Verifying that the uncertainty is 0 for a QM state

• PhysicsKush

PhysicsKush

Homework Statement
Verify that ##\Delta S_{x} = \sqrt{\langle S_{x}^{2} \rangle - \langle S_{x}\rangle^{2}} = 0## for the state ##\ket{+x}##.
Relevant Equations
$$\Delta S_{x} = \sqrt{\langle S_{x}^{2} \rangle - \langle S_{x}\rangle^{2}} = 0$$
$$\ket{+x} = \alpha \ket{+z} + \beta \ket{-z}$$
$$\langle S_{x} \rangle = |\alpha|^{2} \left( \frac{\hbar}{2}\right) + |\beta|^{2} \left( \frac{\hbar}{2}\right)$$
By definition , ##\ket{+x} = \alpha \ket{+z} + \beta \ket{-z}.## Therefore we proceed ,
\begin{align*}
\langle S_{x} \rangle &= \lvert \alpha \rvert^{2} \left(\frac{\hbar}{2}\right) + \lvert \beta\rvert^{2} \left(-\frac{\hbar}{2}\right) = (\alpha^{2} - \beta^{2})\left(\frac{\hbar}{2}\right).\\
\left(\langle S_x \rangle \right)^{2} &= \left(\alpha^{2} - \beta^{2} \right)^{2} \left(\frac{\hbar^{2}}{4}\right). \\
\therefore \Delta S_{x} &= \sqrt{\left(\frac{\hbar^{2}}{4}\right) - \left(\frac{\hbar^{2}}{4}\right)(\alpha^{2} - \beta^{2})^{2}} \\
&= \frac{\hbar}{2}\sqrt{1- \left(\alpha^{2} - \beta^{2}\right)^{2}} \\
&= \frac{\hbar}{2}\sqrt{1-(\alpha^{2} - (1-\alpha^{2}))^{2}} \\
&= \frac{\hbar}{2}\sqrt{1-(4\alpha^{4} - 4\alpha^{2} +1)} \\
&= \frac{\hbar}{2}\sqrt{4\alpha^{2} - 4\alpha^{4}} \\
&= \hbar \alpha \beta \neq 0 ?
\end{align*}

I don't understand what I'm missing or where I'm going wrong. Any indications would be appreciated.

By definition , ##\ket{+x} = \alpha \ket{+z} + \beta \ket{-z}.## Therefore we proceed ,
\begin{align*}
\langle S_{x} \rangle &= \lvert \alpha \rvert^{2} \left(\frac{\hbar}{2}\right) + \lvert \beta\rvert^{2} \left(-\frac{\hbar}{2}\right) = (\alpha^{2} - \beta^{2})\left(\frac{\hbar}{2}\right).\\
\end{align*}

That's ##\langle S_{z} \rangle##, surely?

Abhishek11235 and PhysicsKush
That's ##\langle S_{z} \rangle##, surely?
Right, that's a mistake. So ## \langle S_{x} \rangle = \lvert \langle+x | \varphi \rangle\rvert^{2}##, but I don't know what ##\varphi## is. Is it simply ##\langle S_{x} \rangle = \lvert \langle +x |+x\rangle \rvert^{2}## ?

Right, that's a mistake. So ## \langle S_{x} \rangle = \lvert \langle+x | \varphi \rangle\rvert^{2}##, but I don't know what ##\varphi## is. Is it simply ##\langle S_{x} \rangle = \lvert \langle +x |+x\rangle \rvert^{2}## ?
Is there any reason to work in the z-basis? Why not the x-basis? Does the spin-1/2 formalism work just as well in the x-basis?

One problem is that the expectation value for ##S_x## looks wrong. You seem to have used ##S_x \lvert \pm \rangle = (\pm \hbar/2) \lvert \pm \rangle##, but ##\lvert \pm \rangle## are not eigenstates of ##S_x##. Note that you never seemed to use the fact that you're working with an eigenstate of ##S_x## as opposed to an arbitrary linear combination.

Abhishek11235 and etotheipi
Is there any reason to work in the z-basis? Why not the x-basis? Does the spin-1/2 formalism work just as well in the x-basis?
Hello, all we saw in class is that for 2-level systems

\begin{gather*}
\ket{\pm x} = \frac{1}{\sqrt{2}}(\ket{+z} \pm \ket{-z}) \\
\ket{\pm y} = \frac{1}{\sqrt{2}}(\ket{+z} \pm i \ket{-z})
\end{gather*}
and more generally,
$$\ket{\varphi} = \alpha \ket{+z} + \beta \ket{-z}$$
From the equations above, I'm not sure if we saw a way to express ## \ket{+z} = \alpha' \ket{+x} + \beta' \ket{-x}##, i.e., a change of basis to ##\ket{\pm x}##. I've been spending hours and hours on this problem and I can't seem to get it right :(

