The numbers 2 and -2 corresponding to the two points we have considered are the coordinates of these points. The statement \(2+1=3\) can be expressed as a vector addition . The figure below shows the two vectors corresponding to the points at 2 and 1, respectively: Figure 4: The sum can be expressed by placing the vectors together such that the head of one touches the tail of the other. The net length and direction is the answer as shown below: The result is vector ``3 to the right.'' Rather than continuing to specify vectors in this clumsy way, we introduce a notation for them. Generally, vectors are designated by a boldface symbol, e.g., \(v\) . A standard notation is to specify a vector simply by the coordinates of the point it locates. Thus, if the vector \(v\) is ``2 to the right'', then we write \(v=2\) , and if it is ``2 to the left'' it becomes \(v=-2\) . Here, the minus sign allows the direction to be determined unambiguously. In one dimension, directions are easy, as there are only two choices, ``to the right'' or ``to the left''. By contrast, in two dimensions, there is an infinite number of directions. Now two points are required to specify the location of a point in a plane; its \(x\) and \(y\) coordinates. The coordinates are written as an ordered pair \((x, y)\). Thus, the point shown below: has coordinates \((2,1)\). The vector locating this point appears as: and is given simply by the coordinates \(v=(2,1)\) . Another way to express a vector is to introduce unit vectors \(\hat{i}\) and \(\hat{j}\) along the \(x\) and \(y\) axes and write : \[v = x\hat{i} + y\hat{j} \nonumber \] where \(\hat{i}\) is the vector \((1,0)\) and \(\hat{j}\) is the vector \((0,1)\). Although we will not use this notation in the course, you should be aware of it as many other sources employ it. The magnitude of the vector, written as \(\left | v \right |\), is its length, given by the length of the hypotenuse of the right triangle: \(\left | v \right | = \sqrt{5}\). Its direction is, by convention, given by the angle that the vector makes with the \(x\) axis: \(\theta = tan^{-1}(1/2) = 26.57\) degrees. In general, if a vector is given by \(v = (a,b)\), then its length is \[\left | v \right | = \sqrt{a^2+b^2} \nonumber \] and its direction is \[\theta = tan^{-1}\frac{b}{a} \nonumber \] Note that the length and direction of a vector are the same as the polar coordinate representation of the point \((a,b)\) . The numbers \(a\) and \(b\) are called the components of \(v\) . The sum of two vectors, say \(v_1 = (2,1)\) and \(v_2 = (1,2)\) is obtained by adding the components separately: \(v_1 + v_2 = (2+1, 1+2) = (3,3)\). This is the same as placing the vectors end-to-end as in the one-dimensional case: In general, if \(v_1 = (a_1, b_1)\) and \(v_2 = (a_2, b_2)\) , then the sum \(v_3 = v_1 + v_2\) is given by \[v_3 = (a_1+a_2, b_1+b_2 \nonumber \] or if \(v_3 = v_1 - v_2\) , then \[v_3 = (a_1-a_2, b_1 - b_2) \nonumber \] In general, if \(V\) is the sum of \(N\) vectors \(v_i\) , \(i = 1,...,N\) , then \(V\) will be given by \[V = \sum_{i=1}^{N} v_i = \left (\sum_{i=1}^{N}a_i, \sum_{i=1}^{N}b_i \right ) \nonumber \] Also, the product of a vector \(v\) and a constant \(c\) is obtained simply by multiplying the components of \(v\) by \(c\) : \(cv = (ca,cb)\) One other important operation involves finding the angle between two vectors: The angle \(\theta\) between the two vectors \(v_1\) and \(v_2\) is given by \[\theta = \cos^{-1} \left [\frac{v_1\cdot v_2}{|v_1||v_2|} \right ] \nonumber \] where the product \(v_1\cdot v_2\) is