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

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

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

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.

5 thoughts on “Particle in a box (infinite square well).”

Aiman Khan says:

August 29, 2013 at 08:36

Please mention all your references too so that readers can follow up.

- ravimohan says:
  
  August 29, 2013 at 09:21
  
  Thanks for pointing that out Aiman. I have consulted from
  1) http://arxiv.org/abs/quant-ph/9907069 (an excellent paper I would recommend to undergraduates)
  2) Introductory Functional Analysis With Applications [Kreyszig] (an interesting text I would suggest to study in holidays)
  
ravimohan says:

November 2, 2013 at 12:02

Some pointers from dextercioby (from PF http://www.physicsforums.com/showthread.php?p=4559460#post4559460).

1. the infinite sq. well is a non-physical example. There are no infinities in real life systems. Nonetheless, it serves as the perfect example to the fact that a theoretical resolution of a (non-)physical problem should always be done with the maximum knowledge of mathematics (in this case the theory of unbounded operators on Hilbert spaces).

2. the reason that the momentum operator is symmetric and not self-adjoint relies with the physical boundary condition: psi (outside the box) = psi (left edge of the box) = psi (right edge of the box) = 0. If you relax the boundary condition to psi (left edge of the box) = psi (right edge of the box) =/= 0, then the momentum operator becomes self-adjoint. This would be the case for the finite square well, however (which is in fact a physically correct example).

3. it’s interesting to notice that 2 unphysical situations (the free particle on the real line and the free particle in an infinite square well) are usually presented in a non-mathematical manner to students to illustrate that the energy levels of the particle are either continous, or discrete. The real value of these 2 examples is actually to prove that functional analysis is the root of all quantum mechanical calculations.

- Lars D says:
  
  February 25, 2018 at 18:34
  
  How about a particle in a box with periodic boundary conditions? These are stronger boundary conditions than the above mentioned. Like psi(left) = psi(right), psi'(left) = psi'(right)?
  
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Ravi Mohan

If you haven't found something strange during a day, it hasn't been much of a day. – J. A. Wheeler

Particle in a box (infinite square well).

5 thoughts on “Particle in a box (infinite square well).”

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Particle in a box (infinite square well).

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