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The Schrödinger equation is given by
The only simple quantity in this expression is E, which is the energy of a system. Y is the wave function, and H is the so-called Hamiltonian . The Hamiltonian is not just a number. If it were, then we could divide the left- and the right-hand-side by Y, which would give the simple expression H = E. This expression says that the Hamiltonian is just another name for the energy, which is not quite true. Because the Hamiltonian is not a number, you are not allowed to divide by Y, and the Schrödinger equation cannot be simplified in this way.
The Hamiltonian is an example of a so-called operator . An operator works on a wave function and generally changes it into another wave function. (The only exception is the identity operator, which does nothing with a wave function.) Operators are linear . What this means is most conveniently expressed as
| A(c1Y1+c2Y2) = c1AY1+c2AY2, |
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where Y1 and Y2 are arbitrary wave functions, c1 and c2 are arbitrary (complex) numbers, and A is an operator. In words this expression means that if a linear operator works on the sum of two wave functions then the result is the sum of the wave functions that are obtained when the operator works on the separate wave functions, and if the operator works on a wave function that has been multiplied by a number then the result is a wave function obtained by working on the wave function with the operator and then multiplying with the number. (You will probably have to read the previous sentence a couple of times to understand it. Here you see an advantage of mathematics; it is much more succinct than regular languages.)
Examples of linear operators are multiplication with a complex number and taking a derivative. To see that taking a derivative is a linear operator we assume that the wave functions Y1 and Y2 depend on just one coordinate, i.e. Yn = Yn(x), and we take derivatives with respect to x. If we write the operator for taking derivatives D then we get
So we see that D is indeed a linear operator. Adding a number to a wave function is not a linear operation. Suppose we have the operator B, which we define as
where K is a number. We now have
| B(c1Y1+c2Y2) = (c1Y1+c2Y2)+K |
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and
| c1BY1+c2BY2 = c1Y1+K+c2Y2+K. |
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These two expression are not equal to each other when K ¹ 0, so B is not a linear operator.
There are many operators in quantum mechanics. An important point is that there is an operator for each physical quantity. So there is an operator for each coordinate of each particle, there is an operator for each component of the momentum of each particle, there is an operator for each component of an angular momentum, etc. There is also an operator for the energy of a system. This operator is called the Hamiltonian. The reason why this operator is not just called the operator for the energy, but has a special name, is because of the importance of the Schrödinger equation and the special role that the energy has.
The precise form of the Hamiltonian depends on the system. An example will make clear, however, how to obtain the Hamiltonian. Suppose we have a system that consists of just a single particle that moves in just one direction. We use a coordinate x to indicate the position of the particle, and we assume that the particle at x has a potential energy given by the function V = V(x). We start by writing down the total energy of the particle in terms of the position and the momentum of the particle. The total energy consists of kinetic and potential energy. We already have the expression for the potential energy in terms of the position. This is nothing but V(x). The kinetic energy is given by mv2/2, where m is the mass of the particle and v its velocity. This form is, however, not acceptable. We have to rewrite it in terms of (the position and) the momentum. The momentum is given by p = mv. This is equivalent to v = p/m with which we get for the kinetic energy p2/2m. The total energy E is then given by
We can now get the expression for the Hamiltonian by replacing the momentum p by the operation (h/i2p)(d/dx), where i is the number we use to write complex numbers (i2 = -1). Making the replacement yields
| H = - |
h2
8p2m
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d2
dx2
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+V(x). |
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So we see that we get an operator consisting of two terms. The first term takes the second derivative of the wave function, and the second term multiplies the value of the wave function at a certain position by the potential energy at that position. The sum of these operations gives the effect of the Hamiltonian. For more complicated system, e.g. molecules, we get the Hamiltonian in the same way. The only difference is that we have more coordinate (in general an x, y, and z coordinate for each particle) and more components for the momenta. These difference components each give a second derivative of one coordinate.
The strange thing about the Hamiltonian is of course the presence of the second derivative. This originates from the replacement of the momentum by (h/i2p)(d/dx). This expression is the operator p for the momentum. The reason for this can be seen from the wave function of a free particle and the DeBroglie relation. A free particle has no preferred position. (That is why it is called free.) Because of the Copenhagen interpretation the wave function must be of the form exp[i2px/l], where l can be any real number. The probability of finding the particle somewhere is then the same for all positions. Plotting this wave function shows that l is the wave length of the wave function. A free particle also does not feel a potential (otherwise it wouldn't be free) and so moves with a constant velocity. This means that the momentum p is constant too. DeBroglie relation tells us that p is related to the wave length via p = h/l. If we work with (h/i2p)(d/dx) on the wave function we get
We see that we get back the original wave function multiplied by the momentum. This means that the operator must be the operator for the momentum.
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