called the dot product between the vectors \(v_1\) and \(v_2\) . If \(v_1 = (a_1,b_1)\) and \(v_2 = (a_2, b_2)\) , then \[v_1 \cdot v_2 = a_1a_2 + b_1b_2 \nonumber \] Note that the dot product is a number, not a vector! Everything said here, generalizes to three dimensions, except that the vectors now have a third component, e.g., \(v = (a,b,c)\). Thus, for example, the dot product between two vectors becomes \[v_1 \cdot v_2 = a_1a_2 + b_1b_2+c_1c_2 \nonumber \] and the length of a vector is given by \[\left | v \right | = \sqrt{a^2+b^2+c^2} \nonumber \] Trigonometric identities Trigonometry and the identities obeyed by trigonometric functions arise in many of the calculations we will be doing this semester. Hence, it is important for you to be familiar with these. Let us begin by recalling the definitions of the basic functions \(\cos(x)\) and \(\sin(x)\). Here, we will assume \(x \epsilon [0, 2\pi]\), although \since the functions are periodic, it does not matter if we extend the domain of \(x\). However, by restricting \(x\) in this way, we can make the convenient connection of \(x\) to one of the angles in a right triangle (see figure). Let \(c\) be the length of the hypotenuse of the triangle, and let \(a\) and \(b\) be the sides adjacent and opposite to the angle \(x\) . Then, we have the definitions Note that the definitions of \(\cos(x)\) and \(\sin(x)\) imply that for certain values of \(x\) , \(x = 0\) , \(x = \pi/2\) , \(x = \pi\) , \(x = 3\pi/2\) and \(x = 2\pi\) , \[\cos(0) = 1, \sin(0) = 0 \nonumber \] \[\cos(\pi/2)=0, \sin(\pi/2)=1 \nonumber \] \[\cos(\pi) = -1, \sin(\pi)=0 \nonumber \] \[\cos(3\pi/2)=0, \sin(3\pi/2)=-1 \nonumber \] \[\cos(2\pi) = \cos(0) = 1, \sin(2\pi) = \sin(0) = 0 \nonumber \] A number of other identities concern trigonometric functions whose arguments are sums or differences of angles \(x\pm y\): \[\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y), \cos(x-y) = \cos(x)\cos(y)+\sin(x)\sin(y) \nonumber \] \[\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y), \sin(x-y) = \sin(x)\cos(y)-\cos(x)\sin(y) \nonumber \] In fact, the difference formulas can be derived from the sum formulas u\sing the simple symmetry relations \(\cos(-x)= \cos(x)\) and \(\sin(-x) = -\sin(x)\) . In the next section, we will introduce a very simple technique for proving these relations. Now, if we set \(x = y\), we can derive a useful identity for \(\cos^{2}(x)\). Setting \(x=y\) in the \(\cos(x+y)\) and \(\cos(x-y)\) identities, and adding the two identities together, we obtain \[\cos(2x) = \cos^{2}(x)-\sin^{2}(x) \nonumber \] \[\cos(0) = 1 = \cos^{2}(x)+\sin^{2}(x) \nonumber \] \[1+\cos(2x) = 2\cos^{2}(x) \nonumber \] \[\cos^{2}(x) = \frac{1}{2}[1+\cos(2x)] \nonumber \] If we subtract the identities instead of adding them, we can derive a similar identity for \(\sin^{2}(x)\) : \[\sin^{2}(x) = \frac{1}{2}[1-\cos(2x)] \nonumber \] Finally, we note that \(\cos(x)\) and \(\sin(x)\) are expressible as infinite power series (also known as Taylor series) as \[\cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...=\sum_{k=0}^{\infty}(-1)^{k}\frac{1}{(2k)!}x^{2k} \nonumber \] \[\sin(x) = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+... = \sum_{k=0}^{\infty}(-1)^{k}\frac{1}{(2k+1)!