Collection of Solved Problems in Physics

Linear combination of eigenstates of L² and L_z

Task number: 4580

• Hint 1

  Recall or look up the relation for calculating the expectation value of a physical observable \(F\) in a state described by the normalised wavefunction \(\psi\).

• Solution to hint 1

  The expectation value of a physical observable \(F\) in a state described by the normalised wavefunction \(\psi\) can be determined from \[\langle F \rangle_{\psi} = \left \langle \psi \, \Big | \, \hat F \psi \right \rangle = \psi^{\dagger} \hat F \psi \, ,\] where \(\hat F\) is the operator associated with the physical observable \(F\) and the dagger denotes the Hermitian adjoint, i.e., the transpose of the matrix followed by complex conjugation.

• Hint 2

  Recall or look up the form of the basis states \(|1, \, m\rangle\) in the matrix representation, as well as the form of the angular momentum component operators expressed in this basis.

• Solution to hint 2

  The basis states \(|1, \, 1\rangle, \, |1, \, 0\rangle, \, |1, \, -1\rangle\) have the following matrix representation

  \[ |1, \, 1\rangle \equiv \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \, , \qquad |1, \, 0\rangle \equiv \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \, , \qquad |1, \, -1\rangle \equiv \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \, . \]

  In this basis, the angular momentum component operators have the form

  \[ \hat L_x = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \, , \qquad \hat L_y = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \, , \] \[ \hat L_z = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \, . \]

  Note: The derivation of these matrices can be found in problem Matrix formalism for l = 1, Solution b).

• Solution a)

  First, we express the state \(|\psi\rangle\) in the matrix representation (see Hint 2)

  \[ |\psi\rangle = a |1, \, 1\rangle + b |1, \, 0\rangle + c |1, \, -1\rangle = \] \[ = a \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + b \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + c \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \, . \]

  To compute the expectation value, we use the general formula, substitute the matrix representation, and simplify

  \[ \langle L_x \rangle_\psi = \left \langle \psi \, \Big | \, \hat L_x \psi \right \rangle = \psi^\dagger \hat L_x \psi = \begin{pmatrix} a^* & b^* & c^* \end{pmatrix} \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \] \[ = \frac{\hbar}{\sqrt2} \begin{pmatrix} a^* & b^* & c^* \end{pmatrix} \begin{pmatrix} b \\ a+c \\ b \end{pmatrix} = \frac{\hbar}{\sqrt2} (a^*b + ab^* + b^*c + bc^*) \, . \]

  Note: Since this is the expectation value of a physical observable, the expression in parentheses must be real. Indeed, it is, because

  \[ a^*b + ab^* + b^*c + bc^* = 2 \, \mathrm{Re} (ab^*) + 2 \, \mathrm{Re} (bc^*) \, . \]

  The calculation for \(\langle L_y \rangle_\psi\) and \(\langle L_z \rangle_\psi\) is completely analogous. The resulting values are

  \[ \langle L_y \rangle_\psi = \frac{i\hbar}{\sqrt2} (a{b^*} - {a^*}b + b{c^*} - {b^*}c) \, , \] \[ \langle L_z \rangle_\psi = \hbar^2 (|a|^2 + |c|^2) \, . \]

  These expectation values are also real. For \(\langle L_z \rangle_\psi\) this is obvious, and for \(\langle L_y \rangle_\psi\) we can rewrite the term in the parentheses as

  \[ a{b^*} - {a^*}b + b{c^*} - {b^*}c = 2 i \, \mathrm{Im}(a{b^*}) + 2 i \, \mathrm{Im}(b{c^*}) \, . \]

  Multiplying this by the \(i\) from the numerator of the fraction in front of the parentheses gives a real expression.

• Solution b)

  Here we proceed similarly to the previous section, with the difference that we first compute the squares of the angular momentum operators for each component

  \[ \hat L_x^2 = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \frac{\hbar^2}{2} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix} \, , \] \[ \hat L_y^2 = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} = \frac{\hbar^2}{2} \begin{pmatrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{pmatrix} \, , \] \[ \hat L_z^2 = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} = \hbar^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \, . \]

  Next, we substitute into the general formula for the expectation value in the “matrix” representation and simplify

  \[ \langle L_x^2 \rangle_\psi = \left \langle \psi \, \Big | \, \hat L_x^2 \psi \right \rangle = \psi^\dagger \hat L_x^2 \psi = \begin{pmatrix} a^* & b^* & c^* \end{pmatrix} \frac{\hbar^2}{2} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \] \[ = \frac{\hbar^2}{2} \begin{pmatrix} a^* & b^* & c^* \end{pmatrix} \begin{pmatrix} a+c \\ 2b \\ a+c \end{pmatrix} = \frac{\hbar^2}{2} (aa^* + a^*c + 2bb^* + ac^* + cc^*) = \] \[ = \frac{\hbar^2}{2} (|a|^2 + 2|b|^2 + |c|^2 + {a^*}c + a{c^*}) \, . \]

