Energy operator and the Hamiltonian operator: Are they same?

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  • #2
PeterDonis
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@arpon please use LaTeX and ordinary text in your posts, instead of an image. The image makes it impossible to quote portions of your post. Help on the PF LaTeX feature can be found here:

https://www.physicsforums.com/help/latexhelp/

The answer to your question is no, they are not the same operator. Your equation (3) is wrong because it takes a time-dependent equation, (1), and equates it to a time-independent equation, (2). That's not valid.
 
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  • #3
Physics Footnotes
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I find it best when discussing Schrodinger's Equation to use uppercase ##\Psi=\Psi(x,t)## to capture the entire space-time behavior of a system, and lowercase ##\psi=\psi(x)## to denote the spatial behavior of a system at a particular time (possibly with a subscript, ##\psi_t##, to incorporate a specific time in the label of the function).

When we talk about operators corresponding to observables in the standard Quantum Mechanical formalism, we are really talking about operators on the ##\psi## functions, which form a Hilbert Space. The ##\Psi## functions, on the other hand, do not form a Hilbert Space, and operators on that space do not represent observables.

In other words, an operator like ##E:=i\hbar\frac {\partial}{\partial t}## acts on the ##\Psi## functions and is a completely different object to one like ##H:=\frac {d^2}{dx^2}## which acts on the ##\psi## functions, and so you can't compare them directly.
 
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  • #4
vanhees71
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The ##\Psi## of course form a Hilbert space, namely ##\mathrm{L}^2##. Time in QT is a parameter and not an observable, because otherwise it would be the canonical conjugate to the Hamiltonian, which then would not be bounded from below, and you'd not have a stable world. This argument goes back to Pauli's famous article on wave mechanics.
 
  • #5
dextercioby
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Technically, the time-evolution defines a differentable mapping ## \Psi (t,.) : I \subseteq \mathbb{R} \mapsto \mathcal{M} ##, where ##\mathcal{M}## can be given a certain structure as follows: if the Hamiltonian H is a self-adjoint linear operator with pure point spectrum in a complex separable Hilbert space, then ##\mathcal{M}## is the linear manifold of all functions ##e^{i E_n t} \psi_n (.)##, where ##E_n## and ##\psi_n## are the solutions of the spectral equation for H in that particular Hilbert space. Simple example: the harmonic oscillator in one dimension.
 
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In other words, an operator like ##E:=i\hbar\frac {\partial}{\partial t}## acts on the ##\Psi## functions and is a completely different object to one like ##H:=\frac {d^2}{dx^2}## which acts on the ##\psi## functions, and so you can't compare them directly.
Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?
 
  • #7
vanhees71
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Sure, the Schrödinger equation reads
$$\mathrm{i} \hbar \partial_t \Psi(t,\vec{x})=\hat{H} \Psi(t,\vec{x})=\left [-\frac{\hbar^2}{2m} \Delta + V(\vec{x}) \right]\Psi(t,\vec{x}).$$
 
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  • #8
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The ##\Psi## of course form a Hilbert space, namely ##\mathrm{L}^2##.
In my notation, ##\Psi:=\Psi (x,t):{\mathbb R}^2 \to \mathbb C## is not a wavefunction in the quantum-mechanical sense, but rather a function capturing the entire space-time dependence of a system. To see that these functions do not form (or generally even belong to) ##L^2(\mathbb R^2,\mathbb C)## you need only consider the simple case of a stationary state of a system corresponding to an energy value ##E##, say, for which the space-time function has the form $$\Psi (x,t):(x,t)\mapsto e^{-iEt}\phi(x)$$A simple calculation shows $$\parallel \Psi \parallel^2:=\iint_{\mathbb R^2}|e^{-iEt}\phi(x)|^2dxdt=\infty$$In other words, even though the ##\Psi## form a Vector Space, due to the linearity of the Schrodinger Equation, they do not form an ##L^2## Hilbert Space.
Time in QT is a parameter and not an observable
Yes. And I capture that fact by incorporating time ##t## into the label of my wavefunctions (i.e. ##\psi_t##), each of which does belong to the Hilbert Space ##L^2(\mathbb R, \mathbb C)##.
Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?
Yes, of course. BUT... when you are interpreting ##H## to act on ##\Psi## functions you should really use a different symbol to when you are interpreting ##H## to act on ##\psi## functions, as the latter is an observable while the former is not.

