Standard

Posted by

Posted on

July 17, 2015

Posted under

The holographic view

Comments

Leave a comment

The Twin Paradox (III)

This blog post is a sequel of the blog posts 1 and 2.

Special relativity is very simple and elegant theory of symmetry (ignore the last word if you are not familiar with it). Sometimes, a naive thinking may lead to the contradictions in the relativistic physics. For instance consider a tunnel and a train of same proper length L_0 with some constant relative velocity between them. There is an observer sitting outside the tunnel (observer A) and one sitting in the train (observer B). Now observer A will see the length of the train little shorter than the tunnel and observer B will see the length of tunnel little shorter than train (relativistic length contraction). Let the tunnel be a special one having two doors at the front (from which the train enters the tunnel) and at the end, capable of shutting down simultaneously. Observer A controls the doors. Now since the length of the train is smaller then that of tunnel according to A, she decides to shut down both the doors at that point of time when she sees the train completely inside the tunnel. On the other hand, the observer sitting in the train will deny the fact that the train can be captured by the tunnel as the train is longer than the tunnel according to her. But relativity says that the physics is invariant in any inertial frame therefore both the observers must concur on the facts (whether the train passes or gets captured).

The paradox is resolved by the fact that the shutting of doors is simultaneous in the coordinate chart of observer A and not in the frame of observer B (we have seen in previous blog post, the notion of simultaneity is frame dependent). The observer B sees the door at the end shutting first, which stops the train instantaneously (infinite acceleration), and then the door at front closing. Therefore both the observers will report the same fact, that is, the tunnel captures the train. Here we emphasize again, the events may not be mapped to same coordinates in different coordinate charts. To map the coordinates, of an event, form one inertial coordinate system to another, we must use Lorentz transformation.

Another common paradox that arises is as follows. Consider two twins Alice and Bob living in an inertial coordinate system at same location. Alice starts traveling in a spaceship, moving with a constant relative speed, say .6 meters per light second with respect to Bob along x-axis, for some time (in the frame of the observer B) and then comes back to Bob with same the relative velocity. Now, naively, one can say that according to Bob, the time of Alice goes slower than him, hence when Alice returns, he must age more. According to Alice, Bob’s time goes slowly (moving clocks are slower for Alice too!) and therefore Alice should age more when she returns. But both Alice and Bob should agree on the same fact and here lies the paradox which everyone comes by when studying the relativity for the first time.

We will attack this problem from the perspective of the naive thinking which is based on the assumption that the physical phenomenon is symmetric from the point of view of Alice and Bob. So first let us think how many coordinate charts we require to study the problem?

No, not two but three. In first coordinate chart (of observer B) Bob is at rest at some position (from where Alice starts her journey). In second coordinate chart (of observer A), Alice is at rest at some position when going away from Bob and in third coordinate chart (of observer A’), Alice is at rest at some position when returning back towards Bob. Also note that Alice and Bob are not observers according to our definition in the blog post. So it is Alice who “jumps” from one coordinate chart to another whereas Bob remains in the same frame of reference. Thus the physical phenomenon is not symmetric from the point of view of Alice and Bob.

The word “jumps” has been used because we assume that she accelerated infinitely when changing her direction. Now the postulates of special relativity being valid for the inertial frames doesn’t mean that we can’t study accelerated objects. We can perfectly study the accelerated objects by using the notion of comoving frames. In our example, second and third frames are the comoving frames for Alice. Similarly one can study finitely accelerating objects by considering a series of jumps (of the system) from one comoving frame to another.

The the Minkowski graph of the observer B is (tracking of the coordinate x is not necessary). Different colors represent the time coordinate axis of different frames.BIn this graph, the green line is the world line of Bob and the world line of Alice is represented by the violet line and red line. One can clearly see that Bob ages by 30 light seconds and Alice ages by 24 light seconds.

Now let us look at the graph of the observer AA1
Here one can observe that in the coordinate chart of observer A, Alice ages faster than Bob for half of her trip. For the next half, she ages much slower then Bob (who ages 30 light seconds) such the in the end of the journey, Alice ages less (again 24 light seconds) than Bob.

