# I Hilbert space and conjugate of a wave function

1. Dec 21, 2017

Take a wavefunction $\psi$ and let this wavefunction be a solution of Schroedinger equation,such that:
$i \hbar \frac{\partial \psi}{\partial t}=H\psi$
The complex conjugate of this wavefunction will satisfy the "wrong-sign Schrodinger equation" and not the schrodinger equation,such that $i \hbar \frac{\partial \psi^*}{\partial t}=-H\psi^*$.
This means that the complex conjugate of any wavefunction that satisfy the Schroedinger equation is not a physical wavefunction and it does not belong to the Hilbert space.
But we know also that the complex conjugate of that function is square integrable since it satisfies $\int _{-∞}^{+∞}\psi(x)\psi^{*}(x)dx=1.$
So can we conclude that if the function is square integrable it does not necessarily belong to hilbert space?

2. Dec 21, 2017

### PeroK

First, you are confusing a function $\Psi(x,t)$, which is a solution to the full Schrödinger equation, with a function $\psi(x)$, which is a square integrable function of a single real variable.

Second, the Hilbert space of square integrable functions is not directly connected to any PDE and it's solutions.

3. Dec 22, 2017

OK.
1-I had proved in my first post that the complex conjugate of any wavefunction $\Psi(x,t)$ (which is a solution to Schroedinger equation) doesn't satisfy Schroedinger equation, and so they are not physical right?. If right then they do not belong to the Hibert space, becuase Hilbert space just include physical wave functions.Is this right?
2- I can understand from you that Hilbert space of square integrable functions is not physical because some wavefunctions are square integrable but don't satisfy Schroedinger equation(if your answer to my first question is yes).So can you explain to me briefly what is physical Hilbert space?.Can I define it like this:"It is an infinite dimensional vector space that its elements are square integrable and satisfy the Schroedinger equation"?

4. Dec 22, 2017

### stevendaryl

Staff Emeritus
No, Hilbert space is a mathematical set of functions, including some that obey Schrodinger's equation, and some that don't. In your terms, Hilbert space includes both physical and unphysical wave functions.[/QUOTE]

5. Dec 22, 2017

[/QUOTE]
What about my second question is my definition correct?

6. Dec 22, 2017

### PeroK

Square integrable functions don't satisfy the Schrödinger equation, because the equation is a PDE with both time and spatial derivatives. Square integrable functions are functions of a spatial variable only.

You've got several different concepts all mixed up here.

The first thing you need to get straight is tge difference between a function of two variables $f(x,t)$ and a function of a single variable $g(x)$.

$f(x,t)$ might be the solution of a PDE in $x$ and $t$. E.g. the Schrödinger equation.

$g(x)$ might be the solution of a ODE in $x$, such as the time-Independent Schrödinger equation

$Hg = \lambda g$

In this case, $g$ would be an eigenfunction of the Hamiltonian with eigenvalue $\lambda$.

You could also have a function $h(x)$ which represents the initial state of the function above:

$f(x, 0) = h(x)$

Before you proceed with QM and Hilbert spaces, you need to grasp the basics of differential equations.

Last edited: Dec 22, 2017
7. Dec 22, 2017

All what you have written above are not new to me at all but thanks anyway.
Ok,thanks for noting me this.

Let me rephrase my question:
are the conjugate wave functions $\Psi(x,t)^*$ physical or not?
what characteristics physical hilbert space has so we can say for example these elements belong to the physical hilbert space?

8. Dec 22, 2017

### PeroK

Define "physical". Define "Hilbert Space". Precisely!

9. Dec 22, 2017

### samalkhaiat

Both are Schrodinger equations. If you write $\psi = \psi_{1} + i \psi_{2}$, you find that the Schrodinger equation and its complex conjugate equation lead to the same pair of coupled real equations $$\hbar \partial_{t} \psi_{1} = H \psi_{2} \ , \ \ \ \ \hbar \partial_{t}\psi_{2} = H \psi_{1} \ .$$

10. Dec 22, 2017

I think I made A big mistake.
I will show my mistake tomorrow If I would have time, because I want to sleep now.
Good night.

