Time-energy uncertainty relation

[dyn]

Hi
I have just been looking at the derivation of the uncertainty relationship for non-commutating operators. I have come across the following quote in Quantum Mechanics by Mandl regarding the time-energy relationship. "Time is not an operator ; it is an ordinary parameter which commutes with all observables of a system ; in particular with the energy operator ; ie. the Hamiltonian of the system. Energy , momentum etc. are properties ie. observables of a system ; time is not. We can measure the position or momentum of a particle ; we do not measure the time of a particle. Rather , time is a parameter which identifies the instant at which we specify a property of a system. I have a few questions regarding this quote

1 - If we use the energy operator as iħ(∂/∂t) then t does not commute with that operator. So what is happening there ?

2 - If we set t=0 then surely we can measure t at a later time ? Similar to setting x=0 and then measuring x

3 - If we have a wavefunction Ψ ( x , t ) then are not both x and t independent variables which means they can both be measured ?

4 - This is not from the quote but is related ; the time-energy relationship is given as δEδt and the inequality is given slightly different forms in different book and i have yet to find a rigorous derivation as with the non-commutating observables. It is mostly derived by "hand-wavy" arguements. Does it have a rigorous derivation and an exact form ?

Thanks

 

[PeterDonis]

1 - If we use the energy operator as iħ(∂/∂t) then t does not commute with that operator.

No, the whole idea of $t$, as a parameter, "commuting" with any operator doesn't make sense. I think that Mandl was simply being sloppy when he said that $t$ as a parameter commutes with all operators.

2 - If we set t=0 then surely we can measure t at a later time ?

How do you "set" t = 0? In other words, how do you prepare a system such that t = 0? You can't. All you can do is declare, arbitrarily, that, for example, the time at which you prepared your system and started the experiment is $t = 0$.

You can, of course, "measure" $t$ by looking at a clock on the wall, for example; but that is not a "measurement" in the quantum sense because you are not treating the clock as part of the quantum system you are experimenting on. The clock is just providing "time" labels for various experimental events. That's part of what Mandl means when he says that time is a parameter, not an observable.

Similar to setting x=0 and then measuring x

No, it's not, because you can prepare a system to be at a given position, by, for example, confining it to that position until released. But you can't confine a system to be at a given time; time is going to move forward no matter what you do.

3 - If we have a wavefunction Ψ ( x , t ) then are not both x and t independent variables

No. What $\Psi(x, t)$ actually symbolizes is a continuous series of wave functions $\psi(x)$, parameterized by time $t$. Only $x$ is an independent variable (more precisely, a degree of freedom) of the function.

4 - This is not from the quote but is related ; the time-energy relationship is given as δEδt and the inequality is given slightly different forms in different book and i have yet to find a rigorous derivation as with the non-commutating observables.

There is no such derivation because there is no time observable.

You might find this article by John Baez helpful:

https://math.ucr.edu/home/baez/uncertainty.html

 

[dyn]

Thank you for that answer ; it is very helpful

 

[Demystifier]

1 - If we use the energy operator as iħ(∂/∂t) then t does not commute with that operator. So what is happening there ?

This is not the energy operator. The energy operator is the Hamiltonian such as
$$H=-\frac{\hbar^2\nabla^2}{2m}$$
This operator acts on functions of ${\bf x}$, not on functions of $t$. By "operators", in quantum mechanics we mean operators acting on functions of ${\bf x}$, or more generally, operators acting on vectors in a Hilbert space. When wave functions in quantum mechanics are viewed as functions of ${\bf x}$, then those functions can be interpreted as vectors in a Hilbert space. However, when they are viewed as functions of $t$, then they cannot be interpreted as vectors in a Hilbert space. That's because vectors in Hilbert space must have a finite norm. This works for functions of ${\bf x}$ because $\int d^3x\, |\psi|^2$ can be finite, but does not work for functions of $t$ because
$$\int_{-\infty}^{\infty} dt\, |\psi|^2$$
cannot be finite.

 

[otennert]

Thank you for that answer ; it is very helpful

And on that page, reference to a paper by Mandelstam and Tamm is given. This is in my opinion the most correct derivation and formulation of a time-energy uncertainty principle, which in many places is discussed very confusingly, because -- as mentioned here -- time is not an observable in QM.

Leonid Mandelstam and Igor Tamm, The uncertainty relation between energy and time in nonrelativistic quantum mechanics, Izv. Akad. Nauk SSSR (ser. Fiz.) 9 (1945) 122-128. (English translation: J. Phys. (USSR) 9 (1945), 249–254.)

