/sci/ - Science & Math

[Anonymous No.7661237]

Hey /sci/, I'm just starting some intro QM and I was told that the probability density function is just the wave function multiplied by it's complex conjugate, ei. [eqn] | \psi_{(x,t)} |^2= \psi_{(x,t)} \cdot \psi_{(x,t)}^* [/eqn]
Now my question for all the math/science guys out there is, what is a complex conjugate? All I really know about it is that wherever you see [math] i [/math] you replace it with [math] -i [/math] but what if the wave function has no imaginary parts? Do I just square it to get the probability density? Is the complex conjugate just the same as the original function?

[Anonymous No.7661245]

>>7661237
> what if the wave function has no imaginary parts? Do I just square it to get the probability density? Is the complex conjugate just the same as the original function?

Yep.

Another way to think about it if you were using matrices though, is that the conjugate will flip the matrix, so if it were a column vector it the conjugate is a row vector, and so if you were to do the dot product of those two matrices you'd get a scalar instead of a larger matrix.

[Anonymous No.7661249]

>>7661245
Thanks! Is it usually standard to work with matrices in quantum mechanics? I'm curious as to what's to come.

[Anonymous No.7661254]

>>7661249
operators can be thought of as 'infinite matrices'
so yes, you need them a lot

[Anonymous No.7661262]

>>7661254
interesting. I should go back and try to relearn all that stuff about matrices that my brain blocked out as if it were a traumatic event. Thanks again for the help anon.

[Anonymous No.7661263]

>>7661237
For a wave function f, the complex conjugate is just the function f* such that f*(x)=(f(x))* for all x. The last * is just complex conjugation of a complex number. A complex number has a real part and an imaginary part and conjugating it is just changing the sign on the imaginary part. As you mentioned, this is usually just replacing every instance of i with -i. I say usually because I can't remember if it always holds, I could imagine if you choose a function which is convoluted enough it might not hold.

[Anonymous No.7661269]

>>7661249
Usually you will work with abstract matrices in bra-ket notation. You may have seen notation like:
[math] |\psi> [/math]
which would be analogous to some state represented by a column vector. and
[math] <\phi| [/math]
being a row vector. These may be actual matrices like spin states (up or down) or something totally abstract like "cat alive" and "cat dead".

[math]<\phi |\psi> [/math]

Represents an inner product (dot product). Dimension goes down and would represent how much states phi and psi are alike- if you took the eigenstates from your solved problems like particle in a box this number would be zero unless phi and psi were the same state in which case it would be 1. You can also add in an operator like the position operator:
[math]<\phi |x|\psi> [/math]
Which you have probably done, but only saw in notation using integrals.

[math] |\phi><\psi| [/math]
Represents an outer product (cross product). Dimension goes up and would be interpreted as the space of those two states.

[Anonymous No.7661276]

>>7661269
I don't think I've done any of those tbqhfam. As I said I'm just starting intro QM so the notation and especially the matrix notation has caught me really off guard. I'm officially confused.

[Anonymous No.7661284]

The reason you use the complex conjugate is that it gives the "size" of the function in a convenient way.

Let's just say we have functions of x and t. We'll call them a and b, such that psi(x,t) = a + bi, just understanding that these are functions instead of variables.

Then (a + bi)(a - bi) = a^2 + abi - abi - (bi)^2. Oh, but that's handy. The middle terms cancel, and -(bi)^2 = -(b^2)(-1) = b^2. That gives just a^2 + b^2. If you think of it in terms of plotting psi on a Cartesian grid (x is real, y is imaginary), for any fixed x and t, then all you're doing is measuring the length of the hypotenuse. (Well, the square of it.)

If the function has no imaginary part, then b = 0. Works out just the same.

[Anonymous No.7661287]

>>7661249
Operators in QM are linear maps on an infinite vector space, so they're an infinite-dimensional form of a matrix.

[Anonymous No.7661294]

>>7661287
>infinite vector space
infinite-dimensional even

Polite sage

[Anonymous No.7661302]

>>7661284
Thanks that makes a lot of sense. I also know it's used to find the time independent probability density. [eqn] \psi_{(x,t)} = \psi_{(x)} e^{-iEt/ \hbar } [/eqn] So does that mean that any wave function that is only dependent on x is just the square root of the probability density?

[Anonymous No.7661320]

>>7661294
Thanks for the sage. This secretly was a homework thread, so while I have a few people still here I might as well just post the problem. Squaring the wave function didn't cancel out any terms, so I'm kinda stuck as to where I go from here.

[Anonymous No.7661395]

>>7661302
Well, the time-independent wavefunction may have an imaginary component... and it usually does.

[Anonymous No.7661414]

>>7661320
Honestly, I think you just left the ' off your k'. Otherwise it looks okay.

[Anonymous No.7661449]

>>7661414
Well when I entered into the system it told me the answer doesn't depend on k, m, or [math] \omega [/math] So I tried using the formula [eqn] \hbar \omega = \frac{ \hbar^2 k^2}{2m} [/eqn] to get rid of the omega. Then to get rid of the m I just used the mass of an electron (because I have no idea what else to do) and I switched the k with k'. I ended up getting something like [eqn] e^{2.67*10^{-10} \sqrt{k'^3} } [/eqn] and then it told me my submission wasn't in the correct form. So either I need to multiply [math] e^{ \sqrt{k'^3} } [/math] by e^2.67*10^-10 (which my calculator is telling me is 1), or mastering physics doesn't accept apostrophes in their answers and the k' was supposed to cancel as well. Either way I'm stuck.

[Anonymous No.7661512]

>>7661414
>>7661449
I figured it out. The key term here was "harmonic oscillator" The equation I used was derived from the Energy and momentum of a free particle. [eqn] E=hf= \frac{h}{2 \pi } 2 \pi f= \hbar \omega [/eqn] [eqn] p= \frac{h}{ \lambda } = \frac{h}{2 \pi } \frac{2 \pi }{ \lambda } = \hbar k [/eqn] [eqn] E= \frac{p^2}{2m} [/eqn]

But for harmonic oscillators, [eqn] \hbar \omega = \hbar \sqrt{k'}{m} [/eqn] Thanks for all the help though, You guys actually answered a lot of my questions. I should really read the book before asking /sci/

[Anonymous No.7661643]

In QM there is almost always an imaginary part since it's a wave and you just use eulers.