And finally we see the graph of the observer A’A2

From these three graphs we see that all the observers (A, A’ and B) agree that Bob has aged more than Alice in this trip, by 6 light seconds. It means all the inertial observers observed the same physical phenomenon, as required by the first postulate of special relativity, thus resolving the twin paradox.

Standard

Posted by

Posted on

July 8, 2015

Posted under

The holographic view

Comments

1 Comment

Special relativity in Minkowski graph (II)

In the previous blog post we noted that the Lorentz transformation is the map from one inertial coordinate system to another. This means that the coordinates of a person standing on the platform can be mapped to the coordinates of the person sitting in the train using the transformation, which results from special relativity postulates, and quantitative analysis of the physical phenomenon can be done in both the frames.

We also note that these inertial frames (coordinate systems) are same in nature (that is the reason why we used the same constant c in both the coordinate systems). In other words, no inertial frame is to be given any sort of preference. If you like, you can think a 4 dimensional grid of space and time built from the 4 dimensional version of a unit cube. The 2 dimensional version of this grid is shown in the figure of coordinate chart below.

Now this grid must be exactly same for all the coordinate systems (henceforth, we will talk of inertial coordinate systems only). Therefore the grids in coordinate charts 1, 2 and 3 (of different inertial observers) are congruent. All that differs is the mapping of same spacetime events. For instance the green dots represent the events of aging of Bob. Now this set of spacetime events is mapped differently in different frames as shown.

More precisely, one meter “of” a coordinate system should be equal to one meter “of” any other coordinate system and so must be the case with one second (I am really not sure which preposition to use here, but I think “of” should convey the message). These units of measure which define the grid size, are same for all the frames and are termed as “proper length” and “proper time”. Although, a rod (a set of special spacetime events defined later) of one meter in one frame might not be one meter in other coordinate system. In fact Lorentz transformation makes sure that the length is contracted by a factor of \gamma = 1/\sqrt{1-\frac{v^2}{c^2}} (https://en.wikipedia.org/wiki/Length_contraction). We will also explain this phenomenon later in this post.

The German mathematician Hermann Minkowski utilized the property of coordinate charts (maps from manifold to \mathbb{R}^4) in making a useful geometrical tool called Minkowski graph. The idea is that on a graph paper we have two dimensions at our disposal. And, the Lorentz transformation shows that if relative velocity between two frames is along x-axis then the map of the coordinates y and z to another coordinate system is identity. Therefore we consider the map from the events in the spacetime manifold to \mathbb{R}^2 space (our graph paper) which includes only two coordinates x and t as shown in the figure.

chart

In this coordinate system (frame of reference), the event A has been assigned the coordinates (5,3). Also note that the time axis has been divided by the constant c and, here, we define one light second as the time taken by light to travel one meter (or inverse of c).

Now each point in this graph (coordinate system) represents an event in the spacetime manifold. As the consequence, the evolution of a particle in the spacetime is mapped to a trajectory in the graph (which is drawn by the observer at the origin). This trajectory is called the “world line” of the particle. On carefully examining the transformations, one can note that when the relative speed between two frames is greater than the constant c (which we now set to unity) then the factor \gamma becomes imaginary. Furthermore, the factor approaches infinity when the relative speed approaches unity. In relativistic dynamics the total energy of a free massive particle is directly proportional to \gamma. Hence it would require infinite amount of energy to accelerate a particle near the speed of light.

The slope of the trajectory in the Minkowski graph is the inverse of the speed of particle with respect to the observer (who is drawing the graph). Therefore the slope of the trajectory is always greater than unity in the graph because a massive particle cannot be accelerated to the speed of light. A light ray always follows the trajectory of straight line with unity slope. The Lorentz transformation makes sure that the trajectory of the light remains same in all the graphs, corresponding to various coordinate systems.

The Minkowski graph is quite helpful in comparing the physics from two frames. Consider two observers F and F’ moving with some relative speed.

Lorentz_transforms_2.svg-2

Observer F uses her coordinate system to draw a Minkowski graph as shown in the figure above.

