting point. This doesn't seem physically/mathematically motivated to me. So far my best guess is that it is nothing more than pedagogical way to introduce the concept.

My problem isn't with the approach illustrated by Tong, or even understanding the whole endeavour. My question is rather why some authors simply state $(1)$, to get the answer straight away? I'm not asking why we do this calculation.

I can't see how one could intuitively guess/justify/motivate $(1)$, unless you already new the answer by starting with $\partial_{t} \rho$. If this is the case, I'm fine with that, just wanted to make sure that there isn't some other perspective.

While your answer is a very detailed and an excellent exposition, unless I have missed a subtlety (quite possible), I don't think it answers my specific question.

 

[Haborix]

What I'm about to say is really about conserved quantities etc. One traditional way of solving differential equations, probably taught less often, is by trying to find so-called first integrals. The spirit of the first approach you describe is to construct some equation from known (P)DEs that looks like derivative of single term equals stuff. In other words, you might know the PDE, but you wouldn't know the differentiated "integral." That's a kind of technician's answer.

 

[Demystifier]

My question is rather why some authors simply state (1), to get the answer straight away?

(1) is a simple consequence of the Schrodinger equation. Can you see how?

If you look for a deeper meaning of equation (1), you can see it as the diagonal matrix element in the position basis, of the von Neumann operator equation
$$\frac{\partial\rho}{\partial t}=-\frac{i}{\hbar}[H,\rho]$$
which itself is a quantum version of the classical Liouville equation
$$\frac{\partial\rho}{\partial t}=\{ H,\rho \}$$
Is it too deep for you?

 

[vanhees71]

Thank you very much for the detailed reply.

Indeed, calculating $\partial_{t} \rho$ is by far the most intuitive route to a continuity equation for me -- it just makes physical sense.

The problem for me is that many authours (such as Binney) simply state $$i(\psi^{*} \partial_{t} \psi + \psi \partial_{t} \psi^{*}) = \psi^{*} \hat{H} \psi - \psi\hat{H}^{*}\psi^{*} \tag{1}$$as the starting point. This doesn't seem physically/mathematically motivated to me. So far my best guess is that it is nothing more than pedagogical way to introduce the concept.

My problem isn't with the approach illustrated by Tong, or even understanding the whole endeavour. My question is rather why some authors simply state $(1)$, to get the answer straight away? I'm not asking why we do this calculation.

I can't see how one could intuitively guess/justify/motivate $(1)$, unless you already new the answer by starting with $\partial_{t} \rho$. If this is the case, I'm fine with that, just wanted to make sure that there isn't some other perspective.

While your answer is a very detailed and an excellent exposition, unless I have missed a subtlety (quite possible), I don't think it answers my specific question.

Then it's more a question of didactics, and there's a fine line between not enough didactics and too much ;-). It's for sure not enough didactics to just write down an ansatz without any motivation for it. I tried to give some motivations for the idea to take the time derivative of $\psi^* \psi$ on different levels of sophistication. Whether or not this helps you, I can't know. Just ask for more explanations.

 

[ConjugatedClanger]

Then it's more a question of didactics, and there's a fine line between not enough didactics and too much ;-). It's for sure not enough didactics to just write down an ansatz without any motivation for it. I tried to give some motivations for the idea to take the time derivative of $\psi^* \psi$ on different levels of sophistication. Whether or not this helps you, I can't know. Just ask for more explanations.

Actually I take it back, you did answer my question -- you just didn't spoon feed me (quite right too). It suddenly clicked when I thought about the how we arrive at the conserved current for a for complex scalar field. It's exactly the same thing for the Schroedinger Lagrangian.

So there is a deeper aspect in simply stating $(1)$, but the reason most author's don't go into it is because they'll have to introduce things like Noether's theorem, Lagrangian densities etc. And that's too much detail for an undergraduate Quantum Mechanics course, where those things aren't the main subject matter. So its easier to reduce it all down to a pedagogical trick.

Then it's more a question of didactics, and there's a fine line between not enough didactics and too much ;-).

I now see exactly what you meant by this. Thank you! You got me to the answer without just giving it to me!

 

[vanhees71]

Anyway, it's an important lesson to learn (better sooner than later) that you can understand physics only by doing it yourself. In theoretical physics that means to try to derive at least the important results yourself. In a textbook (let alone in a forum posting) you can never display all the minute detailed steps of such a calculation. The challenging task for the textbook writer (and even more the lecturer) is to find the right balance between just stating "as the reader easily can show,..." (usually meaning that the author is to lazy to write any details) and "spoon feed" the student, such that he thinks he doesn't need to do the calculation himself. That's part of what I meant with the tension between too little and too much didactics.