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, $<p^2>$, 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).

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    $\begingroup$ But I want to tell Mathematica that some of the parameters are real (ie L) and some are integer valued you can us e ComplexExpand it says expands expr assuming that all variables are real, for integers, you can use Assuming[Element[x,Integers],Simplify[....]] $\endgroup$ – Nasser Nov 21 '14 at 22:54
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    $\begingroup$ There's no general way to declare a variable as real, integer etc. Instead, some functions take an Assumptions option which affects that function. Take a look at reference.wolfram.com/language/tutorial/UsingAssumptions.html and also ComplexExpand. There can be a default for the Assumptions option through $Assumptions but it won't affect everything, only functions that know about assumptions. $\endgroup$ – Szabolcs Nov 21 '14 at 22:56

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]}]


enter image description here

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}]

enter image description here

Other simple examples :

1. Example with reals

Re[x + I*y]


-Im[y] + Re[x]

but if x is assumed to be real :

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


x - Im[y]

and if x and y are assumed to be reals :

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



2. Example with integers


gives the same


but if n is assumed to be an integer :

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


| improve this answer | |
  • $\begingroup$ Fantastic! This is exactly what I wanted. Can I tell mathematica that a variable is "globally" real valued, etc? Or do I have to add that Element[] function to every new input line? $\endgroup$ – Matthew Brunetti Nov 22 '14 at 19:38
  • $\begingroup$ After playing around with Refine[] for a bit, it looks like I can just do all of my calculations within the Refine function, add it my constraints at the end, and get a nice looking answer. Am I correct in that assessment? $\endgroup$ – Matthew Brunetti Nov 22 '14 at 21:38
  • $\begingroup$ @MatthewBrunetti Theoretically yes ... It is maybe safer to use assumptions as soon as possible in your calculations. As was shown in the other answer, for example Integrate can use Assumptions. It depends on the problem. Anyway see the docs in particular the tutorials under "Learning Ressources" ;) $\endgroup$ – SquareOne Nov 22 '14 at 22:30
  • $\begingroup$ @MatthewBrunetti Maybe the function Assuming is what you are looking for because it "spreads" the assumptions to all the functions in an expression at the same time. $\endgroup$ – SquareOne Nov 22 '14 at 23:13
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    $\begingroup$ Note that, in place of the assumption Element[t, Reals] you can use the assumption _Symbol ∈ Reals to assume that all explicit variables are real. This allows you to encapsulate this in a function without having to manually pass a list of arguments. For an exhaustive discussion, see here. $\endgroup$ – Jess Riedel Sep 27 '17 at 18:56
$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

| improve this answer | |
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    $\begingroup$ According to the documentation, use [Esc (Escape) key] el [Esc key]. $\endgroup$ – bbgodfrey Jun 13 '17 at 17:38

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:



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
| improve this answer | |
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    $\begingroup$ "which oddly enough is true for the majority of analytic elementary functions" -- This is true for an analytic function iff it has real coefficients in its Taylor series (as can easily be seen by expanding it in said Taylor series). Thus in particular if the function is real on [a segment of] the real axis, and analytic it will work. Be careful with functions with singularities, however. Your statement about logarithms is not in general true. Notice that on Mathematica's branch $\log(-1) = i \pi$ but $\log(\overline{-1}) = i \pi \neq \overline{i\pi}$. You need to cross branches. $\endgroup$ – Sharkos May 19 '16 at 21:05

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