Thus there is unit area under the graph of \u03b4 x which is remarkable given that

# Thus there is unit area under the graph of δ x which

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Thus there is unit area under the graph of δ ( x ), which is remarkable, given that the function vanishes for x negationslash = 0! Although the name of δ ( x ) includes the word ‘function’, this object is not really a function because we cannot assign it a value at the origin. It is best considered to be the limit of a series of functions that all have unit area under their graphs but become more and more sharply peaked around the origin (see Figure 2.1). The analogue of equation (1.31) is integraldisplay d x | ψ ( x ) | 2 = 1 , (2 . 43) which expresses the physical requirement that there is unit probability of finding the particle at some value of x . The analogue of equation (2.2) is I = integraldisplay d x | x )( x | . (2 . 44) 4 The analogy would be clearer if we wrote a ( x ) for ψ ( x ), but for historical reasons the letter ψ is hard to avoid in this context.
2.3 Position representation 25 Figure 2.1 A series of Gaussians of unit area. The Dirac delta function is the limit of this series of functions as the dispersion tends to zero. It is instructive to check that the operator that is defined by the right side of this equation really is the identity operator. Applying the operator to an arbitrary state | ψ ) we find I | ψ ) = integraldisplay d x | x )( x | ψ ) (2 . 45) By equations (2.37) and (2.38) the expression on the right of this equation is | ψ ) , so I is indeed the identity operator. When we multiply (2.45) by ( φ | on the left, we obtain an important formula ( φ | ψ ) = integraldisplay d x ( φ | x )( x | ψ ) = integraldisplay d ( x ) ψ ( x ) , (2 . 46) where the second equality uses equations (2.38) and (2.39). Many practical problems reduce to the evaluation of an amplitude such as ( φ | ψ ) . The expres- sion on the right of equation (2.46) is a well defined integral that evaluates to the desired number. By analogy with equation (2.5), the position operator is ˆ x = integraldisplay d xx | x )( x | . (2 . 47) After applying ˆ x to a ket | ψ ) we have a ket | φ ) = ˆ x | ψ ) whose wavefunction φ ( x ) = ( x | ˆ x | ψ ) is φ ( x ) = ( x | ˆ x | ψ ) = integraldisplay d xx ( x | x )( x | ψ ) = integraldisplay d xxδ ( x x ) ψ ( x ) = x ψ ( x ) , (2 . 48) where we have used equations (2.38) and (2.40). Equation (2.48) states that the operator ˆ x simply multiplies the wavefunction ψ ( x ) by its argument. In the position representation, operators turn functions of x into other functions of x . An easy way of making a new function out of an old one is to differentiate it. So consider the operator ˆ p that is defined by ( x | ˆ p | ψ ) = (ˆ )( x ) = h ∂ψ ∂x . (2 . 49) In Box 2.2 we show that the factor i ensures that ˆ p is a Hermitian operator. The factor ¯ h ensures that ˆ p has the dimensions of momentum: 5 we will find 5 Planck’s constant h = 2 π ¯ h has dimensions of distance × momentum, or, equivalently, energy × time, or, most simply, angular momentum.
26 Chapter 2: Operators, measurement and time evolution Box 2.2: Proof that ˆ p is Hermitian We have to show that for any states | φ ) and | ψ ) , ( ψ | ˆ p | φ ) = ( ( φ | ˆ p | ψ ) ) . We use equation (2.49) to write the left side of this equation in the position representation: ( ψ | ˆ p | φ ) = h integraldisplay d ( x ) ∂φ ∂x .

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