# I Doesn't the superposition in energy violate the conservation of energy

#### ftr

It is said that for a particle in a box the energy is in a superposition. If indeed that is the case what happens when a measurement is made where does the excess energy go. Of course, that is based on my understanding is that superposition is a real physical and not platonic.

• Happiness
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#### Nugatory

Mentor
Our measuring device has to interact with the particle to measure its energy. That interation involves an exchange of energy, and the total energy of the quantum system consisting of the particle and the device is conserved.

• bhobba and Happiness

#### ftr

Our measuring device has to interact with the particle to measure its energy. That interation involves an exchange of energy, and the total energy of the quantum system consisting of the particle and the device is conserved.
Thanks for the reply. Let me clarify my question. When we measure we only get the eigenvalue, but we have said that the system was in SUPERPOSITION of many values, now for position for instants maybe I don't see a serious ramification but for energy its a bit hard for me to understand in regards to conservation of energy.

#### vanhees71

Science Advisor
Gold Member
I think you have a serious misconception about QT. It is important to clearly distinguish between states and observables. The observables are described by self-adjoint operators acting on some Hilbert space. In your case of the particle in a box (with ideally reflecting walls) a convenient choice is the Hilbert space of square integrable complex functions defined on the interval $[0,L]$, where $L$ is the length of the box. The corresponding wave functions must fulfill the boundary conditions $\psi(0)=\psi(L)=0$.

The energy is described by the Hamilton operator. Assuming the usual idealized textbook example you have
$$\hat{H}={\hat{p}^2}{2m}=-\frac{\hbar^2}{2m} \partial_x^2.$$
Now the energy eigenstates are defined by
$$\hat{H} u_E(x)=E u_E(x),$$
where $E$ is the eigenvalue. The equation reads
$$u_E''(x)=-k^2 u_E, \quad k^2=\frac{2 m E}{\hbar^2}.$$
The solution reads
$$u_E(x)=A \sin(k x).$$
To fulfill the boundary conditions you must have
$$k L=n \pi, \quad n \in \mathbb{N}=\{1,2,3,\ldots \}.$$
In the following I write $u_n$ instead of $u_E$, and the eigenvalues are given in terms of $n$ as
$$E_n=\frac{\hbar^2 k_n^2}{2m}=\frac{n^2 \pi^2 \hbar^2}{2m L^2}.$$
We further like to normalize the eigenfunctions
$$\int_0^L \mathrm{d} x |u_n(x)|^2=|A^2| L/2 \; \Rightarrow \; A=\sqrt{2/L}.$$
So we finally have
$$u_n(x)=\sqrt{\frac{2}{L}} \sin(k_n x).$$
Now you can expand any wave function in terms of these eigenfunctions. It's a Fourier series:
$$\psi(x)=\sum_{n=1}^{\infty} \psi_n u_n(x),$$
and
$$\psi_n=\int_0^{L} \mathrm{d} x u_n^*(x) \psi(x).$$
This shows not the energy values are in a superposition (this doesn't make any sense at all to begin with!), but the wavefunction representing the quantum state of the particle is written as a superposition of energy eigenstates!

Now if $\psi$ is normalized, i.e., if
$$\int_0^L \mathrm{d} x |\psi(x)|^2=\sum_{n=1}^{\infty} |\psi_n|^2=1,$$
then according to the rules of QT, when measuring the energy, the system being prepared in the state described by the wave function $\psi$, the probability to measure one of its possible values (which are the above calculated eigenvalues $E_n$, then the probability to find the specific eigenvalue $E_n$ is
$$P_n=|\psi_n|^2.$$
This implies that the energy value is determined if and only if for one $n$ you have $|\psi_n|^2=1$ and then consequently $\psi_{n'}=0$ for all $n' \neq n$. Thus the energy takes a determined value if and only if the state is described of an eigenstate of the Hamiltonian, which represents the energy of the system.

There's no violation of energy conservation, since the operator describing the time derivative of the energy is
$$\mathrm{\hat{H}}=\frac{1}{\mathrm{i} \hbar} [\hat{H},\hat{H}]=0,$$
i.e., the energy is conserved.

• microsansfil

#### ftr

I think you have a serious misconception about QT.
Thanks for the detail reply. Just like a good student most of the time I don't ask too many questions just do the calculations. But once in a while some questions pop into my head (usually not explicitly stated in textbooks) and upon searching for the answer I am relieved that I am not alone.

With your answer and many of these other ones it has become clearer but some unease remains. It was started with Arnold't TI which seems to make it a bit less hard since you only have "one energy" , of course I might be mistaken with my understanding of TI.