Now consider a massive particle at the origin with respect to the observer F’ (i.e x^\prime =0) from time t^\prime = 0 to t^\prime = 1. So how will F draw the world line of the particle in her coordinate system (or Minkowski graph)? As you might have guessed, we will use the Lorentz transformation (because we have to map the information given in F’ to F). In other words we have to find the equation of the particle in coordinates of F subjected to the constraints \Delta x^\prime = 0 and \Delta t^\prime = 1 in the coordinates of F’.

From the Lorentz transformation, the first constraint will give a trejectory equation \Delta t = \frac{\Delta x}{v} which F will interpret as the time axis of the observer F’ (because the line of constant x-coordinate giving t-axis and vice versa is a property of the orthogonal coordinate system in the \mathbb{R}^2 space). Therefore F will draw a straight line in her graph with slope as inverse of the relative speed. Let v=.6 meters per light second. The question now arises is that how will she mark the scale of that time axis in her coordinate system (or graph)? The second constraint gives a result that \Delta t = \frac{1}{\sqrt{1-v^2}} which means that a unit second in the coordinates of F’ is mapped t0 1.25 seconds in the coordinates of F. It means that the two events with a unit time separation at the origin in frame F’, are separated by 1.25 seconds in the frame F. Thus the graph of F will look like

minkowskitime

The violet dots correspond to the events with unit time intervals and zero space displacements in F’ (hence it is the time axis for F’) while green dots in vertical line correspond to the events with unit time intervals and zero space displacements in F. Note that in the graph drawn by the observer F, the 15 units of her time is equal to the 12 units of the time in F’. It is consistent with the fact that \Delta t = 1.25\Delta t^\prime. This is why we say that moving clocks are slower.

Similarly, we can draw the x-axis of F’ in the coordinates of F using the same technique. Consider a situation in which \Delta t^\prime = 0 (simultaneous events) and \Delta x^\prime = 1. Using the Lorentz transformation and first constraint we get \Delta t = v\Delta x. This equation gives the x-axis of the observer F’ in the frame of F. Second constraint gives the relation \Delta x = \frac{1}{\sqrt{1-v^2}}. This means that a unit length (between two simultaneous events) in F’ is mapped to 1.25 meters in F. The graph now looks like

minkowskispacetimeThe violet line with lesser slope represents all the simultaneous events of F’ (with t^\prime = 0) in the coordinates of F where they are having different coordinate t. Thus simultaneity is a relative concept and depends of the frame of reference.

In the end of this post, we will explain length contraction using the Minkowski graph. The length of a rigid rod is defined as the distance (in \mathbb{R}^n space) between its end points at the same time coordinate (convince yourself!). Consider a rod of 2 meters length at rest in the frame F’. Let its first end point be at (4, t^\prime). Therefore, according to the definition of length, the coordinates of other end point will be (6, t^\prime). Now this rod will trace a “world sheet” in the Minkowski graph as shown in the figure below. The red dots on the lower violet line (the x-axis of F’) are the coordinates of the endpoints of the rod which are, according to F’, (4,0) and (6,0). They evolve in the time t^\prime such that after unit second in F’, the coordinates are (4,1) and (6,1) (the couple of next red points on each red line) in F’. And it should be, because the 2 meter rod is at rest in F’.

minklencont

Now how much length will the observer F measure? Let us say that at t=8 light seconds she measures the length. Now according to the definition of the length, she will cut the world sheet of the rod with a line (to obtain simultaneous coordinates in her frame) and measure the distance between space coordinates. In this example the coordinates (pointed by the arrows) are 8 and 9.6. Thus the length of the rod is 1.6 meters in the frame F. This is called length contraction and this is why we say moving trains are shorter.

The main point to keep in mind are that in special relativity, an event in spacetime might not have same set of coordinates in different frames of reference. To map an event from one coordinate chart to another we must always use Lorentz transformation. If this is followed honestly, then all the paradoxes of special relativity can be resolved. One such paradox is “twin paradox” which I will explain in next post.

If you are wondering how I made these cool Minkowski graphs then head over here.

Standard

Posted by

Posted on

June 18, 2015

Posted under

The holographic view

Comments

2 Comments

A modern viewpoint on special relativity (I)

In this series of blog posts, I will explain our current understanding of spacetime using the notions of relativity.