Of course, in practice physicists do not typically make these symbolic distinctions, but it is extremely useful to do so when you are just learning or else you carry misconceptions into the more general Hilbert Space formalism. For example, when you make these careful distinctions, you will not be tricked, as so many people are, into thinking that multiplication by the time variable, ##T:\Psi(x,t) \mapsto t\Psi(x,t)##, should yield a time operator perfectly analogous to the position operator. Such a map ##T## is not an operator in the Hilbert Space, and does not correspond to a quantum observable!
 
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  • #9
atyy
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To the OP:

As you write, Eq 2 is true only if ##|\Psi \rangle## is an energy eigenket.

In contrast, Eq 1 is true for any ##|\Psi \rangle##. For example, in Eq 1, ##|\Psi \rangle## can be a sum of energy eigenkets.
 
  • #10
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To the OP:

As you write, Eq 2 is true only if ##|\Psi \rangle## is an energy eigenket.

In contrast, Eq 1 is true for any ##|\Psi \rangle##. For example, in Eq 1, ##|\Psi \rangle## can be a sum of energy eigenkets.
Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.
 
  • #11
atyy
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Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.
But eigenkets of ##i\hbar \frac{\partial}{\partial t}## are not necessarily eigenkets of H.

For example, you can write the full time-dependent Schroedinger equation (your Eq 1) with several different Hamiltonians (eg. Hamiltonian for the simple harmonic oscillator or Hamiltonian for the hydrogen atom). The eigenkets of each Hamiltonian will be eignekets of ##i\hbar \frac{\partial}{\partial t}##, but eigenkets of the simple harmonic oscillator are not eigenkets of the hydrogen atom.
 
  • #12
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For example, you can write the full time-dependent Schroedinger equation (your Eq 1) with several different Hamiltonians
Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?
 
  • #13
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Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?
Well you have this ordering problem.

In QM operator order matters - in classical physics it doesn't. So when one writes the corresponding quantum Hamiltonian it, on rare occasions, is ambiguous. When that happens only by working through both options and seeing what happens can you decide
https://physics.stackexchange.com/questions/46988/operator-ordering-ambiguities

QM is a tricky devil isn't it. You think you have one thing down pat and it slips from you.

A little project for the advanced - have a read of Chapter 3 of Ballentine, see what he says, and compare it to the above.

Thanks
Bill
 
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  • #14
atyy
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Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?
Do you know how the "separation of variables" where you get from the time-dependent Schroedinger equation to the time-independent Schroedinger equation?
 
  • #15
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Do you know how the "separation of variables" where you get from the time-dependent Schroedinger equation to the time-independent Schroedinger equation?
But eigenkets of ##i\hbar \frac{\partial}{\partial t}## are not necessarily eigenkets of H.

For example, you can write the full time-dependent Schroedinger equation (your Eq 1) with several different Hamiltonians (eg. Hamiltonian for the simple harmonic oscillator or Hamiltonian for the hydrogen atom). The eigenkets of each Hamiltonian will be eignekets of ##i\hbar \frac{\partial}{\partial t}##, but eigenkets of the simple harmonic oscillator are not eigenkets of the hydrogen atom.
Let
$$\Psi(x,t) = A(t) \psi(x)$$
Applying Schrodinger's Time dependent equation:
$$\begin{equation}
i\hbar\frac{\partial}{\partial t}\left(A(t)\psi(x)\right) = H\left(A(t)\psi(x)\right)
\end{equation}$$
Let ##\psi(x)## is an eigenfunction of ##H## with eigenvalue ##E##. So, we get
$$\begin{align}
i\hbar\frac{d}{dt}\left(A(t)\right)\cdot \psi(x) = EA(t)\psi(x)\\
\implies i\hbar\frac{d}{dt}\left(A(t)\right) = EA(t)\\
\implies A(t) = e^{-\frac{iEt}{\hbar}}\\
\implies \Psi(x,t) = \psi (x) e^{-\frac{iEt}{\hbar}}
\end{align}$$

So, the general solution is
$$\Psi(x,t) = \sum \psi _n(x) e^{-\frac{iE_nt}{\hbar}}$$

But the eigenfunctions (of the Hamiltonain operator) ##\psi(x)## are not the same for different systems (i.e for different ##H##).
 
