- #1
Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?Physics Footnotes said: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.
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.vanhees71 said:The ##\Psi## of course form a Hilbert space, namely ##\mathrm{L}^2##.
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)##.vanhees71 said:Time 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.arpon said:Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?
Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.atyy said: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.
arpon said:Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.
Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?atyy said:For example, you can write the full time-dependent Schroedinger equation (your Eq 1) with several different Hamiltonians
arpon said:Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?
arpon said:Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?
atyy said:Do you know how the "separation of variables" where you get from the time-dependent Schroedinger equation to the time-independent Schroedinger equation?
Letatyy said: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.
You should really read a good basic textbook on quantum mechanics!Physics Footnotes said: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!
arpon said:But the eigenfunctions (of the Hamiltonain operator) ##\psi(x)## are not the same for different systems (i.e for different ##H##).
dextercioby said: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.
bhobba said: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?
dextercioby said:So what mathematical model would describe the space of all wavefunctions?
atyy said: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.
bhobba said: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?
atyy said: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.
bhobba said:Sure - but that is still just a wave-function ie Ψ(x,t) what is Ψ (x,t)?
Physics Footnotes said: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.
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.dextercioby said: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.
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.,Physics Footnotes said:@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.
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##.vanhees71 said: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.,
$$\hat{H}=\hat{H}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{S}},t).$$
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.arpon said:
Physics Footnotes said:[...] I was addressing a common confusion over the status of the operator ##\mathrm{i}\hbar \partial_t##.
The energy operator and the Hamiltonian operator are two different mathematical operators used in quantum mechanics to describe the energy of a system. The energy operator, denoted as E, represents the total energy of a system, while the Hamiltonian operator, denoted as H, represents the total energy of a system plus its potential energy. In other words, the Hamiltonian operator takes into account the potential energy of a system, while the energy operator does not.
No, the energy operator and the Hamiltonian operator cannot be used interchangeably. While they both represent the energy of a system, they take into account different factors and have different mathematical representations. The energy operator only considers the kinetic energy of a system, while the Hamiltonian operator considers both kinetic and potential energy.
The energy operator and the Hamiltonian operator are related through the Schrödinger equation. The Schrödinger equation describes the time evolution of a quantum system and includes both the energy operator and the Hamiltonian operator. In other words, the energy operator is a component of the Hamiltonian operator.
The energy operator and the Hamiltonian operator are fundamental to understanding the behavior and properties of quantum systems. They allow us to calculate the energy levels of a system and predict its behavior. These operators are also used in various quantum mechanical equations to describe the dynamics of a system.
Yes, the energy operator and the Hamiltonian operator can be measured in experiments, but they are not directly measurable quantities. In quantum mechanics, physical observables are represented by Hermitian operators, and the energy operator and the Hamiltonian operator are both Hermitian. This means that they can be measured indirectly through other physical observables, such as position or momentum, using mathematical equations and principles.