  The calculations for \(\langle L_y^2 \rangle_\psi\) and \(\langle L_z^2 \rangle_\psi\) are completely analogous. The results are

  \[ \langle L_y^2 \rangle_\psi = \frac{\hbar^2}{2} (|a|^2 + 2|b|^2 + |c|^2 - {a^*}c - a{c^*}) \, , \] \[ \langle L_z^2 \rangle_\psi = \hbar^2 (|a|^2 + |c|^2) \, . \]

  In all three cases, we obtain real values as expected.

  Note: Summing the expectation values of the squares of the individual angular momentum components gives

  \[ \langle L^2 \rangle_\psi = \langle L_x^2 \rangle_\psi + \langle L_y^2 \rangle_\psi + \langle L_z^2 \rangle_\psi = 2 \hbar^2 (|a|^2 + |b|^2 + |c|^2) = 2\hbar^2 \, , \]

  which is consistent with the fact that this is a state with a sharp value of \(\hat L^2\) for \(l = 1\).

• Hint 3

  Recall or look up how the eigenvalues and eigenvectors of matrices are determined.

• Solution to hint 3

  The eigenvalues \(\lambda\) of a matrix \(A\) are obtained by solving the characteristic equation

  \[ \det \left ( A - \lambda \mathbb{E} \right ) = 0 \, , \]

  where \(\mathbb{E}\) denotes the identity matrix of the appropriate order.

  The eigenvector \(\vec u\) corresponding to the eigenvalue \(\lambda\) of the matrix \(A\) is obtained by solving the equation

  \[ \left ( A - \lambda \mathbb{E} \right ) \cdot \vec u = \vec o \, , \]

  where \(\vec o\) denotes the zero vector.

• Hint 4

  Recall or look up how to decompose the state entering the measurement, and how the probabilities of obtaining the individual measurement outcomes can be determined from this decomposition.

• Solution to hint 4

  According to the measurement axiom, the state entering the measurement must be decomposed into a linear combination of the eigenstates of the observable being measured. The probabilities of obtaining the individual measurement outcomes are given by the squared modulus of the corresponding coefficients in this linear combination.

• Solution c)

  After passing through the first Stern–Gerlach apparatus, we select only the particles with the value \(L_z = \hbar\), which corresponds to particles in the state \(|1, \, 1\rangle\). The beam of particles in this state then passes through a Stern–Gerlach apparatus oriented along the \(x\)‑axis. The measurement axiom states that, if we want to determine the probabilities of measuring each of the three possible values, we must express the state entering the measurement as a linear combination of the eigenstates of \(L_x\).

• Solution – eigenvalues and eigenvectors

  To do this, we need to determine the eigenvalues and corresponding eigenvectors of the matrix \(\hat L_x\). This matrix has the form (see Hint 2)

  \[ \hat L_x = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \, . \]

  The eigenvalues \(\lambda\) are obtained by substituting into the characteristic equation \(\det \left ( \hat L_x - \lambda \hat{\mathbb{E}} \right ) = 0\) and simplifying

  \[ \det \left ( \hat L_x - \lambda \hat{\mathbb{E}} \right ) = \det \left ( \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right ) = \] \[ = \det \begin{pmatrix} -\lambda & \frac{\hbar}{\sqrt2} & 0 \\ \frac{\hbar}{\sqrt2} & -\lambda & \frac{\hbar}{\sqrt2} \\ 0 & \frac{\hbar}{\sqrt2} & -\lambda \end{pmatrix} = -\lambda^3 + \frac{\hbar^2}{2} \lambda + \frac{\hbar^2}{2} \lambda = \] \[ = \hbar^2 \lambda - \lambda^3 = \lambda \, \left (\hbar^2 - \lambda^2 \right ) = \lambda \, \left (\hbar - \lambda \right ) \, \left (\hbar + \lambda \right ) = 0 \, . \]

  The eigenvalues are therefore, as expected, \(\lambda_1 = \hbar, \, \lambda_2 = 0, \, \lambda_3 = -\hbar \, \).

