Energy operator and the Hamiltonian operator: Are they same?

  • #26
atyy
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Sure - but that is still just a wave-function ie Ψ(x,t) what is Ψ (x,t)?
Same thing, different notation.
 
  • #27
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@bhobba (my fellow Brisbaneite!), nothing I am saying contradicts standard textbook QM. I am simply choosing a more mathematically careful way of using our ubiquitous ##\psi## friend in two different ways: first in a concrete version of Schrodinger's Equation, considered as just a partial differential equation, and second as a function/vector in a Hilbert Space, corresponding to the more sophisticated and abstract treatment of QM. Maybe a simple example will help clarify.

Consider the following two expressions:
$$\Psi (x,t):\mathbb R^2\ni (x,t)\mapsto e^{-iEt}\phi(x)\in \mathbb C$$ and $$\psi _t(x):\mathbb R \ni x \mapsto e^{-iEt}\phi(x)\in \mathbb C$$
The first function represents a solution to Schrodinger's Equation, but being a function of two variables (configuration space and time) it is not what we mean by a 'quantum state', or a 'quantum wavefunction', since it contains the full space/time dependence; a state is something corresponding to a particular instant in time.

The second expression is a function of one variable, ##x\in\mathbb R##, for any specified instant ##t##. Moreover, for any given time ##t##, this function is a square-integrable element of the Hilbert Space ##L^2(\mathbb R)##, and can rightfully be called a quantum wavefunction/state-vector, or what have you.

The reason it is helpful to distinguish these formulae becomes clear when you are performing operations involving time, such as applying the operator ##
E:=i\hbar\frac {\partial}{\partial t}##. This operator applies properly to the first type of object ##\Psi(x,t)##, meaning that it is not a Hilbert Space operator, and in particular does not correspond to an observable (something countless people get wrong).

Furthermore, lest one be tempted to conceive of ##E:=i\hbar\frac {\partial}{\partial t}## as an observable by reversing the role of ##x## and ##t## (that is, thinking of ##\psi_x(t)## as a function of time for any given fixed value of ##x##), will be met with the cold hard fact that these functions do not form a Hilbert Space, as they are not in general square integrable (as you can see by doing the integration on the above state yourself). Even in the Dirac Equation, space and time are not symmetric in this way.

The more I talk about this, the more complicated I'm making a simple fact sound: The ##\Psi(x,t)## you see in Schrodinger's Equation all the time do not represent quantum wavefunctions/state-vectors; they are quantum world-lines.
 
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  • #28
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Furthermore, lest one be tempted to conceive of ##E:=i\hbar\frac {\partial}{\partial t}## as an observable by reversing the role of ##x## and ##t## (that is, thinking of ##\psi_x(t)## as a function of time for any given fixed value of ##x##), will be met with the cold hard fact that these functions do not form a Hilbert Space, as they are not in general square integrable (as you can see by doing the integration on the above state yourself). Even in the Dirac Equation, space and time are not symmetric in this way.
If that's what you mean then I get it.

Thanks for clarifying.

Thanks
Bill
 
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  • #29
vanhees71
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See, Hendrik, this is where you are wrong because the typical books of QM don't properly address the time-dependence of the wave functions from a mathematical perspective. Technically, in ##\Psi (x,t)## subject to Schroedinger's equation both x and t carry the same mathematical weight, so, depending on the system (i.e. its observables), they are expected to be in ##\mathbb{R}## with "t" necessarily from -infinity to +infinity. So one naively writes the so-called "normalization" condition (which is typical for a true Hilbert space) as you did:

[tex] \int_{\mathbb{R}} dx |\Psi (x,t)|^2 = 1 [/tex]

which is wrong, if "t" "runs". And it must "run" because the Schroedinger equation involves a partial derivative wrt to "t". You can attempt to fix it as

[tex] \int_{\mathbb{R}\times\mathbb{R}} dx {}{}~{} dt ~ |\Psi (x,t)|^2 = 1 [/tex]

which is also incorrect, because there's no reason to expect time conservation of probabilities in a time-varying interaction.

So what mathematical model would describe the space of all wavefunctions? Definitely, for a system with a Hamiltonian independent of time "moving" unrestricted in Euclidean space, one has

[tex] \Psi (x,t): \mathbb{R} \times \mathbb{R} \mapsto \mbox{span} \left(e^{iE_n t} \psi_n (x) \right)[/tex]

I think that geometrically, just as the infinite cylinder in ##\mathbb{R}^3## is the trivial fiber bundle ##\mbox{unit circle} \times \mathbb{R}##, the space of all wave functions - only in this simple scenario of time-independent Hamiltonians - is a trivial fibering of ##L^2(\mathbb{R})## by the unit circle, with a typical fiber being a class of equivalence of functions from ##L^2(\mathbb{R})##. If the Hamiltonian is time dependent, there's no reason to expect a separation of variables (coordinates and time), so from a geometrical perspective, things get really complicated and I don't know of a model to describe this space.
No, here you are wrong. The Schrödinger equation is by construction norm conserving, and you must not integrate over ##t## since ##t## is not an observable. Also ##\hat{H}## can be time dependent. As long as it is self-adjoint, the Schrödinger equation is still norm conserving. In such a case the energy-eigenvectors are time-dependent even in the Schrödinger picture.
 