Special relativity originates from two simple principles but its predictions challenge our day to day experiences. For instance, according to the relativity, a person in the train station will see the the length of a moving train a little smaller than that of stationary train (of same type). Furthermore, the same person will notice that the clocks hanging in the train ticking  little slower than that on her wrist. These phenomena are termed as “length contraction” and “time dilation”. The magnitude of these contractions and dilations depends directly on the relative velocity of the train and the person standing on the platform. Now, commonly, the relative speed of train is so low that one doesn’t see any appreciable changes. Cosmic particles which travel at very high speeds are known to have longer half lives than their laboratory counterparts. But convincing the audience that relativity is a correct theory of the nature is not the purpose of this post. Assuming that relativity is a valid theory, the aim of this post is to describe and discuss the theory.

In my opinion, special relativity is all about the properties of spacetime in fairly small region and how various observers record the events occurring in that spacetime. The words in italics have precise definitions and meanings in physics so let us spend some time in understanding them. When you read the word spacetime you might think of it as a three dimensional space with time t as separate entity running in the background. This is totally a wrong picture of the spacetime. If you are a physics major, and have had a course in relativity, you might think of it as a strange mixed space of \mathbb{R}^3 and t with 4 coordinates in which the invariant path length is given ds^2 = dx^2+dy^2+dz^2-c^2dt^2, where c is the speed of light.

The problem with the last definition is that it does’t capture the essence of spacetime as such. It is like explaining an alien that the person in the red shirt is a boy. Just like the word “boy” is assigned to a specific creature having certain characteristics and is independent of the shirt he wears, the word “spacetime” is assigned to a specific physical entity and is independent of the coordinates assigned to it by the observer (yes, an observer is someone who assigns the coordinates such that her coordinates are fixed  in the coordinate system). Please note that this is not a standard definition of the observer but it is quite helpful in explaining the subject. In relativity, spacetime is a special set, with many interesting properties, which mathematicians call “Manifold”. And the coordinate system is a map from the spacetime manifold (say the set \mathcal{M}) to the set \mathbb{R}^4. The coordinate chart of a N dimensional manifold is shown in the figure below (courtesy Carroll’s notes)

Manifold

Thus, when an observer is assigning coordinates x, y, z and t to an event, he/she is basically mapping the events in \mathcal{M} to a \mathbb{R}^4 space. When an observer assigns a coordinate system, he/she defines a frame of reference. So when we say “the frame of reference of the observer A”, we mean the coordinates assigned by A.

In physics we postulate that all the physical phenomena, comprising of events, happen in this spacetime manifold. In order to quantitatively analyze them, we require an observer who can assign the coordinates  to these events. First postulate of relativity says that the physics doesn’t depend on the coordinate systems of the observers moving with constant relative velocities with respect to each other. We call all such observers as “inertial observers”. It simply means that if we do the physics calculations using the coordinates of any inertial observer we will find the same results. This shouldn’t be hard to accept as the real physical phenomena take place in the spacetime manifold and no matter which map (coordinate system) we use for quantitative analysis, we must find the same physical results. Now be careful! this postulate talks about the maps which are inertial. So we have to be careful when working with accelerated bodies.

Vectors are mathematical  objects (maps, to be precise) which, by definition, remain unchanged under the coordinate transformations. In relativity, we have 4 dimensional vectors living in the spacetime manifold which remain same in all the inertial coordinate systems. We call such objects as invariants. In fact there is a class of invariants called “Tensors” and we assign all sorts of physical objects (energy-momentum, electromagnetic field etc) to the tensors as they, too, don’t depend on the frame of reference. Now events, which are also physical objects, are assigned to 4 vectors. The distance between two events is called the “path length” and is a very good example of an invariant (actually it is a scalar which is a tensor of rank 0). The second postulate gives us a prescription to make scalars, in terms of what is known as metric. According to this prescription the squared path length is mapped to dx^2+dy^2+dz^2-c^2dt^2 in a coordinate system constructed from x, y, z  and ct . Here c is a constant introduced to match the dimensions of the length squared when multiplied by dt^2 . Simple analysis shows that it must have the dimension of \frac{\text{length}}{\text{time}}. In another coordinate system with the coordinates x^\prime, y^\prime, z^\prime and ct^\prime the squared path length is mapped to dx^{\prime 2}+dy^{\prime 2}+dz^{\prime 2}-c^2dt^{\prime 2}. Note the same constant c has been used to maintain equality of the frames. Now, due to the invariance of path length, dx^2+dy^2+dz^2-c^2dt^2=dx^{\prime 2}+dy^{\prime 2}+dz^{\prime 2}-c^2dt^{\prime 2}. This is the starting point of the derivation of the Lorentz transformation in the standard relativity textbooks (although they arrive at this expression using light rays which is misleading sometimes). The Lorentz transformation is a relation between the coordinate systems of two inertial observers, assigned to the same events in the spacetime manifold \mathcal{M}.