11. Dec 23, 2017

Actually I was reading Sakurai's book. Its our textbook in university.
He showed that the Schrodinger equation of unitary time operator is like this
$i\hbar \frac {\partial U(t)}{\partial t}=\hat HU(t)$
Then I reached the point where he said that the eigenstate $U^\dagger|a'>$ satisfies the wrong sign Schroedinger equation,Which is not the Schrodinger equation.
But I read that states that do not satisfy the Schroedinger equation are not physical states.Physical states are states which exist in reality.So I liked to go deep in this.I tried to to know if $U^\dagger$ satisfies the Schroedinger equation, What I got is $-i\hbar \frac {\partial U^\dagger}{\partial t}=U^\dagger \hat H$

My big mistake is that I stated that $i\hbar \frac {\partial \Psi*}{\partial t}=-\hat H\psi$.

In fact according to what @samalkhaiat said, every wave function that satisfies Schrodinger equation its conjugate wave function must also does, which is totally opposite to what I stated in the first post.

Now What I concluded from all of this is that:

1- Any state that satisfy the equation $i\hbar \frac {\partial |α,t>)}{\partial t}=\hat H |α,t>$ or the equation $-i\hbar \frac {\partial <α,t|}{\partial t}=<α,t|\hat H$
are physical states.
Any state that satisfy the equation $i\hbar \frac {\partial |α,t>)}{\partial t}=-\hat H |α,t>$ or the equation $i\hbar \frac {\partial <α,t|}{\partial t}=<α,t|\hat H$
are not physical.
The second ones are not physical because they are states which move inverse of time and so they do not exist thus not physical.
2-Satisfying Schrodinger equation is not a sufficient condition for a state to be real.
Please if what I stated is wrong correct me.

12. Dec 23, 2017

### PeroK

I think you are confusing terminology and concepts here. Suppose a system is in some state $\alpha \rangle$ at some time $t_0$.

The system has no choice but to obey the Schrödinger equation, since the time evolution operator obeys the corresponding operator equation.

There is no question in this context about physical and non-physical states.

Your previous question related to wave mechanics, where the state is expressed as a wave function in position space. This is where the concept of non physical states normally arises. Often in connection with the position and momentum operators, whose eigenstates are not in the Hilbert space of square integrable functions. Hence not normalisable and not physically realisable.

In any case, you need to clear up your up your misunderstanding that some states obey the Schrödinger equation and some do not. The Schrödinger equation governs the time evolution of the state of a system.

Note, finally, in wave mechanics there are solutions of the Schrödinger equation that are not physical. Just take the basis of energy eigenstates with a set of coefficients whose squares do not add to 1.

In general in QM you are only interested in solutions to the Schrödinger that are normalised. In general, there are many other solutions beyond this. This is the opposite, in a way, of your assumption that the Schrödinger equation itself imposes normalisation on its solutions.

13. Dec 23, 2017

### my2cts

That is an important observation.
Indeed, nearly all solutions of the Schrodinger equation are unphysical.
Only an infinitesimal subset is correct.
We know from experiment that we must normalize, but we have no fundamental equation that produces normalized solutions only.

14. Dec 24, 2017

### vanhees71

There is one and only one separable infinite-dimensional Hilbert space (up to equivalence). What you seem to learn at the moment is "wave mechanics", i.e., a special realization of non-relativistic quantum mechanics in terms of square-integrable functions of configuration space (and spins). In another way you can say it's the position-spin representation of non-relativistic quantum mechanics. The square-integrable functions (with using the appropriate equivalence class in saying that functions $\psi_1$ and $\psi_2$ are considered equal iff $\|\psi_1 - \psi_2 \|=0$, where the norm is defined in terms of the scalar product). Of course, if $\psi$ is square integrable then so is $\psi^*$, and thus both are members of the Hilbert space.

The dynamics of the state in wave mechanics is indeed the Schrödinger equation, and the conjugate complex of a solution solves the equation for the time-reflected state. A closer analysis of the representation theory of time-reversal symmetry reveals that it is realized as an antiunitary transformation (in fact the only one in physics, i.e., you can lift all symmetries to unitary representations or a unitary transformation followed by a time-reversal transformation).

Also in general, all square-integrable functions qualify as representing a pure state (in reality it's the ray in Hilbert defined by the wave function, but for the discussion here this subtlety is not that important). You can choose any such function as the initial state and solve the Schrödinger equation to get a corresponding time evolution of the state under the given dynamics (Hamiltonian).

