The energy term in Schroedinger equation

[amjad-sh]

The energy in the time-independent Schroedinger equation H|Ψ>=E|Ψ> is the total energy of the system,which is E=kinetic energy +V where V is the potential energy.
But while solving problems in quantum mechanics,like scattering problems and others,they deal with it as it is just the K.E.
For example in the potential step problem, they take the cases where E>V and E<V, but isn't that wrong since the energy of the system is obviously greater than V and can never be less than V?? because E=K.E+V and in textbooks they meant by E the the energy term in the Schroedinger equation?
What goes wrong here?
thanks.

 

[DrClaude]

You are mixing up a few things here. First, the energy is actually obtained as the expectation value of the Hamiltonian, $\langle E \rangle = \langle \psi | \hat{H} | \psi \rangle$.

Second, what is meant by "E < V" is that the total energy is lower than the barrier height V, not the total potential energy V. It might be less ambiguous if you take it as $E < V_0$, with the potential energy given as $\hat{V} = V(x)$, with $V(x)=0$ almost everywhere, except for $V(x) = V_0$ for $a < x < b$.

Also, scattering problems usually consider the case of an initial plain wave of (kinetic) energy E, since it originates in a region where there is no potential.

 

[amjad-sh]

Second, what is meant by "E < V" is that the total energy is lower than the barrier height V, not the total potential energy V

but when the particle enters the region where V=V0 doesn't this height barrier becomes the total potential energy?
I mean isn't the V(x) in the schroedinger equation is just meant to be the the potential energy of the system?
I'm still confused about it.

 

[DrClaude]

but when the particle enters the region where V=V0 doesn't this height barrier becomes the total potential energy?

No, since energy is conserved. That is why you get tunneling: for a particle to go from the left side of the barrier to the right when E < V, it has to tunnel through the barrier.

I mean isn't the V(x) in the schroedinger equation is just meant to be the the potential energy of the system?

No, it is the potential energy operator. To get the actual potential energy of the system, you need to calculate $E_\mathrm{pot} = \langle \psi | \hat{V} | \psi \rangle$, or equivalently in position representation
$$
E_\mathrm{pot} = \int_{-\infty}^{\infty} \psi^*(x) V(x) \psi(x) \, dx
$$

 

[Orodruin]

You are mixing up a few things here. First, the energy is actually obtained as the expectation value of the Hamiltonian, $\langle E \rangle = \langle \psi | \hat{H} | \psi \rangle$.

This is the energy expectation value for an arbitrary state. The time independent Schrödinger equation, which is mentioned in the OP, is just the eigenvector equation for the Hamiltonian and the E appearing in it is the energy eigenvalue of a particular eigenstate we wish to find.

That being said, an eigenstate of the Hamiltonian does not have a definite potential energy, it makes no sense to, as the OP does, talk about V(x) as the total potential of the system. The potential is an operator which is different from the Hamiltonian and generally does not share the same eigenstates. It also makes little sense to talk about the expectation values for the Hamiltonian or potential for the scattering states as they are not normalisable.

 

[amjad-sh]

@DrClaude @Orodruin
does this mean that the energy term of the Schrodinger equation is just meant to be the kinetic energy of the particle?

The potential is an operator which is different from the Hamiltonian and generally does not share the same eigenstates.

but this operator is included in the Hamiltonian operator.

 

[Orodruin]

but this operator is included in the Hamiltonian operator.

So what? The sum of two operators need not have the same eigenvectors as the operators themselves.

 

[DrClaude]

does this mean that the energy term of the Schrodinger equation is just meant to be the kinetic energy of the particle?

Not, it is potential + kinetic energy.

 

[blue_leaf77]

E<V, but isn't that wrong since the energy of the system is obviously greater than V and can never be less than V??

Well, that's quantum mechanics. A region which is classically forbidden becomes allowed when you go to the quantum world. Classically, the total energy cannot be less than the potential energy, but in quantum mechanics you have the Schroedinger equation which let's you have
$$
\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x) + V \psi(x) = E\psi(x)
$$
with $E<V$, which is classically forbidden.

 

[amjad-sh]

The sum of two operators need not have the same eigenvectors as the operators themselves.

Right.

Not, it is potential + kinetic energy.

Lets take the potential step example,where for x<0 V=0 and for X>0 V=V0.
If the particle is in the region where where V=V0, what is the potential of the particle in this region?