In the end of this post I would like to give the celebrated Lorentz transformations. Consider two observers moving with a constant relative speed along one particular direction (say along x axis), then the relation between the coordinates of the observers is (image courtesy Wikepedia)

          Lorentz_transforms_2.svg-2

Here \gamma = 1/\sqrt{1-\frac{v^2}{c^2}}. When physicists did the experiments they found that the value of the constant c is nearly 3\times 10^8meters per second which is the speed of light. The derivation of these transformations can be found in any standard textbook on relativity or in Wikipedia (https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations).

Standard

Posted by

Posted on

August 6, 2014

Posted under

The holographic view

Comments

3 Comments

Spacetime odyssey

All physical phenomena we see around us take place on something that we call spacetime. Therefore it becomes important to get acquainted with spacetime and understand how it affects the physical phenomena or gets affected by them. I will chronologically introduce the notions associated with spacetime because, usually, everyone gets to learn it in the exact same fashion.

Until 20th century, space and time were thought as different entities. It was thought that there exists a three dimensional euclidean space which acts as a stage for the physical phenomena. This stage was supposed to be rigid, static and flat (rigorous definition will be given later). The space was a three dimensional extension of Euclid’s five axioms for plane geometry which were

  1. Between two points in the space, only one line can be drawn.
  2. A finite line in a space can be extended indefinitely.
  3. A circle with arbitrary radius and center can be drawn in the space
  4. All right angles are equal to one another.
  5. If a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

These axioms were based on the human experience (at the time of Euclid) and considered to be true in every part of the universe.  Even now, based on high-school experience, one doesn’t need enough convincing about the validity of these postulates.

Let us now define our three-dimensional Euclidean space in formal way. It is a set of points with well defined distance between any two points of the set. If you give me two points say P_1(x_1,y_1,z_1) and P_2(x_2,y_2,z_2) belonging to Euclidean space. I will give you the distance between them by computing  d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}.

A couple of interesting points can be noted. First, the definition of distance between two points is static i.e the way to find distance doesn’t change with time. We can represent the definition of distance between two points using matrices. Let us represent the points by column matrices P_1 = \left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) and P_2 = \left(\begin{array}{c}x_2\\y_2\\z_2\end{array}\right) and define a 3\times 3 matrix \eta = \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right). The distance between two points can now be given by d^2 = (P_1-P_2)^T.\eta .(P_1-P_2) (here ‘.’ represents matrix multiplication). Note that the definition of distance depends on the matrix \eta. Physicists and mathematicians call \eta metric of the space. Now we can say that it is the metric which defines the distance between two points of a space. In fact the metric defines the notion of a ‘dot product’ in a space. Take any vector, write its components in column matrix and used the matrix multiplication defined above to find the length of the vector.

The metric of Euclidean space is time independent (build from constant numbers). Although I have used matrix to represent a metric, generally it should be thought as a set of numbers given by \eta^{\i\j} where i,j\in 1,2,3.

Next we note that the metric (of Euclidean space) is independent of space itself. In fact this independence is a direct consequence of Euclid’s axiom 5. This independence implies that the space is flat. I will demonstrate this and provide a formal definition of flatness in a later post.

 

Standard

Posted by

Posted on

July 25, 2013

Posted under

The holographic view

Comments

5 Comments

Particle in a box (infinite square well).