}x^{2k+1} \nonumber \] These power series expressions are actually used in calculators and computer programs to evaluate \(\cos(x)\) and \(\sin(x)\) . Complex numbers Complex numbers play a key role in quantum mechanics, even though anything that can be measured experimentally must ultimately be expressed as a real number. Without quantum theory, it is not possible to understand the nature of the chemical bond, and \since quantum theory is expressed in terms of complex numbers, it is important to review some of the basic facts about complex quantities. Complex numbers are so named because they are composed of two types of more fundamental numbers: real numbers and imaginary numbers. Real numbers are very familiar and are the numbers we deal with on a daily basis. However, when one embarks on the study of nonlinear algebraic equations, it rapidly becomes clear that some equations do not have solutions within the set of real numbers. Take the example of the equation \[x^2+1=0 \nonumber \] There are no real numbers \(x\) that satisfy this equation. The reason is clear if we just naively solve for \(x\) to find \(x=\pm\sqrt{-1}\) . The quantity \(\sqrt{-1}\) is not defined within the real numbers. But mathematicians are never satisfied with the idea that something cannot be defined. In this case, the conundrum can be solved by simply extending the concept of a ``number'' to allow \(\sqrt{-1}\) to be defined. This involves inventing a new set of numbers, called ``imaginary numbers'' (for no better reason than ``imaginary'' is one of the opposites to ``real''), in which \(\sqrt{-1}\) has a definite value, denoted \(i\) . Then, the solution to the above equation is simply \(x=\pm i\) . The number \(i\) is the fundamental ``integer'' unit in the set of imaginary numbers just as \(1\) is the fundamental integer unit in the real numbers. So, just as any real number \(a\) can always be expressed as \(a\cdot 1\) , any imaginary number can be expressed as a multiple of \(i\) . If \(\beta\) is an imaginary number, then we can write \(\beta\) as \(b\cdot i\) , where \(b\) is a real number. The number \(i\) , itself, is really \(1\cdot i\) , and clearly, \(1\) is real. Given imaginary numbers, the most general kind of number we can now imagine has both real and imaginary parts, and such a number is known as a complex number . A complex number \(z\) is expressed as the sum of its real and imaginary parts. If the real part of \(z\) is \(a\) and the imaginary part of \(z\) is \(bi\), then we would write \[z=a+bi \nonumber \] Note that this is similar to expres\sing a two-dimensional vector \(v=x\hat{i}+y\hat{j}\) . Indeed, we can exploit this analogy and use the ordered-pair notation to write a complex number as \(z=(a,bi)\) as an alternate notation. As we will see, there are many such analogies between complex numbers and two-dimensional vectors. For instance, we can think of a complex number as a point in a plane in which the horizontal axis corresponds to the real parts of all complex numbers and the vertical axis corresponds to the imaginary parts. The complex number \(z=a+bi\) would be the point \((a,b)\) in this plane. In order to extract the real and imaginary parts of \(z\) , we introduce the functions \(Re\) and \(Im\) such that \(Re(z) = a\) and \(Im(z)=b\) . We noted above that experimental measurements ultimately yield real numbers, so we need a way of going from a complex number to a real one. For this, we exploit the analogy with two-dimensional vectors even more and define the ``measure'' or magnitude of a complex number \(z=a+ib\), which we denote as \(\left | z\right |\), as \[\left | z \right | = \sqrt{a^2+b^2} \nonumber \] which is the same as the magnitude of a two-dimensional vector \(v = (a,b) = a\hat{i}+b\hat{j}\) . Another important definition is the so-called complex conjugate of \(z\) , denoted \(z^{*}\) , which