Well, that's quantum mechanics. A region which is classically forbidden becomes allowed when you go to the quantum world. Classically, the total energy cannot be less than the potential energy, but in quantum mechanics you have the Schroedinger equation which let's you have
$$
\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x) + V \psi(x) = E\psi(x)
$$
with $E<V$, which is classically forbidden.

I know this, but my question was about the nature of V in the Schrodinger equation.

 

[Orodruin]

If the particle is in the region where where V=V0, what is the potential of the particle in this region?

But in the scattering eigenstates, the particle is not in any particular region. You could find a state which is fully contained in that region but its energy expectation value (it is not going to be an energy eigenstate!) is going to be larger than V. The scattering states have infinite support.

You simply cannot talk about the potential energy of a scattering state, it is not an eigenstate of the potential energy operator.

 

[amjad-sh]

still confused :(

But in the scattering eigenstates, the particle is not in any particular region.

Yes that is because there is a reflection and transmission probabilities.

You could find a state which is fully contained in that region but its energy expectation value (it is not going to be an energy eigenstate!) is going to be larger than V.

I didn't get this part, if you can simplify it more: why it is not going to be an energy eigenstate if I'm sure that it is in this region

−ℏ22md2dx2ψ(x)+Vψ(x)=Eψ(x)

The V in the schrodinger equation, refers to what then?

 

[vanhees71]

I do not understand your problem. The Hamilton operator consists of a term we call kinetic energy and a potential as in classical mechanics. If your have problems to understand the physical meaning of the Hamiltonian you should go back to classical mechanics, before you try to understand quantum mechanics. A solid foundation in classical physics is an indispensable prerequisite to have a chance to get an idea about the meaning of quantum theory! You need to learn about Hamilton's principle in the Hamiltonian formulation, Poisson brackets and symmetry principles related to it (Noether's theorem) before you go back to quantum mechanics!

 

[Orodruin]

I didn't get this part, if you can simplify it more: why it is not going to be an energy eigenstate if I'm sure that it is in this region

If it is it is an eigenfunction of the potential energy operator, but not of the kinetic energy operator (whose eigenstates are $e^{ikx}$). Hence, it is not an eigenstate of the Hamiltonian.

Classically, the total energy cannot be less than the potential energy, but in quantum mechanics you have the Schroedinger equation which let's you have
$$
\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x) + V \psi(x) = E\psi(x)
$$
with $E<V$, which is classically forbidden.

If $V$ here is a constant everywhere, it is also forbidden in quantum mechanics. It is absolutely necessary that $E < V$ in some region, since the kinetic energy operator is positive (semi-)definite.

 

[Orodruin]

I do not understand your problem. The Hamilton operator consists of a term we call kinetic energy and a potential as in classical mechanics. If your have problems to understand the physical meaning of the Hamiltonian you should go back to classical mechanics, before you try to understand quantum mechanics. A solid foundation in classical physics is an indispensable prerequisite to have a chance to get an idea about the meaning of quantum theory! You need to learn about Hamilton's principle in the Hamiltonian formulation, Poisson brackets and symmetry principles related to it (Noether's theorem) before you go back to quantum mechanics!

While I do not disagree, I have often seen students being taught quantum mechanics without the slightest idea about neither the Lagrangian or Hamiltonian formulations of classical mechanics. I do agree that understanding and realising that quantum mechanics is not just some ad hoc assumptions but actually a development from classical mechanics is crucial, but it is not how it is often taught.

 

[blue_leaf77]

If VVV here is a constant everywhere, it is also forbidden in quantum mechanics.

The V in the schrodinger equation, refers to what then?

I should have specified more about $V$, it should be written as $V(x)$ where $V(x)$ is a step potential or barrier or anything which has a portion for some values of $x$ where $V(x)$ is higher than $E$.
I was trying to answer this question:

but isn't that wrong since the energy of the system is obviously greater than V and can never be less than V??

since you forbade $E<V$ in quantum mechanics. It's forbidden in CM but allowed in QM.

 

[vanhees71]

@Orodruin: So, how would you introduce quantum theory? I've no clue, how you'd do that without some ideas about Hamiltonian mechanics, particularly Poisson brackets.

For me, it's the greatest challenge of physics teaching to introduce quantum theory. I've not taught QM 1 (introductory lecture on non-relativistic quantum mechanics) yet. So I've never been in the dilemma to think about it. There are several approaches in the literature. In my opinion there are some no-gos, and these are only very slowly eliminated from the curricula in high schools and at universities (at least in Germany):

One should not start with old quantum theory; it's misleading from the very beginning; particularly to claim that the photoelectric effect "proves the existence of photons" a la Einstein's famous paper of 1905. Even worse is to teach Bohr's model of the atom (which BTW cannot be taught without the Hamiltonian formalism of classical mechanics either!). There's a reason why we don't use old quantum theory anymore but the modern way a la Schrödinger, Heisenberg, Born, Jordan, and (most importantly) Dirac!