  • #30
vanhees71
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@bhobba (my fellow Brisbaneite!), nothing I am saying contradicts standard textbook QM. I am simply choosing a more mathematically careful way of using our ubiquitous ##\psi## friend in two different ways: first in a concrete version of Schrodinger's Equation, considered as just a partial differential equation, and second as a function/vector in a Hilbert Space, corresponding to the more sophisticated and abstract treatment of QM. Maybe a simple example will help clarify.

Consider the following two expressions:
$$\Psi (x,t):\mathbb R^2\ni (x,t)\mapsto e^{-iEt}\phi(x)\in \mathbb C$$ and $$\psi _t(x):\mathbb R \ni x \mapsto e^{-iEt}\phi(x)\in \mathbb C$$
The first function represents a solution to Schrodinger's Equation, but being a function of two variables (configuration space and time) it is not what we mean by a 'quantum state', or a 'quantum wavefunction', since it contains the full space/time dependence; a state is something corresponding to a particular instant in time.

The second expression is a function of one variable, ##x\in\mathbb R##, for any specified instant ##t##. Moreover, for any given time ##t##, this function is a square-integrable element of the Hilbert Space ##L^2(\mathbb R)##, and can rightfully be called a quantum wavefunction/state-vector, or what have you.

The reason it is helpful to distinguish these formulae becomes clear when you are performing operations involving time, such as applying the operator ##
E:=i\hbar\frac {\partial}{\partial t}##. This operator applies properly to the first type of object ##\Psi(x,t)##, meaning that it is not a Hilbert Space operator, and in particular does not correspond to an observable (something countless people get wrong).

Furthermore, lest one be tempted to conceive of ##E:=i\hbar\frac {\partial}{\partial t}## as an observable by reversing the role of ##x## and ##t## (that is, thinking of ##\psi_x(t)## as a function of time for any given fixed value of ##x##), will be met with the cold hard fact that these functions do not form a Hilbert Space, as they are not in general square integrable (as you can see by doing the integration on the above state yourself). Even in the Dirac Equation, space and time are not symmetric in this way.

The more I talk about this, the more complicated I'm making a simple fact sound: The ##\Psi(x,t)## you see in Schrodinger's Equation all the time do not represent quantum wavefunctions/state-vectors; they are quantum world-lines.
That's a clear misconception and in fact contradicts quantum theory. The Hamiltonian is NOT ##\mathrm{i} \partial_t## but a function(al) of some set of fundamental operators (in single-particle non-relativistic QT that's the position, momentum, and spin operators). Sometimes ##\hat{H}## is also explicitly time dependent (e.g., the motion of a charged particle in a time-dependent em. field), i.e.,
$$\hat{H}=\hat{H}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{S}},t).$$
 
  • #31
Physics Footnotes
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That's a clear misconception and in fact contradicts quantum theory. The Hamiltonian is NOT ##\mathrm{i} \partial_t## but a function(al) of some set of fundamental operators (in single-particle non-relativistic QT that's the position, momentum, and spin operators). Sometimes ##\hat{H}## is also explicitly time dependent (e.g., the motion of a charged particle in a time-dependent em. field), i.e.,
$$\hat{H}=\hat{H}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{S}},t).$$
Ummm... is this addressed to me? If so, where did I mention the word 'Hamiltonian'?? I was addressing a common confusion over the status of the operator ##\mathrm{i}\hbar \partial_t##.
 
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  • #32
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The two operators are the same on the space of solutions of the Schrodinger equation. On a more general space of all functions of space and time, those are different operators.
 
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  • #33
dextercioby
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[...] I was addressing a common confusion over the status of the operator ##\mathrm{i}\hbar \partial_t##.
And to be really fair (i.e. to show that I've checked the answers on our competition's website - where both of us already posted on a similar topic), the mathematical sense in which these "quantum worldlines" should be looked at is wonderfully presented in this topic (question properly formulated and the answer to it).

https://physics.stackexchange.com/q...r-as-expressed-in-abstract-hilbert-space-vs-a
 
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  • #34
vanhees71
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Ok, obviously here's a clash of different languages going on. From a physicists point of view it is utmost important to be very clear about the fact that time in quantum theory is NOT an observable but a parameter for the simple reason that otherwise it would be the canonical conjugate to the Hamiltonian, which then due to the analogy with the position-momentum commutation relation would have entire ##\mathbb{R}## as its spectrum and thus there'd be no stable ground state.

In the abstract (representation-free) Hilbert-space formalism the choice of the time dependence is determined only up to a time-dependent unitary transformation, i.e., you can shuffle the time dependence between the operators representing observables and the statistical operator (or state vectors representing the rays standing for the special case of pure states) quite arbitrarily. Of course, this choice of the picture of time evolution doesn't affect any physical outcomes, i.e., probabilities or probability distributions for the outcome of measurements or transition-matrix elements and the like, but I think that's not the issue here.
 

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