  We now focus on the eigenvector corresponding to the eigenvalue \(\hbar\). The eigenvector \(\vec u\) associated with this eigenvalue \(\lambda\) is determined by substituting into the equation \(\left ( \hat L_x - \lambda \hat{\mathbb{E}} \right ) \cdot \vec u = \vec o\) and performing the following simplification

  \[ \left ( \hat L_x - \lambda \hat{\mathbb{E}} \right ) \cdot \vec u = \left ( \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} - \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right ) \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \] \[ = \frac{\hbar}{\sqrt2} \begin{pmatrix} -\sqrt2 & 1 & 0 \\ 1 & -\sqrt2 & 1 \\ 0 & 1 & -\sqrt2 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \, . \]

  We divide this equality by the factor \(\frac{\hbar}{\sqrt2}\) and expand it into its component equations. This yields a system of three linear equations with three unknowns

  \[ -\sqrt2 u_1 + u_2 = 0 \, , \] \[ u_1 - \sqrt2 u_2 + u_3 = 0 \, , \] \[ u_2 - \sqrt2 u_3 = 0 \, . \]

  These equations are linearly dependent. We therefore choose \(u_3 = 1\). From the third equation we then obtain \(u_2 = \sqrt2\), and from the first equation \(u_1 = 1\). In this way, we have determined the direction of the eigenvector corresponding to the eigenvalue \(\hbar\). We now normalise it

  \[ \frac{1}{\sqrt{|u_1|^2 + |u_2|^2 + |u_3|^2}} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \frac{1}{\sqrt{1 + 2 + 1}} \begin{pmatrix} 1 \\ \sqrt2 \\ 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt2 \\ 1 \end{pmatrix} \, . \]

  Thus, we have obtained the normalised eigenvector of the matrix \(\hat L_x\) corresponding to the eigenvalue \(\hbar\)

  \[ \vec u = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt2 \\ 1 \end{pmatrix} \, . \]

  The procedure for the eigenvalues \(0\) and \(-\hbar\) is completely analogous. The calculation of the corresponding eigenvectors can be found at the end of this problem in the section Comment. Here, we present the resulting normalised eigenvectors of the matrix \(\hat L_x\) corresponding to the eigenvalues \(0\)
  and \(-\hbar\)

  \[ \lambda = 0 \, \, \rightarrow \, \, \vec v = \frac{1}{\sqrt2} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \, , \] \[ \lambda = -\hbar \, \, \rightarrow \, \, \vec w = -\frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt2 \\ 1 \end{pmatrix} \, . \]

• Solution – decomposition of the state entering the measurement

  We now express the state \(|1, \, 1 \rangle\) entering the measurement as a linear combination of the eigenstates of the measured observable

  \[ |1, \, 1\rangle = c_1 \vec u + c_2 \vec v + c_3 \vec w \, , \]

  where \(c_1, \, c_2, \, c_3\) are the unknown complex coefficients.

  Substituting the corresponding eigenvectors, we obtain

  \[ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = c_1 \, \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt2 \\ 1 \end{pmatrix} + c_2 \, \frac{1}{\sqrt2} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} - c_3 \, \frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt2 \\ 1 \end{pmatrix} \, . \]

  The sought coefficients can be found using the scalar product

  \[ c_1 = \langle \vec u \, | \, 1, \, 1 \rangle = \vec u^\dagger | 1, \, 1 \rangle = \frac{1}{2} \begin{pmatrix} 1 & \sqrt2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{2} \, , \] \[ c_2 = \langle \vec v \, |\, 1, \, 1 \rangle = \vec v^\dagger | 1, \, 1 \rangle = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{\sqrt2} \, , \] \[ c_3 = \langle \vec w \, | \, 1, \, 1 \rangle = \vec w^\dagger | 1, \, 1 \rangle = -\frac{1}{2} \begin{pmatrix} 1 & -\sqrt2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = -\frac{1}{2} \, . \]

  We now determine the probabilities of measuring the individual values of \(L_x\). The probabilities of obtaining the individual measurement outcomes are given by the squared modulus of the corresponding coefficient, i.e.,

  \[ P_{+\hbar} = |c_1|^2 = \frac{1}{4} \, , \] \[ P_0 = |c_2|^2 = \frac{1}{2} \, , \] \[ P_{-\hbar} = |c_3|^2 = \frac{1}{4} \, . \]

  Thus, we measure the value \(L_x = +\hbar\) with probability \(\frac{1}{4}\), the value \(L_x = 0\) with probability \(\frac{1}{2}\), and the value \(L_x = -\hbar\) with probability \(\frac{1}{4}\).