On the other hand, of course, you cannot just through the postulates in abstract Hilbert-space language to the students and then perform dry calculations of all kinds of applications, but you need some heuristics to get there. I think, a good way is to start with wave mechanics to motivate the Hilbert-space formalism in terms of $L^2$, introducing operators representing observables and Born's rule (minimal interpretation). Then you have a good basis to do the formalism in the representation free (Dirac) way using the modern form of the "correspondence principle" based on symmetry considerations, leading to the operator algebra of observables. To keep the things as simple as possible you can use the Heisenberg algebra, i.e., the position and momentum algebra based on the definition of momentum as generators of spatial translations. Then it's also very natural that the Hamiltonian is the generator for the time evolution (but this must be taken with a grain of salt since time necessarily is not an observable but a parameter in quantum theory). For this you need Poisson brackets from classical mechanics motivating the commutation relations. Even when using no symmetry arguments but canonical quantization (which one should do with some caveat, because it only works for Cartesian coordinates of position and momentum and I find it a bit less physically motivated than to base theoretical physics on symmetry principles from the very beginning).

Later you can introduce also aspects of QT going beyond classical mechanics, using advanced methods of group-representation theory on Hilbert spaces, e.g., the idea of spin by analyzing the Galileo group. I've done this with great success in teaching QM 2 some years ago. The students liked this approach very much, and I think it makes clear in a very convincing way, why non-relativistic quantum theory looks as it does, at least it's more convincing than just telling the formalism and then justify it just by arguing "that it works" ;-).

 

[amjad-sh]

You need to learn about Hamilton's principle in the Hamiltonian formulation, Poisson brackets and symmetry principles related to it (Noether's theorem) before you go back to quantum mechanics!

I know all of these and I have a solid background in classical mechanics. But the problem is that you

I do not understand your problem.

 

[RockyMarciano]

I do not understand your problem... A solid foundation in classical physics is an indispensable prerequisite to have a chance to get an idea about the meaning of quantum theory!...

I believe amjad-sh problem here is precisely that he is holding on too much to the classical foundations, so I don't think it helps to point him to the classical picture. Quantum mechanics departure point is precisely to allow what is forbidden classically, in this particular case with respect to the concepts of potential energy and the Hamiltonian. So amjad-sh, if you don't want to encounter apparent paradoxes between total, kinetic and potential energies from the classical standpoint you basically have to unlearn certain things about the classic Hamiltonian now that the energy operator doesn't commute neither with the potential(except for constant potential) nor the kinetic energy operators. This produces relations between energy and potential energy that are just not compatible with how they were connected in classical physics with a classical Hamiltonian. You just have found one of the many consequences of non-commuting operators.

 

[vanhees71]

Well, quantum theory hinges on the commutation relations of the operators that represent observables. What's new is the probabilistic meaning of the notion of states. So you need the classical foundation to understand the observable algebra following from symmetry principles. Also in classical mechanics the potential describes forces, and the Hamiltonian "doesn't commute" with it, because you want to describe forces.

 

[RockyMarciano]

Sure, I'm not saying that it is not good to have a good grasp of classical physics foundations to work with QM, it is clearly necessary at a certaing level of understanding. What I mean is that you have to unlearn some of it in order to come to terms with QM, in doing this unlearning(basically due to the non-commuting operators) you end up with a probabilistic view of the states, and since what the OP was finding paradoxical was due to sticking literally to the classical view I suggested that as the reason you didn't understand his problem.
What you say about forces and the Hamiltonian is interesting but note that the Hamiltonian picture and the Newtonian forces picture are compatible but somewhat "orthogonal" and thus their "not commuting".

 

[vanhees71]

Well, perhaps I shouldn't participate in this discussion, because I don't understand the problems you are having. The main problem seems to be confusion about the basic foundations, however.

First of all there's nothing orthogonal between "Newtonian forces" and "the Hamiltonian picture". The Hamilton formalism is just another mathematical description of Newtonian mechanics (as long as you stick with Newtonian Hamiltonians of course).