is defined as follows: If \(z=a+ib\) , then \(z^{*}=a-ib\) . In general, all we have to do to obtain the complex conjugate of a number is replace \(i\) by \(-i\) . Note that the magnitude of \(z\) is \[\left | z \right |=\sqrt{z^{*}z}=\sqrt{(a-ib)(a+ib)}=\sqrt{a^2+iab-iab+b^2}=\sqrt{a^2+b^2} \nonumber \] where we have used the fact that \(i(-i)=-i^2=1\) . Thus, a complex number times its complex conjugate is a real number. What happens if we are given a complex number, but it is not expressed in the simple form \(z=a+ib\)? Can we manipulate it so that it finally does have this form? The answer is 'yes' with a little mathematical creativity. Consider, first, a simple product \(z=(u+iv)(x+iy)\). This can be expressed in the canonical form by simply multiplying it out algebraically: \[z=ux+ivx+iuy-vy=(ux-vy)+i(vx+uy) \nonumber \] Next, consider a ratio \(z=(u+iv)/(x+iy)\) . In order to simplify this, we use the fact that \((x+iy)(x-iy)\) is real. Thus, we multiply \(z\) by \(1\) in the form of \((x-iy)/(x-iy)\) : \[z=\frac{u+iv}{x+iy}\frac{x-iy}{x-iy}=\frac{(u+iv)(x-iy)}{(x+iy)(x-iy)}=\frac{ux+yv+i(vx-uy)}{x^2+y^2}=\frac{ux+yv}{x^2+y^2}+i\frac{vx-uy}{x^2+y^2} \nonumber \] One of the most frequently occuring complex number forms is \[z=e^{i\theta} \nonumber \] where \(\theta\) is a real number. The complex conjugate of \(z\) is obtained simply by replacing \(i\) by \(-i\) : \[z^{*}=e^{-i\theta} \nonumber \] and so that magnitude of \(z\) is \[\left | z \right |^{2}=z^{*}z=e^{-i\theta}e^{i\theta}=1 \nonumber \] So, \(z\) is a complex number whose square magnitude is \(1\). On the other hand, if we seek to represent \(z\) in canonical form as \(z=a+ib\) , then it follows that \[\left | z \right |^{2}=a^2+b^2=1 \nonumber \] \since \(a\) and \(b\) are actually functions of \(\theta\) , we can ask what two functions of \(\theta\) are related by \(a^2+b^2=1\) . We have already seen what these are - they are \(\cos(\theta)\) and \(\sin(\theta)\) \since \(\cos^{2}(\theta)+\sin^{2}(\theta)=1\) . Thus, in canonical form, we can let \(a=\cos(\theta)\) and \(b=\sin(\theta)\) and write \[z=e^{i\theta}=\cos(\theta)+i \sin(\theta) \nonumber \] which is known as Euler's formula . The above argument is not a very rigorous one, as there are many other functions \(a\) and \(b\) of \(\theta\) we could have chosen that satisfy \(a^2+b^2=1\) . However, we can actually prove rigorously that \(a\) and \(b\) must be, respectively, \(\cos(\theta)\) and \(\sin(\theta)\) u\sing the fact that the function \(exp(x)\) has the following Taylor series expression: \[e^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+...=\sum_{k=0}^{\infty}\frac{x^k}{k!} \nonumber \] Applying this to \(exp(i\theta)\) , we obtain \[e^{i\theta}=1+i\theta+\frac{(i\theta)^{2}}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}+...=\sum_{k=0}^{\infty}\frac{(i\theta)^k}{k!} \nonumber \] Now the even power terms are purely real, and the odd powers are purely imaginary, so if we wish to put the above power series into the canonical form, we obtain \[e^{i\theta}=1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\frac{\theta^{6}}{6!}+...+i\left ( \theta-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!+...