Yes, I am aware of the ennui caused by this topic. We write schrodinger’s time independent equation, solve it, apply the boundary conditions, find the energy eigen states, normalize them and everybody is happy. If you are a physics major, then you must have done this countless times. But I always had a problem with this system because of its highly un-physical nature (I will come back to this later). That makes it a mathematical curiosity and if it is, then why not explore it to the full extent? Please note that in this post I will use rigorous mathematics and will define terms in proper mathematical language (I hope this post will serve as my notes for future reference). Here one can also appreciate the aesthetics of Linear Algebra.

The question I had been asking myself was, “Why do we always work in position representation? What is wrong in working in momentum representation (for this particular problem)?”. The curiosity to find the answers compelled me to plunge into the depths that are generally not touched and mentioned in physics literature.

The conventional Quantum Mechanics (for this problem) is performed on the Hilbert Space L^2 ([0,1], dx) which I will denote with \mathcal{H}. Here L^2 (read L two) represents a functional space of square integrable functions and [0,1] represents the range of parameter x (position). I am interested in working with L^2 (??, dp) and want to obtain energy eigen states in terms of the wavefunction \varphi(p). I dont know the range of the parameter p (momentum). I can’t take it as whole real line (because of the bounded nature of position). Parseval-Plancherel theorem in not applicable here. I must first find the nature of momentum operator. Then maybe its spectrum might give me the idea about the range of momentum.

Just like a function has a domain, the linear operator on Hilbert space has a domain. The formal definition of the operator is (this definition is for general Hilbert Spaces and not restricted to L two spaces)

An operator on the Hilbert space is a linear map A : \mathcal{D}(A) \mapsto \mathcal{H} such that \psi \mapsto A \psi (where \psi \in \mathcal{D}(A)).

\mathcal{D}(A) represents the linear subspace of \mathcal{H}. This subspace is the domain of the operator. Strictly speaking, Hilbert space operator is the pair (A, \mathcal{D}(A)) which is the specification of the operation with the domain on which the operation is defined. Now two operators are said to be equal if and only if

A \varphi = B \varphi for all \varphi \in \mathcal{D}(A) = \mathcal{D}(B).

Let us consider position operator denoted by Q. The operation is defined as Q \psi (x) = x \psi (x). But given the boundary conditions, the domain is defined as

\mathcal{D}(Q) = \{\psi \in \mathcal{H} \mid Q\psi \in \mathcal{H} and \psi(0) = \psi(1) = 0 \}.

As mentioned earlier, the Quantum Mechanics is performed on the Hilbert Space of square integrable functions (denoted by \mathcal{H}), we certainly would not want a mapping which blows up the square integrability. Hence it is important to put the restriction that the output be Q\psi \in \mathcal{H} which translates into ||Q\psi||^2 = \int_{0}^{1} dx x^2|\psi (x)|^2< \infty. So, in words, the domain of Q is the set of all the square integrable functions such that

  • they satisfy boundary conditions
  • the output of linear map should be finitely square integrable

Now let us define momentum operator P. Quantum Mechanics defines the operation of this operator (on a function) as

P\psi = \frac{\hbar}{\iota} \frac{\partial \psi}{\partial x}

Certainly, we would like to have a set of those functions which besides satisfying boundary condition, also have square integrable derivatives. In formal language

 \mathcal{D}(P) = \{\psi \in \mathcal{H} \mid \psi^{\prime} \in \mathcal{H} and \psi(0) = \psi(1) = 0 \}.

Now let us see the definition of the adjoint of an operator A which means we have to define the operation and domain of the adjoint of operator (let’s denote it by A^{\dagger} ).

\mathcal{D}(A^{\dagger}) = \{\phi \in \mathcal{H} \mid \exists \tilde{\phi} \in \mathcal{H} such that \langle \phi , A\psi\rangle = \langle \tilde{\phi}, \psi\rangle \forall \psi \in \mathcal{D}(A)\}

This definition says that the domain of \mathcal{D}(A^{\dagger}) is set of \phi (square integrable)  such that there exists \tilde{\phi} (again square integrable) which makes the equation \langle \phi , A\psi\rangle = \langle \tilde{\phi}, \psi\rangle true for all the \psi belonging to domain of operator A.

For a given \psi , \tilde{\phi} depends on A and \phi . The operation of  A^{\dagger} is defined as

A^{\dagger} \phi = \tilde{\phi}.