The most simple case is a particle moving in an external field given by a potential, e.g., a point mass moving in the gravitational field of a very heavy other mass that you can treat to sit at rest in the origin:
$$H=\frac{\vec{p}^2}{2m} -\frac{G m M}{|\vec{x}|}.$$
The Hamilton equations of motion read
$$\dot{\vec{x}}=\frac{\partial H}{\partial \vec{p}}=\frac{\vec{p}}{m}, \quad \dot{\vec{p}}=-\frac{\partial H}{\partial \vec{x}} = -\frac{G m M}{|\vec{x}|^3} \vec{x},$$
which are precisely the Newtonian equations of motion and nothing else.

 

[RockyMarciano]

Well, perhaps I shouldn't participate in this discussion, because I don't understand the problems you are having. The main problem seems to be confusion about the basic foundations, however.

First of all there's nothing orthogonal between "Newtonian forces" and "the Hamiltonian picture". The Hamilton formalism is just another mathematical description of Newtonian mechanics (as long as you stick with Newtonian Hamiltonians of course).

The most simple case is a particle moving in an external field given by a potential, e.g., a point mass moving in the gravitational field of a very heavy other mass that you can treat to sit at rest in the origin:
$$H=\frac{\vec{p}^2}{2m} -\frac{G m M}{|\vec{x}|}.$$
The Hamilton equations of motion read
$$\dot{\vec{x}}=\frac{\partial H}{\partial \vec{p}}=\frac{\vec{p}}{m}, \quad \dot{\vec{p}}=-\frac{\partial H}{\partial \vec{x}} = -\frac{G m M}{|\vec{x}|^3} \vec{x},$$
which are precisely the Newtonian equations of motion and nothing else.

I'm not having any problems, I was referring to the OP, and your problem understanding him as claimed by yourself.

When I wrote "orthogonal" with scare quotes(just after writing that they are of course compatible) I was referring to your "not commuting" that was also with scare quotes. Maybe I didn't understand what you meant by not commuting in your post, what was it?

 

[Louis Fry]

The energy in the time-independent Schroedinger equation H|Ψ>=E|Ψ> is the total energy of the system,which is E=kinetic energy +V where V is the potential energy.
But while solving problems in quantum mechanics,like scattering problems and others,they deal with it as it is just the K.E.
For example in the potential step problem, they take the cases where E>V and E<V, but isn't that wrong since the energy of the system is obviously greater than V and can never be less than V?? because E=K.E+V and in textbooks they meant by E the the energy term in the Schroedinger equation?
What goes wrong here?
thanks.

I think the problem here is that the time-independent Schrödinger equation is simply not a direct analog of the conservation of energy, it's probably not helpful to go in thinking of it like that. The case where you look at E<V of course breaks the conservation of energy the way you understand it there, this is part of why quantum physics is not intuitive at all.. In the context of the problems you are looking at, you use the wave-mechanical description of an electron (which describes it's state in a probabilistic framework as vanhees71 points out to you), rather than a particle with a definite position and energy that can be described at all times.

 

[amjad-sh]

Hi all!
Sorry for my late reply, I have been busy these two days.
I think I got it.
take the case where x<0 we have V=V0 and for x>0 we have V=V1.

If the particle is in region x<0, the Schroedinger equation in position representation is:
$-\hbar^2/2m \frac{\partial^2\psi(x)}{\partial x^2}+V0\psi(x)=E\psi(x)$
So $\hat V(x)=V_{0}$ here is the potential energy operator.
E is the allowed energy eigenvalue ant it is equal to E=K.E(kinetic energy of the particle) +Potential of the particle(and it may be not equal to$V_{0}$)
Here I have a question: what is the potential of the particle here and from where it came?
when the particle enters the region x>0,suppose here that E< $V_{1}$,the energy of the particle conserved (<E>=cst) the particle has a probability to exist in this region for a finite distance but it may rebounce back, because the transmission coefficient is zero.( I analysed it by myself is what I said correct?)

 

[blue_leaf77]

$V(x)$, where x<0 we have V=V0 and for x>0 we have V=V1 as you defined it, is the potential energy operator.
Bu to answer the question

what is the potential of the particle here

is like asking what is the momentum of the particle or what is the position of the particle? When you add "of whatever particle" following a quantity, you are talking about the measurement result of that quantity done on the said particle. And the most general answer is, as governed by QM, we don't know until a measurement is done. What you can do is calculating the average potential energy, but that requires you first solving for $\psi(x)$.

 

[Louis Fry]

What you can do is calculating the average potential energy, but that requires you first solving for $\psi(x)$.