There's only one subtle exception to this statement since physics, particularly the symmetries underlying a model (especially spacetime symmstries, i.e., in our case the symmetries of Galileo-Newton spacetime of classical mechancis), sometimes enforce you to forbid certain superpositions. E.g., in non-relativistic physcis there cannot be any superposition of states belonging to systems of different mass. In both non-relativistic and relativistic physics it is not allowed to use superpositions of states of half-integer and integer angular momentum; in QED it's forbidden to use superpositions of states of different electric charge etc. This is called superselection rules.

15. Dec 25, 2017

Ok.
I didn't really notice this.Yes it is true that satisfying the Schroedinger equation doesn't mean that the wavefunction is physical.
I agree with this.
But I think that if a wavefunction doesn't obey to Schroedinger equation this will certainly mean that this wave function is not physical.right?
I will correct the Schroedinger equations that I had written in my last post:
$-i\hbar \frac{\partial |α,t>}{\partial t}=\hat H|α,t>$
In position representation this will be:
$i\hbar \frac {\partial \Psi(x,t)}{\partial t}={-\hbar}^2/2m \frac {{\partial}^2 \Psi(x,t)}{\partial x^2} +V(x,t)\Psi(x,t)$
$i\hbar \frac {\partial <α,t|}{\partial t}= <α,t|\hat H$
In position representation this will be:
$i\hbar \frac {\partial \Psi(x,t)}{\partial t}={-\hbar}^2/2m \frac {\partial^2 \Psi(x,t)}{\partial x^2} +V(x,t)\Psi(x,t)$
The two above equations are the same the different is the state ket.
$-i\hbar \frac{\partial <α,t|}{\partial t}= <α,t|\hat H$ or $i\hbar \frac{\partial |α,t>}{\partial t}=-\hat H|α,t>$ are the inverse sign Schroedinger equation and in position presentation they will be:

$-i\hbar \frac{ \partial \Psi(x,t)}{\partial t}={-\hbar}^2/2m \frac {{\partial}^2 \Psi(x,t)}{\partial x^2} +V(x,t)\Psi(x,t)$
Now this is the inverse Schroedinger wave equation.

So the wave function that obeys the inverse sign Schroedinger equation and not the right Schrodinger equation is certainly not physical.

Take for example the state ket $U^\dagger |α,t>$,as $U(t)$ always obeys the Schroedinger equation and $U^\dagger(t)$ always obeys the wrong sign Schroedinger equation this state will obeys the wrong sign Schroedinger equation and thus not physical.
The reason behind the non-physicality of this state is maybe it is moving inverse of time and such states do not exist in reality.

I feel that what I have wrote is pretty logic, is there any mistakes?

16. Dec 25, 2017

### vanhees71

Again, any square-integrable wave function generally defines a pure quantum state of a particle. The Schrödinger equation tells you how it develops in time, given the Hamiltonian, i.e., you can start with any wave function $\psi_0(\vec{x})$. Then the wave function at any later point in time is determined by solving the Schrödinger equation's initial-value problem,
$$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}), \quad \psi(t=0,\vec{x})=\psi_0(\vec{x}).$$
Since $\hat{H}$ is self-adjoint, the Schrödinger equation guarantees that time evolution is unitary, i.e., with $\psi_0$ square integrable (and normalized) the wave function always stays square integrable (and normalized) at any time.

17. Dec 26, 2017

### PeroK

You seem to be taking a lot of time, effort and maths to convince yourself that:

a) a system evolves under the influence of the time evolution operator.

b) this produces a specific continuous set of state kets, varying with time.

c) any other set of state kets does not represent the system.

d) this includes the case where you reverse the time evolution - in which case you get a set of valid, physical state kets, but in the wrong order.

In principle this is no different from reversing the time in a classical physics problem. A particle with velocity in one direction would be moving in the wrong direction. Each instantaneous state of the particle ( position and velocity) would be perfectly valid, but the state is evolving the wrong way.

You seem to be looking for some special significance here that is distracting you from an understanding of the time evolution operator.

18. Dec 26, 2017

### vanhees71

Instead of saying "time reversal" one should say something like "reversal of motion".

First of all it is important to keep in mind that in the QT formalism time is simply a parameter that labels the causal sequence of events but not an observable. It's not an operator, and thus the Hamiltonian must not be interpreted as the canonical conjugate of time.