It makes a statement on the concurrent measurement of energy $H$ and another observable $A$. If $\langle A\rangle (t)$ changes by $\Delta A$ after a characteristic time $\tau_A$, then $$\Delta H\cdot\tau_A\geq \frac\hbar 2.$$

If $A$ commutes with $H$, then $\tau_A$ is not defined, and there essentially is no time-energy uncertainty relation in this case.

 

[vanhees71]

See App. B in

https://arxiv.org/abs/2207.04898

 

[vanhees71]

No, the whole idea of $t$, as a parameter, "commuting" with any operator doesn't make sense. I think that Mandl was simply being sloppy when he said that $t$ as a parameter commutes with all operators.

In QT time is a real parameter, which of course commutes with all operators that represent observables (time is not an observable in QT, and $\mathrm{i}\hbar \partial_t$ does not represent an oservable either). Energy is represented by the Hamiltonian, $\hat{H}$, not by $\mathrm{i} \hbar \partial_t$.

 

[martinbn]

In QT time is a real parameter, which of course commutes with all operators that represent observables (time is not an observable in QT, and $\mathrm{i}\hbar \partial_t$ does not represent an oservable either). Energy is represented by the Hamiltonian, $\hat{H}$, not by $\mathrm{i} \hbar \partial_t$.

His point was that it makes no sense for a parameter and an operator to commute. What does it mean! How do you multiply them?

 

[dyn]

This is not the energy operator. The energy operator is the Hamiltonian such as
$$H=-\frac{\hbar^2\nabla^2}{2m}$$

In QT time is a real parameter, which of course commutes with all operators that represent observables (time is not an observable in QT, and $\mathrm{i}\hbar \partial_t$ does not represent an oservable either). Energy is represented by the Hamiltonian, $\hat{H}$, not by $\mathrm{i} \hbar \partial_t$.

Here is part of my confusion. In QM 2nd Ed. by Bransden , QM 2nd Ed. by Robinett and QM 3rd Ed. by Zettili the energy operator is given by iħ∂/∂t.

 

[otennert]

This is on p. 102 on Bransden/Joachain 2nd ed., correct? I can confirm you cite correctly, but I also say this is an absolutely misleading statement (or simply: wrong). As stated elsewhere, the energy operator is the Hamilton operator $H$. And the statement in the book refers to the fact that for Hamilton operators that do not explicitly depend on time, a separation ansatz $\psi(r,t)= \phi(r)\exp(-iE/\hbar t)$ delivers a stationary Schrödinger equation, which is an eigenvalue equation for $H$, and has the original $i\hbar d/dt$ replaced by $E$, if you will say so.

Therefore, Bransden/Joachain are also sloppy in their textbook at this point. However: in this context, again $E$ is a number, not an operator.

In any case, in QM there is no time operator, no matter how confusingly written some texts are. And the time-energy uncertainty relationship surely is one of the most often misleadingly formulated relations.

 

[topsquark]

I will add a quick note to otennert's comment: Most textbooks I've seen do this wrong.

-Dan

 

[dyn]

It is also quoted on P59 in Bransden , P66 in Robinett , P193 in Zettili.

I have always been confused why there was 2 energy operators in QM ; the energy operator iħ∂/∂t and the Hamiltonian operator.

It also seems that iħ∂/∂t is not Hermitian and thus can't represent an observable ? Is that correct ?

 

[otennert]

It is surely hermitian in a broader sense, but I am inclined to say it is not an operator at all. However, I may not get the reasoning waterproof here in just a couple of lines.

Beware: we are talking about operators in the Hilbert space of physical states, not as general mathematical entities in a general context!

How would $i\partial /\partial t$ or $i d/dt$ as an operator be defined on a general state $|\psi\rangle$ at a certain point in time, with nothing to start with?

The difference to $p=-i d/d x$ for example is: if you have $p$ act on a state, the canonical commutation relations eventually provide you with a meaningful result. But this is not so with $id/dt$...

 

[dyn]

I cannot see how iħ∂/∂t is Hermitian. To prove an operator is Hermitian such as p_x you usually perform an integral over all space or for p_x from x= +∞ to x = -∞. I don't see how this can apply to ∂/∂t

 

[otennert]

As I said: I think this is barking up the wrong tree. The issue is not whether $id/dt$ is hermitian or not. Rather it is whether it is a well-defined operator on Hilbert space at all! I say: no it isn't, but I may need more time to reflect...