I don't understand this so much myself but would that be defined over a specific region of space? So having solved for $\psi(x)$ in all regions, it would be a measurement of the "potential expectation value"?

$$<V> = \int_{x-domain} \psi(x)V(x)\psi^*(x)\,dx$$

 

[blue_leaf77]

I don't understand this so much myself but would that be defined over a specific region of space? So having solved for $\psi(x)$ in all regions, it would be a measurement of the "potential expectation value"?

$$<V> = \int_{x-domain} \psi(x)V(x)\psi^*(x)\,dx$$

The integration must be over the entire x axis.

 

[amjad-sh]

$V(x)$, where x<0 we have V=V0 and for x>0 we have V=V1 as you defined it, is the potential energy operator.
Bu to answer the question

is like asking what is the momentum of the particle or what is the position of the particle? When you add "of whatever particle" following a quantity, you are talking about the measurement result of that quantity done on the said particle. And the most general answer is, as governed by QM, we don't know until a measurement is done. What you can do is calculating the average potential energy, but that requires you first solving for $\psi(x)$.

In the case here we are talking about free particle, which is the particle that is moving in a constant potential V0, there is no restriction on energy for the free particle, but here I raise a question : why the potential of the particle is changing here? If we are forcing the particle to move from left to right and we are applying a constant potential energy operator which is V0 on the particle, should the total energy of the particle change in this case?

 

[vanhees71]

A constant potential doesn't do anything to the dynamics. It just shifts the ground-state energy from 0 to $V_0$, but absolute values of energy have no physical meaning (except in the general theory of relativity, which we don't consider here).

 

[blue_leaf77]

In the case here we are talking about free particle

I thought in post#25 you were considering a potential step, which is not normally called as free space.

why the potential of the particle is changing here?

You ask why the potential changes? In post #25, you invited us to consider a step potential. So.. isn't it you who defined the potential to change such that the overall profile looks like a step? Otherwise, do I understand you wording correctly?

 

[amjad-sh]

I thought in post#25 you were considering a potential step, which is not normally called as free space.

yes, you are right.

So.. isn't it you who defined the potential to change such that the overall profile looks like a step? Otherwise, do I understand you wording correctly?

yes but I meant just the case where x<0(before the particle reaches x>0 region), and I wanted to know if the particle while moving to x>0 if its (K.E or potential or the total energy(K.E+its potential )) changes or not.( I think they will not change)

 

[vanhees71]

For a free particle the energy is the kinetic energy, and it is conserved.

 

[Mark Harder]

the energy is actually obtained as the expectation value of the Hamiltonian, ⟨E⟩=⟨ψ|^H|ψ⟩⟨E⟩=⟨ψ|H^|ψ⟩\langle E \rangle = \langle \psi | \hat{H} | \psi \rangle.

The whole thread confuses me and I suspect others. Please forgive me, DrClaude, if I pick on you . In this case, the confusion is a failure to qualify what things the energies belong to: the particle, the barrier, or the sum of the 2. E<V when E, the particle's energy (E= U + K, to use "U" for potential of the particle and "K" for K.E. of the particle) is less than the potential energy required to surmount the barrier, which is the quantity V.

what is meant by "E < V" is that the total energy is lower than the barrier height V(x)=0 almost everywhere, except for V(x) = V₀ for a<x<b .

Also, I don't understand the geometry. From the above, it sounds like V(x) is a "bump" in potential over a finite interval, and zero potential everywhere else. Or is U(x) = U(a < x <b) <0, in other words the potential of the particle is negative while it is between the walls of zero potential? I think what is intended is a 1-D particle-in-a-box model, in which U₀ < V(barrier). In the PIB model, a particle can tunnel through a barrier of finite width. But as I (we?) defined it, the barrier is infinite in extent outside of (a,b), so that tunneling into, but not through, the barrier can occur. Is that clear, or am I completely lost?

 

[DrClaude]

I think what is intended is a 1-D particle-in-a-box model, in which U₀ < V(barrier).

The OP clearly mentioned a potential step, and in my post I discussed a potential barrier, which would give the same conceptual problem.

n this case, the confusion is a failure to qualify what things the energies belong to: the particle, the barrier, or the sum of the 2. E<V when E, the particle's energy (E= U + K, to use "U" for potential of the particle and "K" for K.E. of the particle) is less than the potential energy required to surmount the barrier, which is the quantity V.

Yes, that was the basic confusion. I and others have couched it in QM terms, discussing the difference between operators and expectation values. By the way, "the sum of the 2" doesn't make sense in this context.