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  • #16
vanhees71
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In my notation, ##\Psi:=\Psi (x,t):{\mathbb R}^2 \to \mathbb C## is not a wavefunction in the quantum-mechanical sense, but rather a function capturing the entire space-time dependence of a system. To see that these functions do not form (or generally even belong to) ##L^2(\mathbb R^2,\mathbb C)## you need only consider the simple case of a stationary state of a system corresponding to an energy value ##E##, say, for which the space-time function has the form $$\Psi (x,t):(x,t)\mapsto e^{-iEt}\phi(x)$$A simple calculation shows $$\parallel \Psi \parallel^2:=\iint_{\mathbb R^2}|e^{-iEt}\phi(x)|^2dxdt=\infty$$In other words, even though the ##\Psi## form a Vector Space, due to the linearity of the Schrodinger Equation, they do not form an ##L^2## Hilbert Space.
Yes. And I capture that fact by incorporating time ##t## into the label of my wavefunctions (i.e. ##\psi_t##), each of which does belong to the Hilbert Space ##L^2(\mathbb R, \mathbb C)##.
Yes, of course. BUT... when you are interpreting ##H## to act on ##\Psi## functions you should really use a different symbol to when you are interpreting ##H## to act on ##\psi## functions, as the latter is an observable while the former is not.

Of course, in practice physicists do not typically make these symbolic distinctions, but it is extremely useful to do so when you are just learning or else you carry misconceptions into the more general Hilbert Space formalism. For example, when you make these careful distinctions, you will not be tricked, as so many people are, into thinking that multiplication by the time variable, ##T:\Psi(x,t) \mapsto t\Psi(x,t)##, should yield a time operator perfectly analogous to the position operator. Such a map ##T## is not an operator in the Hilbert Space, and does not correspond to a quantum observable!
You should really read a good basic textbook on quantum mechanics!

Again: The Schrödinger equation reads (for 1D motion and with ##\hbar=1##)
$$\mathrm{i} \partial_t \Psi(t,x)=\hat{H} \Psi(t,x)=-\frac{1}{2m} \partial_x^2 \Psi(t,x)+V(x) \Psi(t,x).$$
The wave function ##\Psi## as a function of ##x## at fixed ##t## MUST be square integrable to be a representative of a pure state (i.e., defining a unit ray in Hilbert space, representing this state). Since ##\hat{H}## is self adjoint this means that the initial condition
$$\Psi(t=0,x)=\psi_0(x)$$
must be a square-integrable function. Then ##\Psi## at any ##t## is square integrable and
$$\int_{\mathbb{R}} \mathrm{d} x |\Psi(t,x)|^2=\int_{\mathbb{R}} \mathrm{d} x |\psi_0(x)|^2=\text{const}.$$
The constant is usually fixed to 1, so that ##|\Psi(t,x)|## is the probability distribution for the position at time ##t##.

If ##\psi_0## is an eigenfunction of ##\hat{H}##,
$$\hat{H} \psi_0(x)=E \psi_0(x)$$
then you can immediately integrate the Schrödinger equation to give
$$\Psi(t,x)=\exp(-\mathrm{i} E t) \psi_0(x),$$
and of course
$$\int_{\mathbb{R}} \mathrm{d} x |\Psi(t,x)|^2=\int_{\mathbb{R}} \mathrm{d} x |\psi_0(x)|^2=1.$$
These are the most basic facts about QT, and you really have to learn this in a good textbook like, e.g.,

J.J. Sakurai, Modern quantum mechanics, 2nd edition.
 
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  • #17
Physics Footnotes
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@vanhees71 you are completely missing the point of my notation. Of course for fixed ##t=t_0## the function ##\Psi (t_0,x)## is a square-integrable function, considered now as a function over ##x\in \mathbb R##. However, as I made very clear, I use the notation ##\Psi## to denote the entire functional dependence ##\mathbb R^2\to\mathbb C##. This makes my ##\Psi## the quantum equivalent of a world-line, rather than a state. I reserve the notations ##\psi## for a quantum state, and ##\psi_t## for a parametrised family of quantum states, each of which belongs to ##L^2(\mathbb R)## obviously.

If you are interpreting the ##t## in ##\Psi(x,t)## as fixed, making it a function of a single variable, then you are technically in error writing something like ##\mathrm{i} \partial_t \Psi(x,t)##, as you did above, since ##t## is no longer a variable.

Your rehash of basic QM is both irrelevant and oddly condescending (given that the misunderstanding is your own). I am using a careful notation (not your or Sakurai's notation) to distinguish functions over space-time, which do not form a Hilbert Space, from functions on space (more correctly 'configuration space') at a particular time which do form a Hilbert Space.