 

[topsquark]

I cannot see how iħ∂/∂t is Hermitian. To prove an operator is Hermitian such as p_x you usually perform an integral over all space or for p_x from x= +∞ to x = -∞. I don't see how this can apply to ∂/∂t

Is a real number Hermitian? Technically, but it's not a matrix.

The equation $H \psi = E \psi$ is an eigen-equation. The LHS is an operator acting on a wavefunction and the RHS is a scalar multiplied by the wavefunction. $H \psi = i \hbar \partial _t \psi$ is a more explicit form of the eigen-equation. The RHS is an operator, yes, but it is a "scalar operator."

 

[dyn]

The momentum operator is not a matrix but it is an Hermitian operator and represents an observable.
iħ∂/∂t does not seem to be Hermitian and so shouldn't represent an observable but many textbooks and university lecture notes call it the energy operator

 

[Demystifier]

I cannot see how iħ∂/∂t is Hermitian. To prove an operator is Hermitian such as p_x you usually perform an integral over all space or for p_x from x= +∞ to x = -∞. I don't see how this can apply to ∂/∂t

It can't. One could try to re-define it as hermitian by performing the integral over all time, but then the partial integration creates a boundary term that doesn't vanish. It is this non-vanishing boundary term that makes it non-hermitian.

As a side remark, recently I discovered that sometimes even the integration over space may create a non-vanishing boundary term, which has surprising physical implications: https://arxiv.org/abs/2207.09140

 

[topsquark]

It can't. One could try to re-define it as hermitian by performing the integral over all time, but then the partial integration creates a boundary term that doesn't vanish. It is this non-vanishing boundary term that makes it non-hermitian.

As a side remark, recently I discovered that sometimes even the integration over space may create a non-vanishing boundary term, which has surprising physical implications: https://arxiv.org/abs/2207.09140

I am not going to argue your point...it makes sense and I've never gone that far with the concept. However, I now have a question. If the operator side of the equation is Hermitian and the other side isn't what do we Mathematically do to require that the eigenvalues of, in particular, energy to be real? A Hermitian operator must have real eigenvalues; a non-Hermitian matrix does not have that property. The only way I know how to do that is to presume that both sides are Hermitian or at least that one side is a scalar operator.

Or does this set a condition on form of the wavefunction?

-Dan

 

[vanhees71]

His point was that it makes no sense for a parameter and an operator to commute. What does it mean! How do you multiply them?

Since the operators representing observables in QT are linear operators defined on a Hilbert space, they commute by definition with all complex or real-valued quantities, and time is just such a real-valued parameter within QM.

Unfortunately it's not discussed in detail in most QM textbooks. The argument goes back to Pauli's famous "Handbuchartikel" about wave mechanics (a masterpiece by the way!): If time were an observable it would have to obey $[\hat{H},\hat{t}]=\mathrm{i} \hbar \hat{1}$ since $\hat{H}$ by definition generates the time evolution. Up to a conventional sign that's analogous to the commutation relation between position and momentum operators, and thus can be used to derive that then the spectrum of both $\hat{H}$ and $\hat{t}$ would always be entire $\mathbb{R}$, which is in clear contradiction to the discrete spectra of atoms. It is also in contradiction to the stability of matter since the Hamiltonian were not bounded from below, and there's be no stable ground state for any quantum-mechanical model. That's why time must be a parameter, represented by a "c-number" (in Dirac's language).

NB: In relativistic QFT also the position enters in terms of parameters first, i.e., as the spatial part of spacetime arguments of the local quantum fields. In this case it turns out that one can construct position operators/observables only for massive particles out of the corresponding irreducible representations of the proper orthochronous Poincare group. For massless particles you have a position operator only for spin-0 and spin-1/2 particles. For those with higher spin there's no position operator in the strict sense at all.

 

[martinbn]

Since the operators representing observables in QT are linear operators defined on a Hilbert space, they commute by definition with all complex or real-valued quantities, and time is just such a real-valued parameter within QM.

This is still unclear. What do you mean by complex or real valued quantity? Do you mean functions, defined on what? Or do you mean just complex or real numbers? Or something else? And for two things to comute you need to have an operation in order to say that the order in which you do it doesn't change the result.

What I find puzzling is why, in so many textbooks or forum posts, they cannot just say exactly what they mean!