• Answer

  a) The expectation values of the components of the angular momentum in the state \(|\psi\rangle\) are

  \[ \left \langle L_x \right \rangle = \frac{\hbar}{\sqrt2} (a^*b + ab^* + b^*c + bc^*) \, , \] \[ \left \langle L_y \right \rangle = \frac{i\hbar}{\sqrt2} (ab^* - a^*b + bc^* - b^*c) \, , \] \[ \left \langle L_z \right \rangle = \hbar^2 (|a|^2 + |c|^2) \, . \]

  b) The expectation values of the squares of the components of the angular momentum in the state \(|\psi\rangle\) are

  \[ \left \langle L_x^2 \right \rangle = \frac{\hbar^2}{2} (|a|^2 + 2|b|^2 + |c|^2 + {a^*}c + a{c^*}) \, , \] \[ \left \langle L_y^2 \right \rangle = \frac{\hbar^2}{2} (|a|^2 + 2|b|^2 + |c|^2 - {a^*}c - a{c^*}) \, , \] \[ \left \langle L_z^2 \right \rangle = \hbar^2 (|a|^2 + |c|^2) \, . \]

  c) After the beam of particles with projection \(L_z = \hbar\) passes through a Stern–Gerlach apparatus oriented along the \(x\)‑axis, the values of \(L_x\) are measured with the following probabilities

  \[ L_x = +\hbar \, \, \rightarrow \, \, P_{+\hbar} = \frac{1}{4} \, , \] \[ L_x = 0 \, \, \rightarrow \, \, P_0 = \frac{1}{2} \, , \] \[ L_x = -\hbar \, \, \rightarrow \, \, P_{-\hbar} = \frac{1}{4} \, . \]

• Comment – calculation of the other eigenvectors

  Now we determine the eigenvector \(\vec v\) corresponding to the eigenvalue \(\lambda = 0\) by substituting into the equation \(\left ( \hat L_x - \lambda \hat{\mathbb{E}} \right ) \cdot \vec v = \vec o\) and performing the following simplification

  \[ \left ( \hat L_x - \lambda \hat{\mathbb{E}} \right ) \cdot \vec v = \left ( \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} - 0 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right ) \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \] \[ = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \, . \]

  We divide this equality by the factor \(\frac{\hbar}{\sqrt2}\) and expand it into its component equations. This yields a system of three linear equations with three unknowns

  \[ v_2 = 0 \, , \] \[ v_1 + v_3 = 0 \, , \] \[ v_2 = 0 \, . \]

  These equations are linearly dependent. We immediately see that \(v_2 = 0\). Let us choose \(v_1 = 1\) and therefore \(v_3 = -1\). This determines the direction of the eigenvector corresponding to the eigenvalue \(0\). We now normalise it

  \[ \frac{1}{\sqrt{|v_1|^2 + |v_2|^2 + |v_3|^2}} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \frac{1}{\sqrt{1 + 0 + 1}} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \frac{1}{\sqrt2} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \, . \]

  Thus, we have obtained the normalised eigenvector of the matrix \(\hat L_x\) corresponding to the eigenvalue \(0\)

  \[ \vec v = \frac{1}{\sqrt2} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \, . \]

  When determining the eigenvector \(\vec w\) corresponding to the eigenvalue \(-\hbar\), we proceed in a completely analogous way

  \[ \left ( \hat L_x - \lambda \hat{\mathbb{E}} \right ) \cdot \vec w = \left ( \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} + \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right ) \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} = \] \[ = \frac{\hbar}{\sqrt2} \begin{pmatrix} \sqrt2 & 1 & 0 \\ 1 & \sqrt2 & 1 \\ 0 & 1 & \sqrt2 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \, . \]

  We divide this equality by the factor \(\frac{\hbar}{\sqrt2}\) and expand it into its component equations. This yields a system of three linear equations with three unknowns

  \[ \sqrt2 w_1 + w_2 = 0 \, , \] \[ w_1 + \sqrt2 w_2 + w_3 = 0 \, , \] \[ w_2 + \sqrt2 w_3 = 0 \, . \]

  These equations are linearly dependent. According to the established convention, let us choose \(w_3 = -1\). From the third equation we then obtain \(w_2 = \sqrt2\), and from the first equation \(w_1 = -1\). This determines the direction of the eigenvector corresponding to the eigenvalue \(-\hbar\). We now normalise it

  \[ \frac{1}{\sqrt{|w_1|^2 + |w_2|^2 + |w_3|^2}} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} = \frac{1}{\sqrt{1 + 2 + 1}} \begin{pmatrix} -1 \\ \sqrt2 \\ -1 \end{pmatrix} = -\frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt2 \\ 1 \end{pmatrix} \, . \]

  Thus, we have obtained the normalised eigenvector of the matrix \(\hat L_x\) corresponding to the eigenvalue \(-\hbar\)

  \[ \vec w = -\frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt2 \\ 1 \end{pmatrix} \, . \]