- #1

- I
- Thread starter arpon
- Start date

- #1

- #2

- 31,131

- 10,047

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.

- #3

Physics Footnotes

Gold Member

- 37

- 81

When we talk about operators corresponding to

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.

- #4

- 15,972

- 7,277

- #5

- 13,001

- 550

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.

Last edited:

- #6

- 236

- 16

Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?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.

- #7

- 15,972

- 7,277

$$\mathrm{i} \hbar \partial_t \Psi(t,\vec{x})=\hat{H} \Psi(t,\vec{x})=\left [-\frac{\hbar^2}{2m} \Delta + V(\vec{x}) \right]\Psi(t,\vec{x}).$$

- #8

Physics Footnotes

Gold Member

- 37

- 81

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 aThe ##\Psi## of course form a Hilbert space, namely ##\mathrm{L}^2##.

Yes. And I capture that fact by incorporating time ##t## into theTime in QT is a parameter and not an observable

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.Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?

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

- #9

atyy

Science Advisor

- 14,042

- 2,333

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

- 236

- 16

Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.

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.

- #11

atyy

Science Advisor

- 14,042

- 2,333

But eigenkets of ##i\hbar \frac{\partial}{\partial t}## are not necessarily eigenkets of H.Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.

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

- 236

- 16

Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?For example, you can write the full time-dependent Schroedinger equation (your Eq 1) with several different Hamiltonians

- #13

bhobba

Mentor

- 9,483

- 2,572

Well you have this ordering problem.Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?

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

Last edited:

- #14

atyy

Science Advisor

- 14,042

- 2,333

Do you know how the "separation of variables" where you get from the time-dependent Schroedinger equation to the time-independent Schroedinger equation?Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?

- #15

- 236

- 16

Do you know how the "separation of variables" where you get from the time-dependent Schroedinger equation to the time-independent Schroedinger equation?

LetBut 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.

$$\Psi(x,t) = A(t) \psi(x)$$

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##).

- #16

- 15,972

- 7,277

You should really read a good basic textbook on quantum mechanics!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 aVector Space, due to the linearity of the Schrodinger Equation, they donotform an ##L^2##Hilbert Space.

Yes. And I capture that fact by incorporating time ##t## into thelabelof my wavefunctions (i.e. ##\psi_t##), each of whichdoesbelong 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 atime operatorperfectly analogous to theposition operator. Such a map ##T## is not an operator in the Hilbert Space, and does not correspond to a quantum observable!

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.

- #17

Physics Footnotes

Gold Member

- 37

- 81

If you are interpreting the ##t## in ##\Psi(x,t)## as

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

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.

- #18

- 13,001

- 550

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.

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

[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.

Last edited:

- #19

atyy

Science Advisor

- 14,042

- 2,333

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.But the eigenfunctions (of the Hamiltonain operator) ##\psi(x)## are not the same for different systems (i.e for different ##H##).

See Physics Footnotes's and dextercioby's posts for a more mathematically careful explanation of what is going on.

- #20

bhobba

Mentor

- 9,483

- 2,572

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.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.

So exactly what is

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

Science Advisor

- 14,042

- 2,333

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

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.

- #22

bhobba

Mentor

- 9,483

- 2,572

I will byte - a RHS.So what mathematical model would describe the space of all wavefunctions?

Thanks

Bill

- #23

bhobba

Mentor

- 9,483

- 2,572

So what you are saying theBasically, 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.

Thanks

Bill

- #24

atyy

Science Advisor

- 14,042

- 2,333

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.So what you are saying is theΨis, 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?

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

bhobba

Mentor

- 9,483

- 2,572

Sure - but that is still just a wave-function ie Ψ(x,t) what isLet'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.

Thanks

Bill

- Replies
- 4

- Views
- 8K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 557

- Last Post

- Replies
- 15

- Views
- 2K

- Replies
- 4

- Views
- 18K

- Last Post

- Replies
- 2

- Views
- 933

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 4K

- Replies
- 0

- Views
- 287