} \right )=\cos(\theta)+i \sin(\theta) \nonumber \] which proves Euler's formula. Euler's formula can be used to simplify proofs of trigonometric identities. For example, we know that \(z^{*}z=1\). But from Euler's formula, \(z^{*}z=\cos^{2}(\theta)+\sin^{2}(\theta)\), so \(\cos^{2}(\theta)+\sin^{2}(\theta)=1\). If we use the fact that \[z^{*}=e^{-i\theta}=\cos(\theta)-i \sin(\theta) \nonumber \] Thus, adding \(z\) to \(z^{*}\) , we obtain \[\cos(\theta)=\frac{1}{2}\left ( e^{i\theta}+e^{-i\theta}\right ) \nonumber \] Subtracting \(z\) from \(z^{*}\) yields \[\sin(\theta)=\frac{1}{2i}\left ( e^{i\theta}-e^{-i\theta}\right ) \nonumber \] Now consider \(\cos^{2}(\theta)\) . From the above formula, we have \[\cos^{2}(\theta)=\frac{1}{4}\left ( e^{i\theta}+e^{-i\theta}\right )^{2}=\frac{1}{4}\left ( e^{2i\theta}+2e^{i\theta}e^{-i\theta}+e^{-2i\theta}\right ) \nonumber \] \[=\frac{1}{4}\left ( 2+e^{2i\theta}+e^{-2i\theta}\right )=\frac{1}{2}+\frac{1}{4}\left ( e^{2i\theta}+e^{-2i\theta}\right )=\frac{1}{2}\left [1+\cos(2\theta)\right ] \nonumber \] Finally, note that if we multiply Euler's formula by a real number \(A\) , we have a complex number of the form \[z=A e^{i\theta} \nonumber \] In canonical form \[z=A \cos(\theta)+i A \sin(\theta) \nonumber \] and \(\left | z \right |^{2}=A exp(-i\theta)*A exp(i\theta)=A^2\) , so \(\left | z \right | = A\) . When we write \(z=A exp(i\theta)\) , The angle \(\theta\) is called the phase of \(z\) . Derivative of a function A line, described by the equation \[y=mx+b \nonumber \] has a constant slope \(m\) defined by \[m=\frac{\Delta y}{\Delta x} \nonumber \] where the above equations means compute the change, \(\Delta y\) in \(y\) for a given change, \(\Delta x\) in \(x\) as shown in the figure: For a general function, such as that shown below: the slope is not a constant. However, it is possible to define the slope of the curve at each point by taking the slope of the line tangent to the curve at each point. The slope will also be a function of \(x\) and is called the derivative of the function \(f(x)\) . To see how the slope is computed, consider a line connecting two points of the function corresponding to two points \(x\) and \(x+\Delta x\) , where \(\Delta x\) is a small number: The slope of this line is \[m=\frac{f(x+\Delta x)-f(x)}{\Delta x} \nonumber \] To find the slope at the \single point \(P\) , we need to let \(\Delta x\) go to 0, so that the line intersects the curve at the point \(P\) , where it will be tangent to the curve. Thus, by taking the limit as \(\Delta x \rightarrow 0\) , we obtain the derivative of \(f(x)\) at the point \(P\) . But \since the point \(P\) is arbitrary, we may perform this operation at any point and, thus, obtain the derivative of \(f(x)\) at any point. The general expression is \[f'(x) = \frac{df}{dx}=lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \nonumber \] where \(f'(x)\) and \(df/dx\) are standard notations for the derivative. As an example, consider the function \(f(x) = x^2\). According to the formula: \[f(x)=ln x, f'(x)=\frac{1}{x} \nonumber \] \[f(x)=\sin x, f'(x)=\cos x \nonumber \] \[f(x) = \cos x' f'(x) = -\sin x \nonumber \] Other important rules of differentiation are the product rule : if \(h(x)=f(x)g(x)\) then \(h'(x)=f'(x)g(x)+f(x)g'(x)\) and the chain rule : if \(h(x)=f(g(x))\), then \(h'(x)=f'(g(x))g'(x)\) Integral of a function It is often necessary to calculate the area under a function, \(f(x)\) between two points \(x_1\) and \(x_2\). Such an operation is necessary, for example, in mechanics to compute the work performed by an agent exerting a force on an object. An analogous situation arises in thermodynamics in computing the work done by an agent exerting an external pressure on a system. Consider the area under the curve designated below: One way to calculate the area approximately would be to cover the area with narrow rectangles or width \(\Delta x\) :