 

[vanhees71]

A linear operator, $\hat{O}$, is linear, because for any $\lambda_1,\lambda_2 \in \mathbb{C}$ and any Hilbert-space vectors $|\psi_1\rangle$ and $\psi_2$ you have
$$\hat{O}[\lambda_1 |\psi_1 \rangle + \lambda_2 |\psi_1 \rangle]=\lambda_1 \hat{O} |\psi_1 \rangle + \lambda_2 \hat{O} |\psi_1 \rangle.$$
This implies that the operator commutes with any complex (and thus also real) number, and time is a real-valued parameter in QT for the reasons given in my previous posting, i.e., $t \hat{O}=\hat{O} t$.

 

[Demystifier]

If the operator side of the equation is Hermitian and the other side isn't what do we Mathematically do to require that the eigenvalues of, in particular, energy to be real? A Hermitian operator must have real eigenvalues; a non-Hermitian matrix does not have that property. The only way I know how to do that is to presume that both sides are Hermitian or at least that one side is a scalar operator.

You can do at least 3 things with vectors to get new vectors.
1. You can act on vector with an operator.
2. You can multiply the vector with a scalar.
3. You can add another vector to it.
In a computation of a time derivative you are essentially doing 3, because the time derivative is proportional to a sum of two vectors, namely $\psi(t)$ and $-\psi(t-dt)$.

 

[PeroK]

It is surely hermitian in a broader sense, but I am inclined to say it is not an operator at all. However, I may not get the reasoning waterproof here in just a couple of lines.

Beware: we are talking about operators in the Hilbert space of physical states, not as general mathematical entities in a general context!

How would $i\partial /\partial t$ or $i d/dt$ as an operator be defined on a general state $|\psi\rangle$ at a certain point in time, with nothing to start with?

The difference to $p=-i d/d x$ for example is: if you have $p$ act on a state, the canonical commutation relations eventually provide you with a meaningful result. But this is not so with $id/dt$...

It is an operator on a different space of functions. The Hilbert space in basic QM is a subset of the square integrable functions (leaving aside the issues of eigenfunctions of the position and momentum operators). In any case, these are functions of space only. The SDE describes how the state changes over time, with time now being an additional parameter. To view the SDE as an operator equation on a function space we need to consider functions of time and space. To talk about an operator being Hermitian on this function space, you would first have to define the inner product. And taking the integral over all space and time doesn't give a finite number and won't work.

The momentum operator is not a matrix but it is an Hermitian operator and represents an observable.
iħ∂/∂t does not seem to be Hermitian and so shouldn't represent an observable but many textbooks and university lecture notes call it the energy operator

It's the SDE that equates the complex time derivative with the Hamiltonian. Which is the sense in which it's the "energy operator". In this case, these are operators on the space of time-dependent functions, not on the space of square-integrable spatial wavefunctions.

 

[otennert]

"It is an operator on a different space of functions. The Hilbert space in basic QM is the a subset of the square integrable functions (leaving aside the issues of eigenfunctions of the position and momentum operators). In any case, these are functions of space only. The SDE describes how the state changes over time, with time now being an additional parameter. To view the SDE as an operator equation on a function space we need to consider functions of time and space. To talk about an operator being Hermitian on this function space, you would first have to define the inner product. And taking the integral over all space and time doesn't give a finite number and won't work."

Yes, this I would agree with! After a little research, I think some formal level of mathematics on this can be found in the context of a Bochner-Lebesgue integral.

 

[martinbn]

A linear operator, $\hat{O}$, is linear, because for any $\lambda_1,\lambda_2 \in \mathbb{C}$ and any Hilbert-space vectors $|\psi_1\rangle$ and $\psi_2$ you have
$$\hat{O}[\lambda_1 |\psi_1 \rangle + \lambda_2 |\psi_1 \rangle]=\lambda_1 \hat{O} |\psi_1 \rangle + \lambda_2 \hat{O} |\psi_1 \rangle.$$
This implies that the operator commutes with any complex (and thus also real) number, and time is a real-valued parameter in QT for the reasons given in my previous posting, i.e., $t \hat{O}=\hat{O} t$.

If $t$ is a fixed number, sure. But when $t$ stands for time, it is not just a single number, it chahnges. If you meant it as a fixed moment of time, what does $tO$ mean? Say 5seconds multiplied by the opreator $O$ or $O$ multiplied by 5seconds. What are those? So we can say that they are the same?

 

[PeroK]

If $t$ is a fixed number, sure. But when $t$ stands for time, it is not just a single number, it chahnges. If you meant it as a fixed moment of time, what does $tO$ mean? Say 5seconds multiplied by the opreator $O$ or $O$ multiplied by 5seconds. What are those? So we can say that they are the same?

