Verifying that the uncertainty is 0 for a QM state

In summary: I was thinking about ##\alpha \ \text{or} \ \beta## as being real numbers and I was wondering how to get the imaginary part of ##\langle S_{x}\rangle##. Thank you again; this was a tough one for me, but I learned a lot !In summary, the conversation discusses the calculation of the expectation value and uncertainty of the x-component of spin for a two-level system. The key steps involve defining the state ##\ket{+x}## in terms of the basis states ##\ket{+z}## and ##\ket{-z}##, and then using the formula for the expectation value and uncertainty of ##S_x## in this state. There is also a
  • #1
PhysicsKush
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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.
 
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  • #2
Mihail Anghelici said:
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?
 
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  • #3
PeroK said:
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}## ?
 
  • #4
Mihail Anghelici said:
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?
 
  • #5
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.
 
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  • #6
PeroK said:
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 :/
 
  • #7
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.
 
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  • #8
Mihail Anghelici said:
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!
 
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  • #9
PeroK said:
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!
 

FAQ: Verifying that the uncertainty is 0 for a QM state

1. What is "uncertainty" in quantum mechanics?

Uncertainty in quantum mechanics refers to the inherent unpredictability of certain physical properties of a quantum system. This is due to the fundamental nature of quantum mechanics, which states that the exact position and momentum of a particle cannot be known simultaneously.

2. How is uncertainty calculated for a quantum state?

Uncertainty is calculated using the Heisenberg uncertainty principle, which states that the product of the uncertainties in position and momentum must be greater than or equal to a constant value. This can be represented mathematically as ΔxΔp ≥ h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant.

3. Why is it important to verify that the uncertainty is 0 for a quantum state?

Verifying that the uncertainty is 0 for a quantum state is important because it allows us to accurately predict the behavior of a quantum system. If the uncertainty is not 0, it means that there is a limit to our knowledge of the system and we cannot make precise predictions about its properties.

4. How do scientists verify that the uncertainty is 0 for a quantum state?

Scientists use various experimental techniques, such as the double-slit experiment, to verify that the uncertainty is 0 for a quantum state. They also use mathematical models and equations, such as the Schrödinger equation, to calculate the uncertainty and confirm that it is equal to 0.

5. Can the uncertainty ever be exactly 0 for a quantum state?

No, according to the Heisenberg uncertainty principle, the uncertainty can never be exactly 0 for a quantum state. This is because the principle states that there will always be a minimum amount of uncertainty in a quantum system, regardless of how precise our measurements or calculations may be.

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