I have also attempted the follwing approach:
Let ## \ket{\varphi} = \alpha \ket{+z} + \beta \ket{-z}##, then ## P_{x} = \lvert \langle+x | \varphi \rangle\rvert^{2}##. Now since ## \ket{+x} = \frac{1}{\sqrt{2}}(\ket{+z} + \ket{-z}),## it follows that
\begin{align*}
\lvert \langle+x | \varphi \rangle\rvert^{2} &= \lvert \alpha \frac{1}{\sqrt{2}} + \beta \frac{1}{\sqrt{2}} \rvert^{2} = \frac12 (\alpha + \beta)^2\\
\therefore \langle S_{x} \rangle &= \frac12 (\alpha+\beta)^2 \left( \frac{\hbar}{2}\right) + \left( 1- \frac12 (\alpha+\beta)^2 \right)\left( -\frac{\hbar}{2}\right) \\
&= \left( \frac{\hbar}{2}\right) \left((\alpha+\beta)^{2} -1 \right) \\
\implies \langle S_{x} \rangle^{2} &= \left( \frac{\hbar}{2}\right)^{2} \left((\alpha+\beta)^{2} -1 \right)^{2}
\end{align*}
Now since ##\alpha^{2} + \beta^{2} =1 \implies 1-\alpha^{2} = \beta^{2}##, the expression reduces to
\begin{align*}
\langle S_{x} \rangle^{2} &= \left( \frac{\hbar}{2}\right)^{2} 4\alpha^{2} \beta^{2} \\
\therefore \Delta S_{x} &= \sqrt{\frac{\hbar^{2}}{4} - \frac{\hbar^{2}}{4} 4\alpha^{2}\beta^{2}} \\
\Delta S_{x} &= \frac{\hbar(2\alpha^{2} -1 )}{2}
\end{align*}
Again with no success ! Could someone provide me a hint I went through all avenues that my mind can phantom with this problem :/

Before going through your working, some observations.

1) z is just an arbitrary direction. There must be no physical difference between ##|+z\rangle## and ##|+x\rangle##, other than different orientation of the axes in the lab.

2) Any measurement of ##S_x## in the ##|+x\rangle## state must return a value of ##+\frac {\hbar} 2##, by definition.

3) In your case ##\langle +x|\varphi \rangle = \langle +x|+x \rangle = 1##. You don't need to mess about with ##\alpha## and ##\beta##.

That said, it is a good exercise to calculate ##\langle S_x \rangle## for an arbitrary state ##|\varphi \rangle = \alpha |+z \rangle + \beta|-z \rangle##. So let me post where you went wrong below.

Last edited:
PhysicsKush
I have also attempted the follwing approach:
Let ## \ket{\varphi} = \alpha \ket{+z} + \beta \ket{-z}##, then ## P_{x} = \lvert \langle+x | \varphi \rangle\rvert^{2}##. Now since ## \ket{+x} = \frac{1}{\sqrt{2}}(\ket{+z} + \ket{-z}),## it follows that
\begin{align*}
\lvert \langle+x | \varphi \rangle\rvert^{2} &= \lvert \alpha \frac{1}{\sqrt{2}} + \beta \frac{1}{\sqrt{2}} \rvert^{2} = \frac12 (\alpha + \beta)^2\\
\therefore \langle S_{x} \rangle &= \frac12 (\alpha+\beta)^2 \left( \frac{\hbar}{2}\right) + \left( 1- \frac12 (\alpha+\beta)^2 \right)\left( -\frac{\hbar}{2}\right) \\
&= \left( \frac{\hbar}{2}\right) \left((\alpha+\beta)^{2} -1 \right) \\
\implies \langle S_{x} \rangle^{2} &= \left( \frac{\hbar}{2}\right)^{2} \left((\alpha+\beta)^{2} -1 \right)^{2}
\end{align*}
Now since ##\alpha^{2} + \beta^{2} =1 \implies 1-\alpha^{2} = \beta^{2}##, the expression reduces to
\begin{align*}
\langle S_{x} \rangle^{2} &= \left( \frac{\hbar}{2}\right)^{2} 4\alpha^{2} \beta^{2} \\
\therefore \Delta S_{x} &= \sqrt{\frac{\hbar^{2}}{4} - \frac{\hbar^{2}}{4} 4\alpha^{2}\beta^{2}} \\
\Delta S_{x} &= \frac{\hbar(2\alpha^{2} -1 )}{2}
\end{align*}
Again with no success ! Could someone provide me a hint I went through all avenues that my mind can phantom with this problem :/
One problem here is that you don't treat ##\alpha## and ##\beta## as complex numbers. All your working is correct as long as ##\alpha## and ##\beta## are real. In general, you should get:
$$P_{x+} = \frac 1 2 + Re(\alpha \beta^*)$$
$$\langle S_x \rangle = \hbar Re(\alpha \beta^*)$$
If, however, we pick up from where you gave up and set ##\alpha = \beta = \frac 1 {\sqrt 2}##, then:
$$\Delta S_{x} = \frac{\hbar|(2\alpha^{2} -1)|}{2} = 0$$
You were nearly there!

PhysicsKush
One problem here is that you don't treat ##\alpha## and ##\beta## as complex numbers. All your working is correct as long as ##\alpha## and ##\beta## are real. In general, you should get:
$$P_{x+} = \frac 1 2 + Re(\alpha \beta^*)$$
$$\langle S_x \rangle = \hbar Re(\alpha \beta^*)$$
If, however, we pick up from where you gave up and set ##\alpha = \beta = \frac 1 {\sqrt 2}##, then:
$$\Delta S_{x} = \frac{\hbar|(2\alpha^{2} -1)|}{2} = 0$$
You were nearly there!
Wow thank you so much ! Indeed, I completely dropped the absolute values and I did not think of setting ##\alpha \ \text{or} \ \beta = \frac{e^{i\delta_{\pm}}}{\sqrt{2}}##. Cheers!