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# Math Test at Interview?

Discussion in 'Prop Firms' started by TradingBillions, Mar 3, 2007.

Have any of you ever had a math test for your interview at a prop firm? I'm getting one and am not sure exactly what will be asked of me. Any ideas?

3. ### CalScholar

Who are you interviewing with, if you don't mind me asking. You can PM if you like. In exchange, I've got a piece of advice for you...

4. ### Wayne Gibbous

Those tests are no big deal.

This is the one I had:

A particle with mass meff is confined to move in one dimension under the potential V(x) = {0 if -L/2 < x < L/2, Â¥ otherwise}. Schroedinger's equation for this system is

Use the variation principle to find approximate eigenvalues and eigenfunctions for a trial function having the form of a polynomial summation.

Dimensionless Form
With a little effort the Schroedinger equation can be transformed into an equivalent but simpler equation without units. The new dimensionless Schroedinger equation applies to all PIB systems regardless of the particle mass or the length of the box. First, remove the units of length by defining x=yL/2 for the potential is zero when -1<y<1 and infinte outside this range. Thus, the box is referenced to dimensionless coordinate y and has been standardized to length 2 with the center at y=0. This change of coordinates induces a change of the wave function: u(y) = y(yL/2). Second, introduce the "natural" or characteristic energy and divide both sides of Schroedinger's equation by this amount. The result is the following dimensionless Schroedinger equation:
,
where e=E/Enatural.

Notice that the dimensionless Schroedinger equation contains no memory of the particle mass or the box length - this information is contained in the coordinate transformation x=yL/2 and the natural energy. The goal will be to solve the dimensionless form of the problem and then restore units for the particular mass and box at hand.

Choosing the Basis Functions
It is crucial to satisfy the boundary conditions: any basis functions used must satisfy the boundary conditions. With this in mind, some possible variational trial functions that might be part of a basis are f1(y)=(y-1)(y+1), f2(y)=cos(py/2), and f3(y)=cosh(y)-cosh(1). In order to perform the linear variation we require a large set of linearly independent trial functions.

Example
As an example, here is such a set generalized from f1:
, with n=0,1,...,M.
For instance, when M=2 the three basis functions are f0=(y-1)3(y+1), f1=(y-1)2(y+1)2, and f2=(y-1)(y+1)3. Ultimately, the basis functions are chosen for reasons of simplicity, convenience, and aesthetics.

Basis functions are only useful if the required overlap and hamiltonian matrix elements (integrals) can be readily computed. We illustrate using the example basis.

Overlap and Hamiltonian Matrices
Elements of the overlap matrix are computed from the definition .

For the polynomial basis functions of the example, this is straightforward though tedious. Both basis functions can be expressed in a binomial expansion the the powers of y can be integrated term by term. From the definition, it follows that the overlap matrix has to be symmetric: Si,j = Sj,i for our example.

Elements of the hamiltonian matrix are computed from . This time, the second derivative of basis functions has to be computed and again integrals performed. For our example, the derivatives are polynomials in y so that the matrix element evaluation is straightforward. As a check, the hamiltonian matrix also is symmetric: Hi,j = Hj,i.

Numerical values of S and H are shown in the complete solution file.

Secular equation and secular determinant
Eigenvalues are found from the secular determinant: |H-WS|=0, a M-th order polynomial with M roots. Denote these roots by W0<W1<...WM. For the ground state, the smallest root, W0, is used. Higher roots are bounds for excited states.

For the example, with M=3 (four basis functions) the lowest root is 1.00001471 (an error of only ~15 ppm from the exact ground state energy).

Wave functions are found by solving the secular equation: (H-WiS)ai=0, and substituting the eigenvector into the basis expansion: . Note that the vector ai is labelled for its eigenvalue and its elements also carry a label for the basis function. The collection of ai,j comprise a square matrix the columns of which are the the eigenvectors belonging to Wi.

For the example, see the calculated S and H matrices

Heh. Just kiddin.

5. ### neke

What an interesting diversion? I am sure even you do not understand what you wrote. Where did you dub this from?

6. ### Wayne Gibbous

I probably could understand most of it with enough work. But I'm way too lazy and it's been 25 years since I was in EE school...

It's some Quantum Mechanics stuff dealing with particle motion. I'd rather trade Nat Gas futures!

No. They usually don't give you a math test. It would be unusual if they did. They dont' care if you know how to add and subtract. They only care if you have the guts to buy and sell 1 million plus shares a day. The more shares you trade, the more money they make off you in commissions. Your answer is, NO, there should be no math test.

LOL. He's kidding. LOL.

Umm hopefully you read what I wrote because YES I am getting a math test

10. ### Hydroblunt

He is interviewing with a REAL prop firm (aka futures/options prop) not the fake prop (aka US equities prop).

So a futures shop or options shop. Or maybe credit derivatives or whatever. Those places will give math & logic tests, sometimes the problems are not easy and require quick thinking.

Like Susquehanna is known to give math some other places as well. I've even heard where they give the test over the phone, think I had one once.

#10     Mar 4, 2007
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