We have a space of functions defined on four parameters: $(t, x, y, z)$. Each of those parameters can be mapped to an operator which multiplies the function by that parameter:
$$t \to \hat t: \ \hat t [\Psi(t, x)] = t\Psi(t,x)$$That's the usual formalism. And if observables are represented by operators that are functions of $\hat x$ and $\frac {\partial}{\partial x}$ etc., then $\hat t$ commutes with these operators.

 

[martinbn]

We have a space of functions defined on four parameters: $(t, x, y, z)$. Each of those parameters can be mapped to an operator which multiplies the function by that parameter:
$$t \to \hat t: \ \hat t [\Psi(t, x)] = t\Psi(t,x)$$That's the usual formalism. And if observables are represented by operators that are functions of $\hat x$ and $\frac {\partial}{\partial x}$ etc., then $\hat t$ commutes with these operators.

So, it is possible to write it clearly in two sentences.

 

[vanhees71]

If $t$ is a fixed number, sure. But when $t$ stands for time, it is not just a single number, it chahnges. If you meant it as a fixed moment of time, what does $tO$ mean? Say 5seconds multiplied by the opreator $O$ or $O$ multiplied by 5seconds. What are those? So we can say that they are the same?

This is really simple standard quantum theory. Time is a c-number parameter and not an operator. That's why it commutes with all linear operators by definition. I don't see, where your problem is coming from to multiply an operator with a scalar parameter. You need it, e.g., for the formal solution of the Schrödinger equation. With a time-independent Hamiltonian you get
$$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t/\hbar) |\psi(0) \rangle.$$

 

[dyn]

We have a space of functions defined on four parameters: $(t, x, y, z)$. Each of those parameters can be mapped to an operator which multiplies the function by that parameter:
$$t \to \hat t: \ \hat t [\Psi(t, x)] = t\Psi(t,x)$$That's the usual formalism. And if observables are represented by operators that are functions of $\hat x$ and $\frac {\partial}{\partial x}$ etc., then $\hat t$ commutes with these operators.

Is this post now saying that time is operator ? The t here even has a hat over it. If t is an operator then there is still the problem that t and the "energy operator" iħ∂/∂t won't commute.
The above posts seem to agree that iħ∂/∂t is not the energy operator. despite many textbooks calling it so. Is it any form of operator ; it it hermitian and does it represent an observable ?

 

[PeterDonis]

Each of those parameters can be mapped to an operator which multiplies the function by that parameter

I'm not sure what operators you are talking about. For the position operators, which I think is what you mean by the operators corresponding to $x$, $y$, and $z$, those operators only correspond to "multiply by $x$" (or $y$ or $z$) in the position representation. They don't in other representations. $t$ does not work that way; it is a scalar parameter and the only thing you can do with it is multiply by it, regardless of representation.

Also, $x$, $y$, and $z$ are not really parameters; they are labels for degrees of freedom (roughly speaking, dimensions in the Hilbert space). That is not a property that $t$ has.

 

[vanhees71]

$x,y,z$ are not "dimensions in the Hilbert space" but the eigenvalues of the self-adjoint operators that represent the components of the position vector of the particle in the 1st-quantization formalism of (non-relativistic) QT.

 

[PeterDonis]

$x,y,z$ are not "dimensions in the Hilbert space" but the eigenvalues of the self-adjoint operators that represent the components of the position vector of the particle in the 1st-quantization formalism of (non-relativistic) QT.

Strictly speaking, yes, $x$, $y$, and $z$, or more precisely each possible value for each of those, are eigenvalues. But all of the possible eigenvalues, taken together, form a set of 3 continuous parameters that label the possible range of variation in the Hilbert space for a single spinless particle, in the position representation. "Dimension" might not be precisely the right technical term for this, but this is what I was referring to. And none of these things are the same as what $t$ is.

 

[vanhees71]

It's a "generalized basis of eigenvectors" (to put it in the usual physicists' hand-waving slang). The Hilbert space of the 1st-quantization formalism is the separable Hilbert space, i.e., it has a countable basis, isomorphic to $\text{L}^2(\mathbb{R}^3)$. An example are the eigenfunctions of the 3D harmonic oscillator.

As stressed several times before, in QT $t$ is not an observable but a parameter; $t$ thus does also not denote eigenvalues of any operator.

In relativistic QFT $x=(ct,\vec{x})$ are Minkowski-vector valued parameters and also not eigenvalues of observables.