Give an example of a potential in which there is a complete set of bound stationary states of we

Give an example of a potential in which there is a complete...

Key Concepts

Symmetry and Alternative Sets of Eigenstates

In a system with a symmetric potential, the eigenstates can be chosen to have well-defined parity. However, by considering an alternative representation, such as taking specific linear combinations or applying a potential modification (like in the case of a molecule exhibiting inversion, e.g., ammonia), one may obtain a complete set of bound states that do not display this definitive parity. This exemplifies how symmetry properties can depend on the chosen basis and how breaking or modifying symmetry can lead to alternative, physically equivalent representations of the quantum states.

Complete Set of States

A complete set of states forms a basis for the Hilbert space associated with a quantum system, meaning that any physical state of the system can be expressed as a linear combination of these basis states. In the context of bound states, having a complete set ensures that the spectral decomposition of the Hamiltonian is fully captured, whether the individual states have well-defined symmetries or not.

Bound Stationary States

Bound stationary states are the eigenstates of a quantum system’s Hamiltonian that have discrete energy eigenvalues and are spatially confined. These states are important because they represent the stable configurations of particles in a potential well, and their time evolution is simply given by phase factors, making them ideal for constructing a basis for the system's state space.

Parity Operator

The parity operator is a symmetry operator in quantum mechanics that inverts spatial coordinates. When a state is an eigenstate of the parity operator, it exhibits a definite symmetry (even or odd) under spatial inversion. This property is significant because, in systems with reflection symmetry, the Hamiltonian commutes with the parity operator, allowing bound stationary states to be chosen as eigenstates of parity.

Transcript

00:01 Okay, in this question we are going to analyze the bone state of finite square well, where si x is an odd function and the si si x equal to f e minus k x, where x is greater than a and d sine l x, and x is greater than minus a and less than a.

00:38 And f e k x where x is less than minus a this is the move in function in order to check whether the function is odd we have a property of the function to be also f of minus x should be equal to minus f of x then our function is set to be ought so now we have to find out the constant d and f so we also know that the here k is equal to minus 2m e out of h -cat and l is equal to undr 2m e divide by h -cat first we need to consider the fact that xx need to be continuous at this limit that is a x is greater than minus a and less than a and the derivative of this above well function should also be continuous so the derivative is given as respect to x is equal to sorry we can write this equation si x x x x here we can run f minus k a equal to d sine l x this is equation number one and the derivative of the wave function should also be continued so therefore d sye by d x are equal to minus k f e minus k x here x is greater than a and l d cos l x here x is greater than a minus a and a less than a so this should also be continued so therefore we can write it as minus k f e minus k a equal to l d cos into l a this is equation number two so dividing this equation two by equation or one we get f e minus k a divide by d sine and a so therefore we get minus k minus k equal to l caught la and i'm multiplying the this equation by a we get a k equal to la court la put z is equal to la in a bow equation we get we get minus a k equal to z caught z so for the new equation we can write it is a square into k square plus l square are equal to a square into minus 2m e divide by h cut plus 2m e minus v0 divide by h cut scale and this can be written as 2m v0 out of h cut square and this is equal to z not square and and therefore z not equal to a out of h cut and root 2 m v not so therefore a square k square equal to z not square minus a square l square and this a square k square equal to z a square k square equal to z not square minus z square so ak equal to unrude z not square minus z square and this can be written as so putting in the equation in the in the voe equation we get unrude z nard square minus z square is equal to z caught z and in this chord z equal to unroot z z not upon z whole square minus 1 and this is equation number are 3 so now we will find out the value of z so a bo equation can specify also to find out the solution for this equation we need to graph it so we have a graph like this where we plot z on x xx and this is the function of z so this is is pi this is 2 pi this is 3 pi and this is called z -not and if we draw a graph then this graph is like this this graph become like this and like this so the value of z -not is here so these points these intersections uh these three intersection at this point give us the value of z not so no we can explore some of the limit so the limits are the first limit is is when the square well is when the well is white and deep if the well is white which is represented by and if it is deep it is represented by v -not so here z is equal to n pi where the value of n is equal to 1, 2, 3 dash, dash.

08:46 So here z is equal to a, l, which is equal to n pi...

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