#### vanhees71

Science Advisor
Gold Member
What do you mean by "one energy"? Indeed there's one energy for any system, given by its Hamiltonian. Don't worry about the TI too much... It's better to learn the physics in the usual ("orthodox") interpretation first. It lets you concentrate on understanding the physics meaning of the theory rather than to scatter you mind in overly philosophical speculations. • bhobba

#### Happiness

It is said that for a particle in a box the energy is in a superposition. If indeed that is the case what happens when a measurement is made where does the excess energy go. Of course, that is based on my understanding is that superposition is a real physical and not platonic.
Based on Nugatory's reply, the short answer is, in your terms, the excess energy goes to the measuring system.
Our measuring device has to interact with the particle to measure its energy. That interation involves an exchange of energy, and the total energy of the quantum system consisting of the particle and the device is conserved.
Suppose the particle is in a superposition of energy E1 and E2. If the measuring device measures the energy of the particle to be E1, the device will have, after measurement, energy E3. If the measuring device measures E2, the device will have energy E4. And E1+E3=E2+E4, so energy is conserved.

• dextercioby

#### Demystifier

Science Advisor
2018 Award
It is said that for a particle in a box the energy is in a superposition. If indeed that is the case what happens when a measurement is made where does the excess energy go. Of course, that is based on my understanding is that superposition is a real physical and not platonic.
When the system is in a superposition of two or more energies, one should think of it as a system that does not have a well defined energy at all. A measurement of energy then gives energy a well defined value. Violation of energy conservation would occur if the system evolved from a state with one value of energy to a state with another value of energy. But here it is not what happens, instead we have an evolution from a state with no value of energy to a state with a value of energy. Hence we cannot say that conservation of energy is violated.

• microsansfil and vanhees71

#### Happiness

When the system is in a superposition of two or more energies, one should think of it as a system that does not have a well defined energy at all. A measurement of energy then gives energy a well defined value. Violation of energy conservation would occur if the system evolved from a state with one value of energy to a state with another value of energy. But here it is not what happens, instead we have an evolution from a state with no value of energy to a state with a value of energy. Hence we cannot say that conservation of energy is violated.
Does the conservation of energy imply that whenever a particle A is created to be in a superposition of energy $E_1$ and $E_2$, another particle B must also be created in a superposition of energy $E_1+c$ and $E_2+c$ such that if A is measured to have energy $E_1$, B's wave function is instantaneously collapsed to the state of $E_2+c$, and if A is measured to have energy $E_2$, B's wave function is instantaneously collapsed to the state of $E_1+c$, so as to keep the total energy of the Universe constant? ($c$ is a constant.)

So is it true that, to use @ftr's term, the "excess energy" of A doesn't go to the measuring device in general, but goes to B instead?

#### HomogenousCow

Only the expectation value of symmetry operators are conserved, and that is only under unitary time evolution. Measurements are not unitary since we’ve ignored the time evolution of the measuring device, hence the apparent “violation”. The energy expectation value of the system plus the measuring device is conserved).

#### Happiness

Measurements are not unitary since we’ve ignored the time evolution of the measuring device, hence the apparent “violation”.
Does this imply that measurements cannot be trusted to give the true energy of a particle? And so the principle of conservation of energy is unfalsifiable?

#### PeterDonis

Mentor
Does the conservation of energy imply that whenever a particle A is created to be in a superposition of energy $E_1$ and $E_2$, another particle B must also be created...
No.

is it true that, to use @ftr's term, the "excess energy" of A doesn't go to the measuring device in general, but goes to B instead?
No.

Does this imply that measurements cannot be trusted to give the true energy of a particle? And so the principle of conservation of energy is unfalsifiable?
No.

What all of these things do imply is that to properly evaluate conservation of energy at all, you have to consider a total system that does not interact with anything else. Trying to look at only parts of a system that are interacting with other parts doesn't work.

#### Happiness

Does the conservation of energy imply that whenever a particle A is created to be in a superposition of energy $E_1$ and $E_2$, another particle B must also be created in a superposition of energy $E_1+c$ and $E_2+c$ such that if A is measured to have energy $E_1$, B's wave function is instantaneously collapsed to the state of $E_2+c$, and if A is measured to have energy $E_2$, B's wave function is instantaneously collapsed to the state of $E_1+c$, so as to keep the total energy of the Universe constant? ($c$ is a constant.)
No.
How do we know that this is not the case? Particle B could be taken as the rest of the Universe apart from the newly created particle A.

#### PeterDonis

Mentor
Particle B could be taken as the rest of the Universe apart from the newly created particle A.
You can't create the rest of the universe in a chosen state. It is what it is. Nor can you measure the energy of the entire universe; you're part of the universe, and such a measurement would require you to be outside it, interacting it with a measuring device also outside it.

#### Vanadium 50

Staff Emeritus
Science Advisor
Education Advisor
Before we go too far down the rabbit hole, I would encourage anyone who thinks energy is not conserved in QM to describe an experiment that would measure this supposed non-conservation. Otherwise we will be chasing ghosts.

