How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc

Question

I'm trying to expedite some quantum mechanical calculations (expectation values etc.) by running them through Mathematica. When I say, for example,

u[x_] := Sqrt[2/L] * Sin[Pi * n * x / L]

and then take the complex conjugate, I get

Sqrt[2] Conjugate[Sqrt[1/L]] Sin[(π Conjugate[n x])/Conjugate[L]]

But I want to tell Mathematica that some of the parameters are real (i.e. $L$) and some are integer valued and real (i.e. $n$). Is there a way to do that? I've tried adapting some syntax that I've seen in other context (but do not strictly know what it means or does) but it hasn't worked. For example,

Conjugate[u[x], Im[n] = 0]
Sin[n*Pi] /. n = Integer

Don't work the way I want them to. Chugging through this, however, when it comes time to compute values (like, in this example, $\langle p^2\rangle$, I get the following:

Integrate[u[x]*(-h^2)*u''[x], {x, 0, L}]
(* => (h^2 n π (n π - 1/2 Sin[2 n π]))/L^2 *)

where the second term there is clearly zero for all integer values of $n$ (but Mathematica doesn't know that).

Answer 1

You can also use Refine with Element :

Refine[Sqrt[2] Conjugate[Sqrt[1/L]] Sin[(Pi* Conjugate[n x])/Conjugate[L]],
{Element[L, Reals], Element[n, Integers]}]

gives

and if you add that L>0:

Refine[Sqrt[2] Conjugate[Sqrt[1/L]] Sin[(Pi* Conjugate[n x])/Conjugate[L]],
{Element[L, Reals], Element[n, Integers], L > 0}]

Other simple examples :

1. Example with reals

Re[x + I*y]

returns

-Im[y] + Re[x]

but if x is assumed to be real :

Refine[Re[x + I*y], Element[x, Reals]]

gives

x - Im[y]

and if x and y are assumed to be reals :

Refine[Re[x + I*y], {Element[x, Reals], Element[y, Reals]}]

gives

x

2. Example with integers

Sin[n*Pi]

gives the same

Sin[n*Pi]

but if n is assumed to be an integer :

Refine[Sin[n*Pi], Element[n, Integers]]

returns

0

Answer 2

$Assumptions = n > 0 && n ∈ Integers && L ∈ Reals && L > 0

You can type ∈ symbol in the form n ∈ Integers using ESC+el+ESC, or by typing \[Element], or you can use the the alternative Element[n, Integers] form.

To delete all global assumptions: $Assumptions = True

Answer 3

Conjugate by default assumes that all symbolic quantities are potentially complex. This may seem annoying at first, but there is a very good reason for it, and one way to see why is to define your own version of Conjugate, and see it fail. For educational purposes, I do that below.

Define $Conjugate as follows:

$Conjugate[x_] := x /. Complex[a_, b_] :> a - I b;

This simply replaces any occurrence of I with -I. For example:

$Conjugate[u[x]]

produces

Sqrt[2] Sqrt[1/L] Sin[(n π x)/L]

which is presumably what you expected.

Beware that this simple definition of $Conjugate is only guaranteed to work on numbers and functions which satisfy $f(\overline{x})=\overline{f(x)}$, which oddly enough is true for the majority of analytic elementary functions, such as Zeta, Gamma, Sin, Log, and many others.

One possible counterexample is SphericalHarmonicY:

$Conjugate[ SphericalHarmonicY[3, θ, ϕ, a + b I]] /. 
                                                    {θ -> 2.3, ϕ -> 1.0, a -> 1.2, b -> 2.5}
$Conjugate[ SphericalHarmonicY[3, θ, ϕ,  a + b I] /. 
                                                    {θ -> 2.3, ϕ -> 1.0, a -> 1.2, b -> 2.5}]  

which produces two different outputs:

-104.193 + 41.8086 I
-0.00105549 - 0.000423525 I

In contrast, replacing $Conjugate with the built-in Conjugate gives correct results. So while Conjugate may give silly-looking outputs, it is guaranteed to be correct; thus a simple substitution $a+bi\rightarrow a-bi$ does not necessarily give correct results!

Meanwhile, to make the assumption that n is an integer in your integral, do this:

Simplify[Integrate[u[x]*(-h^2)*u''[x], {x, 0, L}], Assumptions -> n ∈ Integers]

which returns

(h^2 n^2 π^2)/L^2