The purpose of this page is to show tricks to dealing with the confusing math that sometimes arises from problems in quantum physics. It will not answer your homework, but might help you become unstuck. When a question asks you to normalize a wavefunction, they mean defining the value of some coefficient such that ∫ − ∞ ∞ | Ψ ( x , t ) | 2 d x = 1 {\displaystyle \int _{-\infty }^{\infty }\left|\Psi (x,t)\right|^{2}\,dx=1} For example, consider a function Ψ ( x , t ) = A β ( x , t ) {\displaystyle \Psi (x,t)=A\beta (x,t)} To find A, you must first find the absolute value of | Ψ ( x , t ) | 2 {\displaystyle \left|\Psi (x,t)\right|^{2}} . Because wavefunctions often have complex parts, it is not sufficient to square the wavefunction. You must instead multiply it by its complex conjugate. To learn how to take the complex conjugate of a number, see the page on Complex numbers. | Ψ ( x , t ) | 2 = A 2 β ( x , t ) β ( x , t ) ⋆ {\displaystyle \left|\Psi (x,t)\right|^{2}=A^{2}\beta (x,t)\beta (x,t)^{\star }} Rearranging to solve for A, A = 1 / ∫ − ∞ ∞ β ( x , t ) β ( x , t ) ⋆ d x {\displaystyle A=1/{\sqrt {\int _{-\infty }^{\infty }\beta (x,t)\beta (x,t)^{\star }dx}}} Note that since the limits of integration are infinite, | Ψ ( x , t ) | 2 {\displaystyle \left|\Psi (x,t)\right|^{2}} must go to zero as x goes to infinity or negative infinity. Otherwise, the probability of a particle being anywhere is greater than 1, and the function is not normalizable. There are only a few cases where these conditions are satisfied. One of the most common cases is a function where | Ψ ( x , t ) | 2 {\displaystyle \left|\Psi (x,t)\right|^{2}} is of the form A 2 e − α x 2 {\displaystyle A^{2}e^{-\alpha x^{2}}} . In which case, you end up with an integral: A 2 ∫ − ∞ ∞ e − α x 2 d x = 1 {\displaystyle A^{2}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx=1} But this means integrating ∫ e x 2 d x {\displaystyle \int e^{x^{2}}dx} , a function which is impossible to integrate. Integrating it in Mathematica and Maple will give you a bunch of complex functions which are not easy to handle. Trying to transform it to ∫ e 2 x d x {\displaystyle \int e^{2x}dx} is not the correct method to proceed, remember that there is a difference between ∫ e x 2 {\displaystyle \int e^{x^{2}}} and ∫ ( e x ) 2 {\displaystyle \int (e^{x})^{2}} . However, although the limits of integration are at infinity and negative infinity, the integral is still definite, and there is a solution. Define a new function I = ∫ − ∞ ∞ e − α x 2 d x {\displaystyle I=\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx} . Square the function. I 2 = ∫ − ∞ ∞ e − α x 2 d x ∗ ∫ − ∞ ∞ e − α x 2 d x {\displaystyle I^{2}=\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx*\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx} Use a change of variables, x = y {\displaystyle x=y} , on one integral, to get: I 2 = ∫ − ∞ ∞ e − α x 2 d x ∗ ∫ − ∞ ∞ e − α y 2 d y = ∫ − ∞ ∞ ∫ − ∞ ∞ e − α ( x 2 + y 2 ) d x d y {\displaystyle I^{2}=\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx*\int _{-\infty }^{\infty }e^{-\alpha y^{2}}dy=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-\alpha (x^{2}+y^{2})}dxdy} Do another change of variables into polar coordinates (for an introduction to polar coordinates, see the Calculus wikibook). x = r cos ⁡ ( θ ) {\displaystyle x=r\cos(\theta )} , y = r sin ⁡ ( θ ) {\displaystyle y=r\sin(\theta )} , r 2 = x 2 + y 2 {\displaystyle r^{2}=x^{2}+y^{2}} I 2 = ∫ 0 2 π ∫ 0 ∞ r e − α r 2 d r d θ {\displaystyle I^{2}=\int _{0}^{2\pi }\int _{0}^{\infty }re^{-\alpha r^{2}}drd\theta } Which is a much easier form to integrate. The five most common quantum operators in basic quantum physics problems are: x operator: x ^ = x {\displaystyle {\hat {x}}=x} x 2 {\displaystyle x^{2}} operator: x 2 ^ = x 2 {\displaystyle {\hat {x^{2}}}=x^{2}} momentum operator: p ^ = ħ / i ∇ {\displaystyle {\hat {p}}={\hbar /i}\nabla } momentum squared operator: p 2 ^ = − ħ 2 ∇ 2 {\displaystyle {\hat {p^{2}}}=-\hbar ^{2}\nabla ^{2}} Hamiltonian operator: H ^ = − ħ 2 / 2 m ∇ 2 + V {\displaystyle {\hat {H}}=-\hbar ^{2}/{2m}\nabla ^{2}+V} To use these operators on a wavefunction, sandwich it in the integral. For example, to find the average value of x, also known as the expectation value of x or the first moment of x, take the integral: < x >= ∫ − ∞ ∞ Ψ x ^ Ψ ⋆ d x {\displaystyle =\int _{-\infty }^{\infty }\Psi {\hat {x}}\Psi ^{\star }dx} If a wavefunction is symmetric and centered around the origin, it is even (f(x)=f(-x)). Therefore, the expectation value of x must be found at 0, as the probability that it will be in the region of -x is equal to the probability that it will be in the region of x. A more rigorous proof is as follows: Suppose y ( x ) = | Ψ ( x , t ) | 2 {\displaystyle y(x)=\left|\Psi (x,t)\right|^{2}} is even. y(x) = x is odd (f(-x)=-f(x)). Then y ( x ) = | Ψ ( x , t ) | 2 ∗ x {\displaystyle y(x)=\left|\Psi (x,t)\right|^{2}*x} is an odd function (even*odd = odd). Then ∫ − ∞ ∞ | Ψ ( x , t ) | 2 ∗ x d x = ∫ − ∞ 0 | Ψ ( x , t ) | 2 ∗ x d x + ∫ 0 ∞ | Ψ ( x , t ) | 2 ∗ x d x {\displaystyle \int _{-\infty }^{\infty }\left|\Psi (x,t)\right|^{2}*xdx=\int _{-\infty }^{0}\left|\Psi (x,t)\right|^{2}*xdx+\int _{0}^{\infty }\left|\Psi (x,t)\right|^{2}*xdx} Since y ( x ) = | Ψ ( x , t ) | 2 ∗ x {\displaystyle y(x)=\left|\Psi (x,t)\right|^{2}*x} is an odd function, ∫ − ∞ 0 | Ψ ( x , t ) | 2 ∗ x d x = − ∫ 0 ∞ | Ψ ( x , t ) | 2 ∗ x d x {\displaystyle \int _{-\infty }^{0}\left|\Psi (x,t)\right|^{2}*xdx=-\int _{0}^{\infty }\left|\Psi (x,t)\right|^{2}*xdx} So ∫ − ∞ ∞ | Ψ ( x , t ) | 2 ∗ x d x = − ∫ 0 ∞ | Ψ ( x , t ) | 2 ∗ x d x + ∫ 0 ∞ | Ψ ( x , t ) | 2 ∗ x d x = 0 {\displaystyle \int _{-\infty }^{\infty }\left|\Psi (x,t)\right|^{2}*xdx=-\int _{0}^{\infty }\left|\Psi (x,t)\right|^{2}*xdx+\int _{0}^{\infty }\left|\Psi (x,t)\right|^{2}*xdx=0} We can use the properties of even and odd functions to our advantage, and find the average value of  by inspection. There are many reoccurring integrals to be solved in quantum physics, and these general solutions may help you solve them. ∫ x sin ⁡ ( a x ) d x = ( 1 / a 2 ) sin ⁡ ( a x ) − ( x / a ) cos ⁡ ( a x ) {\displaystyle \int x\sin(ax)dx=(1/a^{2})\sin(ax)-(x/a)\cos(ax)} ∫ x cos ⁡ ( a x ) d x = ( 1 / a 2 ) cos ⁡ ( a x ) − ( x / a ) sin ⁡ ( a x ) {\displaystyle \int x\cos(ax)dx=(1/a^{2})\cos(ax)-(x/a)\sin(ax)} ∫ 0 ∞ x n e − x / a d x = n ! a n + 1 {\displaystyle \int _{0}^{\infty }x^{n}e^{-x/a}dx=n!a^{n+1}} ∫ 0 ∞ x 2 n e − x 2 / a 2 d x = π ( 2 n ) ! n ! ( a 2 ) 2 n + 1 {\displaystyle \int _{0}^{\infty }x^{2n}e^{-x^{2}/a^{2}}dx={\sqrt {\pi }}{\frac {(2n)!}{n! }}({\frac {a}{2}})^{2n+1}} ∫ 0 ∞ x 2 n + 1 e − x 2 / a 2 d x = π a 2 n + 2 {\displaystyle \int _{0}^{\infty }x^{2n+1}e^{-x^{2}/a^{2}}dx={\sqrt {\pi }}a^{2n+2}} Integration by parts is also very useful: ∫ u d v = u v − ∫ v d u {\displaystyle \int udv=uv-\int vdu} Often in a question involving expectation values, you will be expected to find the expectation value and the expectation value squared, so using integration by parts usually means you've already solved a part of the integral for the expectation value squared. Although the math involved in finding expectation values is messy and offers many opportunities for making mistakes, quantum theory comes with it's own error check - the Heisenberg uncertainty principle. σ x σ p ≥ ħ 2 {\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}}} σ x {\displaystyle \sigma _{x}} is the standard deviation of position, σ p {\displaystyle \sigma _{p}} is the standard deviation of momentum. Don't worry, we don't have to do a whole lot of tedious calculations like you might be used to in statistics to find standard deviation. Instead, use the relation: σ x 2 =< x 2 > − < x > 2 {\displaystyle \sigma _{x}^{2}=-^{2}} If you find that your answers don't satisfy the uncertainty principle, either you've made a mistake, or you've been assigned an unphysical question.
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