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• Nugatory and vanhees71

#### Happiness

You can't create the rest of the universe in a chosen state. It is what it is. Nor can you measure the energy of the entire universe; you're part of the universe, and such a measurement would require you to be outside it, interacting it with a measuring device also outside it.
Is QM consistent with the assertion that if a particle A is newly made to be in the superposed state $\psi_A=c_1\psi_{A, E_1}+c_2\psi_{A, E_2}$, then the rest of the Universe (apart from A) can be prescribed the state $\psi=c_2\psi_{E_3}+c_1\psi_{E_4}$, where $E_1+E_3=E_2+E_4$?

In other words, if A is in a superposed state, then the rest of the Universe must also be in a superposed state.

#### PeterDonis

Mentor
Is QM consistent with the assertion that if a particle A is newly made to be in the superposed state $\psi_{A, E_1}+c_2\psi_{A, E_2}$, then the rest of the Universe (apart from A) can be prescribed the state $\psi=c_2\psi_{E_3}+c_1\psi_{E_4}$, where $E_1+E_3=E_2+E_4$?
Mathematically, what you wrote down is of course consistent with QM, but there are an infinite number of states we could write down that are mathematically consistent with QM. That doesn't mean they mean anything physically.

Physically, we have no way of knowing what happens to the rest of the universe if we put a particular system into a superposition of energy eigenstates, since we don't know the state of the whole universe, or even if the whole universe is in some pure state that is an eigenstate of the Hamiltonian operator for the whole universe, even if we don't know exactly which one. And we certainly don't know how to write down explicitly the exact state transformation on the universe that describes the process of putting particle A in the state you describe. So the fact that what you wrote down is mathematically consistent with QM doesn't really tell us anything useful.

#### PeterDonis

Mentor
we don't know the state of the whole universe, or even if the whole universe is in some pure state that is an eigenstate of the Hamiltonian operator for the whole universe, even if we don't know exactly which one
In fact, we don't even know for sure that the concepts "state of the whole universe" and "Hamiltonian operator for the whole universe" are well-defined. All of our experimental knowledge of QM comes from experiments done on limited quantum systems by experimenters and apparatus that are not modeled as part of the quantum system and are considered to be outside it. But, as I said before, there is no way to make measurements or run experiments on the universe from outside it. So we cannot simply take it for granted that we can take the theory developed from measurements "from the outside" and apply it to a system for which there is no "outside".

#### Happiness

Does the conservation of energy imply that whenever a particle A is created to be in a superposition of energy $E_1$ and $E_2$, another particle B must also be created in a superposition of energy $E_1+c$ and $E_2+c$ such that if A is measured to have energy $E_1$, B's wave function is instantaneously collapsed to the state of $E_2+c$, and if A is measured to have energy $E_2$, B's wave function is instantaneously collapsed to the state of $E_1+c$, so as to keep the total energy of the Universe constant? ($c$ is a constant.)

So is it true that, to use @ftr's term, the "excess energy" of A doesn't go to the measuring device in general, but goes to B instead?
Does the principle of conservation of energy (PCoE) rule out the possibility of creating such an entangled pair of particles A and B (where now neither A and B is taken to be the rest of the Universe, to avoid the difficulty in dealing with the rest of the Universe)?

Is it the case that (the existence of) such entangled pair is ruled out because this would cause an instantaneous transfer/teleport of "excess energy" from one part of the Universe (where A is) to another part of the Universe (where B is), violating PCoE locally in both parts of the Universe? Or is it the case that such entangled pair is not rule out because such transfer/teleport of "excess energy" is somehow not actual but only apparent?

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#### microsansfil

When the system is in a superposition of two or more energies,
When the system is described as a superposition of two or more eigenstates of a hamiltonian (operator corresponding to the total energy of the studied system).

Can this eigenstates be interpreted as an energy of the studied system?

Energy is an observable, not a state, isn't it?

/Patrick

#### vanhees71

Science Advisor
Gold Member
In quantum theory an observable, including energy, takes a determined value if and only if the system is prepared in a state, for which this is the case. If it is a pure state it is an eigenstate of the Hamiltonian of the system.

Since the Hamiltonian plays the special role to generate the time evolution of the system, an eigenstate of the Hamiltonian is also a stationary state. In the Schrödinger picture the entire time evolution is carried by the states, and if you prepare the system to be initially in an eigenstate of the Hamiltonian, $|\psi_0 \rangle=|u_{E} \rangle$, then
$$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |u_E \rangle=\exp(-\mathrm{i} E t) |u_E \rangle.$$
This means the state ket depends on time only via a phase factor and thus it always represents the same state,
$$\hat{\rho}(t)=|\psi(t) \rangle \langle \psi(t) | =|u_E \rangle \langle u_E|=\hat{\rho}(0).$$
Any other initial state can be written as a superposition of energy eigenstates. If you superimpose eigenstates with different energy eigenvalues, it's of course not anymore an energy eigenstate and thus it describes a time dependent state, and energy is not determined.