Any ray, defined by an arbitrary vector in Hilbert space represents (with the caveat of possible superselection rules forbidding certain Hilbert-space vectors for some physical reason, usually symmetry constraints, but that's unimportant here) a pure state. Equivalently and more simply to understand you can describe a general state by a statistical operator, and the pure states are distinguished from all other ("mixed") states by the fact that they are represented by projection operators.

The Schrödinger equation tells you how a given initial state evolves in time. Choosing the Schrödinger picture of time evolution, the statistical operator evolves with the full Hamiltonian, via the von Neumann equation
$$\partial_t \hat{\rho}(t)+\frac{1}{\mathrm{i}} [\hat{\rho},\hat{H}]=0,$$
where the first term takes into consideration a possible explicit time dependence.

Now let's consider a pure state of a free non-relativistic particle and use the position representation ("wave mechanics") to analyze the role of complex conjugation. The Hamiltonian is
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}.$$
The only thing you have to know further is the Heisenberg algebra of position and momentum operators, which basically say nothing else than that the momentum operators are the generators of spatial translations, as it must be according to Noether's theorem:
$$[\hat{x}_j,\hat{p}_k]=\mathrm{i} \delta_{jk} \mathbb{1}, \quad [\hat{x}_j,\hat{x}_k]=[\hat{p}_j,\hat{p}_k]=0, \quad j,k \in \{1,2,3\}.$$
Of course, it is most simple to work in the momentum eigenbasis, which form a complete set of generalized orthonormal basis vectors (for a spin-0 particle):
$$\hat{\vec{p}} |\vec{p} \rangle=\vec{p} |\vec{p} \rangle.$$
With help of the commutation relations you can find quite easily the position representation of the momentum eigenvectors and the realization of the momentum operator as the gradient (which again of course is clear since momentum operators are generators of spatial translations):
$$u_{\vec{p}}(\vec{x})=\langle \vec{x}{\vec{p}}=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{x} \cdot \vec{p})$$
and
$$\hat{p} \psi(\vec{x}) := \langle \vec{x}|\hat{p} \psi \rangle=-\mathrm{i} \vec{\nabla} \psi(\vec{x}).$$
The state ket is defined by
$$\hat{\rho}(t)=|\psi,t \rangle \langle \psi,t|,$$
and a clever choice of phases allows you to take as a time evolution equation
$$\mathrm{i} \partial_t |\psi,t \rangle=\hat{H} |\psi,t \rangle.$$
This is the Schrödinger equation in the Schrödinger picture of time evolution in the representation free (Dirac) formalism. It's easy to prove that it indeed leads to a solution of the von Neumann equation for the Statistical Operator.

The state of the system is, of course, specified by the preparation of the system in this state at an initial time, which for convenience I choose as $t=0$. Indeed to solve the Schrödinger equation uniquely, you need an initial condition, i.e., you must have given $|\psi,t=0 \rangle=|\psi_0 \rangle$. For a time independent Hamiltonian it's easy to see that the formal solution of the Schrödinger equation's initial-value problem is
$$|\psi,t \rangle=\exp(-\mathrm{i} t \hat{H}) |\psi_0 \rangle.$$

For free particles, the momenum eigenstates are of course also energy eigenstates, and thus it's simpler to use them instead of solving the Schrödinger equation in position representation directly:
$$\psi(t,\vec{x})=\langle \vec{x}|\psi,t \rangle=\langle \vec{x}| \exp(-\mathrm{i} t \hat{H}) \psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d} \vec{p} \vec{x}|\exp(-\mathrm{i} t \hat{H}) |\vec{p} \rangle \langle \vec{p}|\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d} \vec{p} \vec{x}|\exp(-\mathrm{i} t \hat{H}) \exp(-\mathrm{i} E_{\vec{p}} t) \tilde{\psi}_0(\vec{p}),$$
where $E_{\vec{p}}=\vec{p}^2/(2m)$ is the "on-shell energy" of a free non-relativistic particle and
$$\tilde{\psi}_0(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^{3/2}} \exp(-\mathrm{i} \vec{p} \cdot \vec{x}) \psi_0(\vec{p})$$
is the Fourier transform of the initial position-space wave function.

Now take the complex conjugate, and you'll see that you get a solution of the Schrödinger equation traveling of course forward in time but with the opposite momentum.