# Calculating the amplitude that a particle is spin up

1. Sep 9, 2013

### dEdt

The wavefunction for a spin 1/2 particle is a spinor field of the form $$\psi(\mathbf{x},t)=\left( \begin{array}{cc} \psi_{+}(\mathbf{x},t)\\ \psi_{-}(\mathbf{x},t) \end{array} \right).$$
$\psi_{+}(\mathbf{x},t)$ is the amplitude that the particle is both spin up and located at position x at time t. How can I calculate the amplitude that the particle is just spin up, without any regards for its position? Would it be something like
$$\int \psi_{+}(\mathbf{x,t})d^3 \mathbf{x}?$$

Last edited: Sep 9, 2013
2. Sep 11, 2013

### Ravi Mohan

Here is what I thought of

$$|\Psi\rangle = |\psi\rangle\otimes|\alpha\rangle$$

$$\int dx |x\rangle\langle x| \otimes \left(|\uparrow\rangle\langle \uparrow |+ |\downarrow\rangle\langle \downarrow |\right) = \mathbb{I}$$

Thus we have the expansion
$$\left[\int dx |x\rangle\langle x| \otimes \left(|\uparrow\rangle\langle \uparrow |+ |\downarrow\rangle\langle \downarrow |\right)\right] |\Psi\rangle = \int dx \langle x|\psi\rangle |x\rangle \otimes \left(\langle \uparrow |\alpha\rangle |\uparrow\rangle + \langle \downarrow |\alpha\rangle |\downarrow\rangle\right)$$

Take the inner product with spin up state vector. [I dont know if this is valid]
$$\langle \uparrow |\Psi\rangle = \int dx \langle x|\psi\rangle \langle \uparrow |\alpha\rangle |x\rangle$$

The norm of this should give us the desired number but this number can be obtained by taking inner product of $|\psi\rangle\otimes |\uparrow\rangle$ with the state vector written in first equation.

I am myself confused. How do we write $\psi_{+}$ in form of innerproduct?
Can we write
$$\psi_{+} = \langle x|\psi\rangle \langle \uparrow |\alpha\rangle ?$$

Last edited: Sep 11, 2013
3. Sep 13, 2013

### Ravi Mohan

any correction/suggestion will be helpful :)

I think the correct answer is (in terms of probability)
$$\int |\psi_{+}(\mathbf{x,t})|^2d^3 \mathbf{x}$$
I want to know if the mathematics I have done is correct. Thanks

4. Sep 13, 2013

### The_Duck

There's no such quantity. You can calculate the *probability* that the particle is spin-up without regards for its position, and your expression for this probability in #3 is correct. But in general it doesn't make sense to talk about the *amplitude* for the particle to be spin-up, regardless of position. You can't just do $\int d^3 x \psi_+(x)$ because squaring this will not give you the correct probability that you wrote down in #3.

5. Sep 13, 2013

### Ravi Mohan

In quantum computation we do it all the time. We dont care at what position our nucleus is (in case of NMR) but we do care about the amplitudes of nucleus in spin up and down states.

6. Sep 13, 2013

### The_Duck

Sure, I think it makes sense to define such an amplitude if $\psi_+(x)$ has essentially the same phase everywhere in space. Then I think we can define a sensible amplitude $e^{i \phi} \sqrt{\int d^3 x |\psi_+ (x)|^2}$ where $e^{i \phi}$ is the constant global phase of $\psi_+(x)$. But if $\psi_+(x)$ is not coherent in this way I don't think you can define an amplitude to be spin up independent of position. We are tracing out the position degree of freedom and we end up with a density matrix, whose entries are probabilities and not amplitudes.