Energy conservation means that energy doesn't change with time. If $\hat{A}$ is a self-adjoint operator (that is not explicitly time-dependent) representing an observable $A$, the operator
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i}} [\hat{A},\hat{H}]$$
represents the time derivative $\dot{A}$ of the observable $A$.

Thus the energy, represented by the Hamiltonian $\hat{H}$, is conserved, provided $\hat{H}$ is not explicitly time dependent.

In relation with states the conservation law implies the time-independence of the state, as we have already seen above, i.e., no matter what happens to the system, the system once prepared in an energy eigenstate, its state stays in this energy eigenstate, and thus the initially determined total energy of the system stays determined at all times, and the determined value of the energy doesn't change.

If the state is not an energy eigenstate, then energy conservation still holds for the average value of the energy. This is a special case of Ehrenfest's theorem. If the system is prepared in a pure state with the state ket $|\psi(t) \rangle$ The expectation value of the energy is given by
$$\langle E \rangle=\langle \psi(t)|\hat{H}|\psi(t) \rangle.$$
The expectation value for the time derivative of the energy is
$$\langle \dot{E} \rangle = \langle \psi(t)|\mathring{\hat{H}}|\psi(t) \rangle=0.$$
Infamously Bohr and Kramer shortly before the correct modern quantum theory was discovered hypothesized that energy conservation holds only on average in general.

That was famously disproven by Bothe with his ingenious invention of coincidence counting, where he could meausure the energy (and momenta) of the photon and the electron in Compton scattering, showing that total energy and momentum are conserved on an event-by-event basis, i.e., in any scattering event the energy-momentum conservation holds strictly true and not only on average (1924).

https://arxiv.org/abs/1106.1365

It's quite fitting that Bothe got his Nobel prize for his coincidence-counter method together with Born for his probabilistic interpretation of the QT, which clearly resolved the puzzle about the conservation laws on the microscales in the sense given above.

• microsansfil

#### microsansfil

$$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |u_E \rangle=\exp(-\mathrm{i} E t) |u_E \rangle.$$
What is mathematically E in this mathematical term $\exp(-\mathrm{i} E t)$? A linear operator?

If the system is prepared in a pure state with the state ket $|\psi(t) \rangle$ The expectation value of the energy is given by
$$\langle E \rangle=\langle \psi(t)|\hat{H}|\psi(t) \rangle.$$
I would have written this instead : $\langle \hat{H}\rangle_\psi =\langle \psi(t)|\hat{H}|\psi(t) \rangle.$

/Patrick

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#### vanhees71

Science Advisor
Gold Member
No, $E$ is the eigenvalue of $\hat{H}$. I never ever write an expectation value of an operator. It's the expectation value of an observable, represented by an operator. The former is, as the name says, an observable, the latter is a mathematical description of it.

#### PeterDonis

Mentor
Does the principle of conservation of energy (PCoE) rule out the possibility of creating such an entangled pair of particles A and B (where now neither A and B is taken to be the rest of the Universe, to avoid the difficulty in dealing with the rest of the Universe)?
No, why would it?

Is it the case that (the existence of) such entangled pair is ruled out because this would cause an instantaneous transfer/teleport of "excess energy" from one part of the Universe (where A is) to another part of the Universe (where B is), violating PCoE locally in both parts of the Universe? Or is it the case that such entangled pair is not rule out because such transfer/teleport of "excess energy" is somehow not actual but only apparent?
There is no "transfer/teleport of excess energy". Again you are incorrectly limiting your viewpoint to just the two particles, but any measurement on either particle will involve an interaction with a measuring apparatus, and conservation laws will only be satisfied for the full system of particles + apparatus, not for the particles alone, because the particles are not isolated. Conservation of energy (or indeed any conservation law) can only be expected to hold for an isolated system that doesn't interact with anything else.

#### Demystifier

Science Advisor
2018 Award
Does the conservation of energy imply that whenever a particle A is created to be in a superposition of energy $E_1$ and $E_2$, another particle B must also be created in a superposition of energy $E_1+c$ and $E_2+c$ such that if A is measured to have energy $E_1$, B's wave function is instantaneously collapsed to the state of $E_2+c$, and if A is measured to have energy $E_2$, B's wave function is instantaneously collapsed to the state of $E_1+c$, so as to keep the total energy of the Universe constant? ($c$ is a constant.)

So is it true that, to use @ftr's term, the "excess energy" of A doesn't go to the measuring device in general, but goes to B instead?
That's true if the initial state before those particle were created had a well defined energy. But in general this is not true.

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