So we, now, have a full working definition of the adjoint of a operator. Let’s find the adjoint of P. The inner product equation \langle \phi , P\psi\rangle = \langle \tilde{\phi}, \psi\rangle must hold. So

\int_{0}^{1} \overline{P^{\dagger}\phi}(x) \psi (x) dx = \int_{0}^{1} \overline{\phi}(x) \frac{\hbar}{\iota} \frac{\partial \psi}{\partial x}(x) dx .

On integrating by parts, the RHS

\int_{0}^{1} \overline{P^{\dagger}\phi}(x) \psi (x) dx = \frac{\hbar}{\iota}\left[_{0}^{1} \overline{\phi}(x) \psi (x)\right] -\frac{\hbar}{\iota} \int_{0}^{1}\overline{\frac{\partial \phi}{\partial x}}(x) \psi (x) dx.

For this equation to be true we must have

\overline{P^{\dagger}\phi}(x) = -\frac{\hbar}{\iota}\overline{\frac{\partial \phi}{\partial x}}(x)

or

P^{\dagger}\phi(x) = \frac{\hbar}{\iota}\frac{\partial \phi}{\partial x}(x).

The most curious and interesting point is that the function \phi(x) does’t need to satisfy boundary condition (of infinite square well) since \left[_{0}^{1} \overline{\phi}(x) \psi (x)\right] = \left[ \overline{\phi}(1) \psi (1) - \overline{\phi}(0) \psi (0) \right] is zero as \psi(x) already satisfies the boundary condition. Clearly the \phi(x) has no restriction save square integrability which shows that the set \mathcal{D}(P^{\dagger}) has more elements than set \mathcal{D}(P). The formal definition of \mathcal{D}(P^{\dagger}) is

\mathcal{D}(P^{\dagger}) = \{\phi \in \mathcal{H} \mid \phi^{\prime} \in \mathcal{H}\}.

It can be seen that  \mathcal{D}(P) \subset \mathcal{D}(P^{\dagger}). Thus P^{\dagger} \neq P i.e the momentum operator is not self-adjoint. On the other hand it is quiet simple to show that Q^{\dagger} = Q i.e the position operator is self-adjoint.

Now let us see the definition of Hermitian operators.

The operator A on \mathcal{H} is Hermitian if  \langle \phi , A\psi\rangle = \langle A\phi,\psi\rangle for all \phi, \psi \in \mathcal{D}(A)

Since A operates on both the functions, needless to say, they must belong to the domain of the operator. Let us use the definition of adjoint operators. Operator A^{\dagger} is such that \langle \phi , A\psi\rangle = \langle A^{\dagger}\phi,\psi\rangle which implies \langle A\phi,\psi\rangle = \langle A^{\dagger}\phi,\psi\rangle . Thus if the operation of the operator is same as that of its adjoint, then it is Hermitian operator. The domain of both the operators need not be same.

We have seen that the specification of both P and P^{\dagger} is same \frac{\hbar}{\iota}\frac{\partial}{\partial x}. The momentum operator is Hermitian for this case but not self-adjoint. It is easy to note that all the self-adjoint operators are Hermitian but vice-versa is not true.

Now the spectral theorem of linear operators states that

If the Hilbert space operator A is self-adjoint, then its spectrum is real and the eigenvectors associated to different eigenvalues are mutually orthogonal; moreover, the eigenvectors together with the generalized eigenvectors yield a complete system of (generalized) vectors of the Hilbert space.

This property does not hold for only Hermitian operators.

As the momentum operator, here, is only Hermitian its eigen vectors won’t form complete basis. Thus there is no use in solving this problem in momentum space. Moreover, L^2 (??, dp) space wont have continuous momentum (because of quantization imparted by boundary conditions).

Normally one assumes that the eigen vectors of an observable forms a complete set of basis. This means that observables must be represented by self-adjoint operators. From above we note that momentum operator is not an observable for this problem! 

Absurd, isn’t it? Well, the boundary condition itself gives us the hint of the highly non-physical nature of the problem. The “non self-adjointness” of momentum operator further shows us how the infinite potential square well is a mathematical curiosity with tenuous nexus to real physical systems.