Taking first semester Quantum Mechanics and our first assignment is some pretty basic stuff related to the braket...

Taking first semester Quantum Mechanics and our first assignment is some pretty basic stuff related to the braket notation.

I'm realizing I was never very fluent in my complex conjugates and how to deal with absolute values (yikes!).


I'm wondering if there are some good examples anywhere online regarding normalization of a wave state? Or even somewhere I can check my work to make sure I'm correct.

For example, here is one of my problems:

[math]\mid \psi \rangle = 3\mid + \rangle - e^{i \pi /3} | - \rangle [/math]

[math]1=\langle \psi \mid \psi \rangle[/math]

[math]1= C^* ( 3\langle+| + e^{i\pi/3} \langle-|)C(3\langle+| - e^{i\pi/3} \langle-|)[/math]
And in previous problems I end up with |C|^2= -1/7, which doesn't make sense.

Other urls found in this thread:

mathpages.com/home/kmath638/kmath638.htm
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Bras and kets are complex conjugates of each other, so just take the complex conjugate of the ket then slam them together and solve.

I know. But what am I doing wrong here? How can the absolute value of C^2 be negative??

Your bra has picked up a minus on the 4, but 4 isn't complex. [eqn] | \psi \rangle = 3 | + \rangle + 4 | - \rangle \\ \implies \langle \psi | = 3 \langle + | + 4 \langle - | [/eqn]

But I thought the bra was the conjugate? Hmm, something I'm not getting here. I mean it makes sense, so you only switch the signs if the 'b' term in the complex number is imaginary?

What about this, pic realted.

When you take the complex conjugate of something the sign attached to the [math] i [/math] changes, so [eqn] \left ( e^{ i \pi / 3 } \right ) ^{*} = e ^ { -i \pi / 3 } [/eqn]

mathpages.com/home/kmath638/kmath638.htm

|+> and |-> are orthonormal. =0, =1, etc.

=9+(-exp(-ipi/3))(-exp(ipi/3))=9+1=10

So the normalized state is |psi>/sqrt(10)

Okay so... this!

Don't know if this is useful at all, but here goes

[math] \displaystyle
\\ z_1 = x_1+y_1i \; \; \; \; \; z_2 = x_2+y_2i
\\ z_1^* = x_1-y_1i \; \; \; \; \; z_2^* = x_2-y_2i
\\ | z_1 | = \sqrt{x_1^2+y_1^2} \; \; \; \; \; | z_2 | = \sqrt{x_2^2+y_2^2}
\\ z_1+z_2 = x_1+x_2 + (y_1+y_2)i
\\ \left | z_1+z_2 \right |^2 = \left ( \sqrt{(x_1+x_2)^2+(y_1+y_2)^2} \right ) ^2 = (x_1+x_2)^2 +(y_1+y_2)^2
\\ z_1z_2^* = x_1x_2 -x_1y_2i +y_1ix_2 -y_1iy_2i = x_1x_2+y_1y_2 +(x_2y_1-x_1y_2)i
\\ z_1^*z_2 = x_1x_2 +x_1y_2i -y_1ix_2 -y_1iy_2i = x_1x_2+y_1y_2 +(x_1y_2-x_2y_1)i
\\ z_1z_2^* + z_1^*z_2 = 2(x_1x_2+y_1y_2) = \text{2Re}(z_1z_2^*) = \text{2Re}(z_1^*z_2)
\\ |z_1|^2+|z_2|^2 + z_1z_2^* + z_1^*z_2 = x_1^2+y_1^2 + x_2^2 + y_2^2 + 2(x_1x_2+y_1y_2)
\\ = (x_1^2 + 2x_1x_2 + x_2^2) + (y_1^2 + 2y_1y_2 + y_2^2) = (x_1+x_2)^2 +(y_1+y_2)^2
\\
[/math]

Fuck, apologies for the sideways shit. Didn't appear that way on my device.

Thanks!


Much thanks! My complex conjugation skills are questionable. Never formally learned it or practiced it. I know how to deal with i, but not conjugation and such. So thanks.

Yes, that's it.

What do [math]| - \rangle[/math] and [math]\mid + \rangle[/math] denote?

My understanding is that brakets just denote the standard hermitian scalar product, no?

Awesome.

Lastly it is asking me to find the probabilities of finding spin up or spin down along all THREE cartesian axes. Use braket notation for the whole thing.


I don't see what differences would be between the axes? I can do the calculation for the z-axis, but what would the difference be for the x/y axes?

Bra-ket notation is such cancer.

In this context, it's the spin state of a particle.

Bumping this.

Consider the complex conjugate to just mean that i becomes (-i)

brilliant!

why would this be brilliant? its the definition of complex conjugate.

Because I never viewed it that way. Only the dumb a+\-bi form.

Also, this