Moreover, I have also clearly justified my notation, both in clarifying the question posed by the OP, and in avoiding common misconceptions people have about the role played by time in the standard QM formalism.

You are of course welcome to dislike my use of the symbols ##\Psi##, ##\psi##, and ##\psi_t##, however you need to keep in mind that I wasn't talking to you; I was talking to @arpon.
 
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  • #18
dextercioby
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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.
 
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  • #19
atyy
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But the eigenfunctions (of the Hamiltonain operator) ##\psi(x)## are not the same for different systems (i.e for different ##H##).
OK, why don't you try something concrete to see what I mean. Write the Hamiltonian of the harmonic oscillator, and the Hamiltonian of the hydrogen atom. Then get the eigenfunctions of the respective Hamiltonians. You will see that the eigenfunctions of both Hamiltonians are eigenfunctions of ##i\hbar \frac{\partial}{\partial t}## because of the ##\exp{(iEt)}## term. However, the eigenfunctions of the harmonic oscillator are not eigenfunctions of the hydrogen atom. So roughly speaking, ##i\hbar \frac{\partial}{\partial t}## acts on the time part of the wave functions, while the Hamiltonians act on the space part.

See Physics Footnotes's and dextercioby's posts for a more mathematically careful explanation of what is going on.
 
  • #20
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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.
I must say I am not getting this. The state is either an abstract vector or if you expand it in eigenfunctions of position its a wave-function - which is just a representation of the actual vector - not the vector itself. That's basic linear algebra with a bit of RHS's stitched on to get the continuous eigenvectors.

So exactly what is ψ and ψ? I am pretty sure ψ is the wave-function and I thought ψ was the actual vector but I got totally confused when it was said it was a function. What exactly is going on?

BTW I understand and agree completely with what Dextercioby wrote above - but some of this other stuff has me - well confounded. Can anyone enlighten me?

I am getting the feeling my fellow Brisbaneite, Physics Footnotes, like Vanhees thinks, may have some kind of misunderstanding, but this does not gel with other stuff I have seen him write that shows a good understanding of QM - like I said for me its really confounding. Hopefully enlightenment will come soon. Maybe I am the one confused - it wouldn't be the first nor will it be the last time that happens.

Thanks
Bill
 
  • #21
atyy
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So exactly what is ψ and ψ? I am pretty sure ψ is the wave-function and I thought ψ was the actual vector but I got totally confused when it was said it was a function. What exactly is going on?
Basically, to use QM we usually fix a preferred frame, with its notion of simultaneity (a global instant of time). The Hilbert space of wave functions refers to wave functions at any fixed instant of time. Maybe it's a little easier to see this in the Heisenberg picture.

If you allow time, then for an eigenfunction you have exp(iEt) which clearly is not square integrable, so not part of what we usually mean by Hilbert space.
 
  • #23
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Basically, to use QM we usually fix a preferred frame, with its notion of simultaneity (a global instant of time). The Hilbert space of wave functions refers to wave functions at any fixed instant of time. Maybe it's a little easier to see this in the Heisenberg picture.
So what you are saying the Ψ is, the actual vector and its dependent of (x,t)? I get dependent on t, but x? Whats the idea there? Is this relativistic frame jumping?

Thanks
Bill
 
  • #24
atyy
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So what you are saying is the Ψ is, the actual vector and its dependent of (x,t)? I get dependent on t, but x? Whats the idea there? Is this relativistic frame jumping?
Let's say you do the simple harmonic oscillator. From separation of variable you can get a spatial eigenfunction of the Hamiltonian, say ##\psi(x)##. Then the time evolution will be given by ##\Psi(x,t) = \psi(x) exp(iEt)##. Now is ##\Psi(x,t)## a member of a Hilbert space? It is if you fix t. But if t is not fixed, the ##exp(iEt)## is not square integrable.

It is right to fix t, because the normalization guarantees that the particle can be found somewhere at any particular time (any fixed t).
 
  • #25
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Let's say you do the simple harmonic oscillator. From separation of variable you can get a spatial eigenfunction of the Hamiltonian, say ##\psi(x)##. Then the time evolution will be given by ##\Psi(x,t) = \psi(x) exp(iEt)##. Now is ##\Psi(x,t)## a member of a Hilbert space? It is if you fix t. But if t is not fixed, the ##exp(iEt)## is not square integrable.
Sure - but that is still just a wave-function ie Ψ(x,t) what is Ψ (x,t)?

Thanks
Bill
 

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